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NCERT Solutions For Class 11 Chemistry Chapter 3 Classification Of Elements And Periodicity In Properties Long Question And Answers

Question 1. Arrange according to the instructions given in the bracket:

• O, Te, Se, S (Increasing order of electronegativity)
• Na, Cu, Zn (Increasing order of electropositive character)
• I, F, Br, CI (Increasing order of metallic character)
• I, F, Br, Cl (Decreasing order of electron affinity)
• Na, K, F, Cl, Br (Increasing order of atomic radius)
• Mg, AI, Si, Na (Increasing order of ionization potential)
• PbO, MgO, ZnO (increasing order of basic character)
• Na+, Mg²⁺, Al³⁺ (Decreasing order of size)
• Cu, S, C (graphite) (Increasing order of electrical conductivity)
• Be, C, B, N, O (Increasing order of electron affinity)
• Cl, Mg, C, S (Increasing order of electronegativity)
• A1₂O₃, P₂O₅, Cl₂O₇, SO₃ (Increasing order of acidic property)
• MgO, ZnO, CaO, Na₂O, CuO (Increasing order of basic property)
• Na⁺, F-, O^2-, Mg²⁺, N^3- (Increasing order of ionic radii)
• B —Cl, Ba—Cl, Br —Cl, Cl —Cl (Increasing order of bond polarity)
• Br, F, Cl, I (Increasing order of oxidizing property)
• Na, Cs, K, Rb, Li (Increasing order of atomic volume)
• Sb₂O₃, N₂Og, AS₂O₃ (Increasing order of acidic property)

Answer:

• Te < Se < S < O
• Cu< Zn< Na
• F < Cl < Br <I
• Cl > F > Br >I
• F < Cl < Br < Na < K
• Na<Al<Mg<Si
• ZnO < PbO < MgO
• Na⁺ > Mg²⁺ > Al³⁺
• S < C(graphite) <Cu
• Be<N<B<C<0
• Mg < C = S < Cl
• Al₂O₃ < P₂O₅ < SO₃ < Cl₂O₇
• CuO < ZnO < MgO < CaO < Na20
• Mg²⁺ < Na⁺ < F^– < O^2-– < N^3-
• Ba —Cl > B —Cl > Br —Cl > Cl —Cl
• I < Br < Cl < F
• Li < Na < K < Kb < Cs
• Sb₂O₃ < AS₂O₃ < N₂O₅

Question 2. The atomic numbers of elements A, B, and C are 10, 13, & 17 respectively.

1. Write their electronic configurations.
2. Which one of them will form a cation and which one an anion?
3. Mention their valencies.

Answer:

1. Electronic configuration of ₁₀A: ls²2s²2p⁶

Electronic configuration of ₁₃B: ls²2s²2p⁶3s³3p¹

Electronic configuration of ₁₇C: ls²2s²2p⁶3s²3p⁵

2. The element, A belonging to group 18, is an inert gas.

So it will form neither a cation nor an anion. The element B, belonging to group 13, is a metal. It will readily form a cation by the loss of 3 electrons from its valence shell (3rd shell). The element C will readily gain one electron in its outermost 3rd shell to attain inert gas electronic configuration (Is²……3s²3p⁶). So, C will form an anion.

3. Valency of A = 0 (it has a complete octet of electrons in the outermost shell). Valency of B = 3 (by the loss of 3 electrons from the 3rd shell it will attain stable inert gas electronic configuration). Valency of C = 1 (because by the gain of the electron, it can attain stable inert gas configuration).

Question 3. A, B, and C are three elements with atomic numbers group (8 + 2) = 10. 17, 18, and 20 respectively. Write their electronic configuration. Which one of them is a metal and which one is a non-metal? What will be the formula of the compound formed by the union of A and C? What may be the nature of valency involved in the formation of the above compound?
Answer:

Electronic configuration of ₁₇A: ls²2s²2p⁶3s²3p⁵

Electronic configuration of ₁₀B: ls²2s²2p⁶3s²3p⁶

Electronic configuration of ₂₀C: ls²2s²2p⁶3s²3p⁶4s²

Element C is a metal as it can easily form a dipositive ion by the loss of two electrons from 4s -orbital. ElementA is an anon-metal as it can achieve inert gas configuration by accepting one electron in a 3p- subshell.

As already mentioned, the element C can easily, form a dipositive cation (C²⁺), while the element A readily forms a uninegative anion (A^–).

So the elements A and C can combine to form the compound CA₃.

The above-mentioned compound is electrovalent because it will be formed by the union of two A^– ions with one C²⁺ ion.

Question 4. Outer electronic configuration of 4 elements is as follows: 3d°4s1 3s²3p⁵ 4s²4p⁶ Electronic configuration of 10A: ls²2s²2p⁶ 3d⁸4s². Find their positions in the periodic table
Answer:

This element (3d°4s1) is an s -block element. So it is an element of period group 1.

This element (3s²3p⁵) is a p -p-block element containing (2 + 5) or 7 electrons in the valence shell (n = 3). Soitis an element ofthe 3rd periodin group (10 + 2 + 5) = 17.

This element (4s²4p⁶) is a p -p-block element containing (2 + 6) or 8 electrons in the valence shell (n = 4). So it is an element of the 4thperiodin group (10 + 2 + 6) = 18.

This element (3d⁸4s²) is a d-block element containing 8 electrons in the d – d-orbital of the penultimate shell (n = 3) and 2 electrons in the s – s-orbital of the valence shell (n = 4). So, it is an element of the 4th period in group (8+2)=10

Question 5. Write the electronic configuration of the element with atomic number 35. What will be the stable oxidation states of the element?
Answer:

Electronic configuration: ls²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁵. The most stable oxidation state is -1 because it can accept one electron to achieve inert gas configuration (Is²… 3d¹⁰4s²4p⁶).

Again in an excited state, it can also exhibit oxidation number +3 or +5 by forming a covalent bond by using its 3 or 5 odd electrons in its outermost shell.

Question 6. The ionization potential of O is less than that of N—explain.
Answer:

The reason for such a difference may be explained based on their electronic configurations filled, its electronic configuration is highly stable.

So, a large amount of energy is required to form N+ ion by removal of 2pelectron.

On the other hand, the formation of 0+ ion by removal of one electron from a partially filled 2p -orbital requires less energy, since the 2p -orbital of 0+ion is half-filled, the electronic configuration assumes stability. Hence, oxygen has a lower ionization potential than nitrogen.

Question 7. Explain why the ionization potentials of inert gas are very high while that of alkali metals are very low. ses are
Answer:

Outermost shells of inert gases contain octets of electrons. Besides this, each of the inner shells of inert gas elements is filled.

• Such configuration is exceptionally staMe conversion of a neutral inert gas atom into its ions try removal of an electron from the outermost shell requires large energy.
• As a result, they have high ionization potentials.
• The configuration of the outermost and penultimate shell of alkali metals is (n-1)s²(n-1)p⁶nsl (except Li ).
• Thus loss of 1 electron from their outermost shell brings about a stable configuration of inert gases.
• Hence, the conversion of alkali metals to their ions requires comparatively less amount of energy. As a result, alkali metals have low values of ionization potential.

Question 8. Which member in each of the following pairs has a lower value of ionization potential? F, Cl S, Cl Ar, K O Kr, Xe Na, Na⁺.
Answer:

Cl has lower ionization enthalpy than F because electrons of 2p-orbital are more strongly attracted by the nucleus than the 3p-electrons in Cl.

• (Note that effective nuclear charge on the outermost electrons is nearly the same for both and Cl).
• S has lower ionization enthalpy than Cl because the size of S is greater than that of Cl and also the nuclear charge of is less than that of Cl
• K has a lower ionization potential than Ar as the outermost shell is filled with electrons in Ar. On the other hand, K can attain a stable configuration like the inert gas Ar by the loss of only 1 electron.
• Xe has lower ionization enthalpy than Kr because ionization enthalpy decreases on moving down a group in the periodic table.
• Na (ls²2s²2p²⁶3s¹) has lower ionization enthalpy’ than Na⁺(ls²2s²2p⁶), because the former can attain inert gas-like electronic configuration by loss of 1 electron from its outermost shell, whereas the latter attains unstable electronic configuration (Is²2s²2p⁵) by loss of one electron from its outermost shell.

Question 9. A, B, C, and D are four elements of the same period, of which A and B belong to s -block. B and D react together to form B+D^–. C and D unite together to produce a covalent compound, CD₂.

1. What is the formula of the compound formed by A and D?
2. What is the nature of that compound?
3. What will be the formula and nature of the compound formed by the union of B and C

Answer:

Since A and B are s-block elements of the same period, one of them is an alkali metal group-1A while the other is an alkaline earth metal of group-2A. B and D react to form anionic compound B+D.^– Therefore, B is a monovalent alkali metal of group 1A, and D is a monovalent electronegative element of group 4A.

Hence, the other element A of the s -the block is a bivalent alkaline earth metal of group-2A. C and D combine to produce the covalent compound CD₂. Hence, C is a bivalent electronegative element belonging to group 6A.

1. The formula of the compound formed by the combination of electropositive bivalent element A with electronegative monovalent element D is AD₂
2. The compound is ionic or electrovalent.
3. A compound formed by reactions of electropositive monovalent element B with electronegative bivalent element C will have the formula B₂C. It is an electrovalent or ionic compound.

Question 10. What changes in the following properties are observed while moving from left to right along a period & from top to bottom in a group? Atomic volume, Valency, Electronegativity, Oxidising, and reducing powers.
Answer:

On moving from left to right across a period, atomic volume first decreases and then increases. In a group, atomic volume increases with an increase in atomic number down a group.

1. Oxygen-based valency goes on increasing from left to right over a period but does not suffer any change down a group.
2. Electronegativity increases gradually from left to right across a period while it decreases down a group with an increase in atomic number.
3. Oxidizing power increases from left to right across a period and decreases from top to bottom in a group.
4. On the other hand, reducing power decreases from left to right in a period but it increases from top to bottom in a group

Question 11. Which property did Medeleev use to classify the elements in his periodic table? Did he stick to that?
Answer:

Mendeleev classified th<? elements based on their increasing atomic weights in the periodic table. He arranged almost 63 elements in order of their increasing atomic weights placing together elements with similar properties in a vertical column

• He observed that while classifying elements in the periodic table according to increasing atomic weight, certain elements had different properties than those elements belonging to the same group.
• For such cases, Mendeleev prioritized the properties of the element over its atomic weight.
• So, he placed an element with a higher atomic weight before an element with a lower atomic weight.
• For example, iodine [I (126.91)] with a lower atomic weight than tellurium [Te (127.61)] is placed after tellurium in group VII along with elements like fluorine, chlorine, etc., due to similarities in properties with these elements.
• Thus, Mendeleev did not stick to his idea of classifying elements only according to the increasing atomic weights.

Question 12. What is the basic difference in approach between Mendeleev’s Periodic Law & Modern Periodic Law?
Answer:

According to Mendeleev’s periodic law, the physical and chemical properties of elements are a periodic function of their atomic weights.

• On the other hand, the modern periodic table states that the physical and chemical properties of the elements are a periodic function of their atomic numbers.
• Thus the basic difference in approach between Medeleev’s periodic law and modern periodic law is the change in the basis of the classification of elements from atomic weight to atomic number.
• Based on quantum numbers, justify that the sixth period of the periodic table should have 32 elements. In the modern periodic table, each period begins with the filling of a new principle energy level. Therefore, the sixth period starts with the filling of the principal quantum number, n = 6.
• In the sixth period elements, the electron first enters the 6s -orbital, and then from left to right across a period the electrons enter the 4f, 5d, and 6p orbitals of the elements.

Filling of electrons in orbitals in the case of6th period continues till a new principal energy level of quantum number, n = 7 begins, i.e., for elements of the sixth period, electrons fill up the 6s, 4f, 5d, and 6p orbitals a total number of orbitals in this case =1 + 7 + 5 + 3 = 16. Since each orbital can accommodate a maximum of two electrons, there can be 16 × 2 or 32 elements in the sixth period.

Question 13. What is the significance of the terms—’ isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron-gain enthalpy?
[Hint: Requirements for comparison purposes]
Answer:

The ionization energy of an element is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of an isolated gaseous atom existing in its ground state to form a cation in the gaseous state.

• Electron-gain enthalpy is defined as the enthalpy change involved when an electron is added to an isolated gaseous atom in its lowest energy state (ground state) to form a gaseous ion carrying a unit negative charge.
• The force with which an electron gets attracted by the nucleus of an atom is influenced by the presence of other atoms in the molecule or the neighborhood.
• Thus, to determine the ionization enthalpy, the interatomic forces should be minimal. Interatomic forces are minimal in the case of the gaseous state as the atoms are far apart from each other.
• Consequently, the value of ionization enthalpy is less affected by the surroundings. Similarly, for electron affinity, the interatomic forces of attraction should be minimal for the corresponding atom.
• Tor Tills reason, the term ‘Isolated gaseous atom’ Is used while defining Ionisation enthalpy and electron-gain enthalpy.
• Ground state means the state at which the atom exists In Its most stable state. If the atom Is In the excited state, then the amount of heat applied to remove an electron or the amount of heat liberated due to the addition of an electron is low.
• So, for comparison, the term ‘ground state’ Is used while defining ionization enthalpy and electron-gain enthalpy.

Question 14. The energy of an electron in the ground state of the Hatom is -2.18× 10^-18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J . mol^-1. [Hint: Apply the idea of the mole concept.]
Answer:

Amount of energy required to remove an electron from a hydrogen atom at the ground state

= E_∞-E₁= 0 – E₁

= -(-2.18 ×10^-18)J

= 2.18 × 10^-8 J

Ionization enthalpy atomic hydrogen per mole = 2.18 × 10^-18 × 6.022 × 10²³

= 1312.8 × 10³ J.mol^-1 .

Question 15. How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Answer:

Evident that in both atoms, the valence electrons enter the 3s orbital. However, the nuclear charge of the Mg atom (+12) is greater than that of the Na atom (+11).

Again, the 3s orbital of the Mg atom being filled is more stable than half-filled.

• 3s -orbital of Na atom. Thus, the first ionization enthalpy of sodium is lower than that of magnesium.
• On the other hand, the removal of one electron from the valence shell of the Na atom leads to the formation of the Na⁺ ion whose electronic configuration is highly stable (similar to inert gas, neon).
• So high amount of energy is required to remove the second electron because it disturbs the stable electronic configuration. However, the electronic configuration of Mg⁺ is not as stable as that of Na⁺, but the electronic configuration of Mg²⁺ is more stable as it is similar to the electronic configuration of the inert gas, neon.
• So, less amount of energy is required to remove an electron from Mg⁺. Thus, the second ionization enthalpy of sodium is higher than that of magnesium.

Question 16. First ionisation enthalpy values (in kjmol^-1) of group-13 elements are: B = 801, Al = 577, Ga = 579, In = 558 and Tl = 589. How would you explain this deviation from the general trend?
Answer:

On moving down group-13 from B to Al, ionization enthalpy decreases due to an increase in atomic size and shielding effect which jointly overcome the effect of an increase in nuclear charge.

• However, ionization enthalpy increases slightly on moving from Al to Ga (2 kj.mol^-1).
• This is because due to poor shielding of valence electrons by 3d -electrons effective nuclear charge on Ga is slightly more than Al.
• On moving from Ga to In, the shielding effect of all the inner electrons overcomes the effect of the increase in nuclear charge. Thus, the ionization enthalpy of In is lower than Ga.
• Again, on moving from Into Tl, there is a further increase in nuclear charge which overcomes the shielding effect of all electrons present in the inner shells including those of 4f- and 5d -orbitals. So, the ionization enthalpy of Tl is higher than In.

Question 17. Which of the given pairs would have a more negative electron-gain enthalpy: O or F F or Cl?
Answer:

O and F both belong to the second period. As one moves from O to F, atomic size decreases and nuclear charge increases.

• Due to these factors, the incoming electron when enters the valence shell, and the amount of energy liberated in the case of F is more than that of O.
• Again, Fatom (ls²2s²2p⁵) accepts one electron to form F^– ion (ls²2s²2p⁶) which has a stable configuration similar to neon.
• However, O-atom when converted to O- does not attain any stable configuration.
• Thus energy released is much higher going from F to F^– than in going from O to O^–.
• So, the electron-gain enthalpy of is much more negative than that of O

Question 18. Would you expect the second electron-gain enthalpy of 0 as positive, more negative, or less negative than the first? Justify your answer.
Answer:

There are several valence electrons in oxygen and it requires two more electrons to complete its octet. So, the O^– atom accepts one electron to convert into an O^– an ion and in the process liberates energy. Thus, the first electron-gain enthalpy of oxygen is negative.

⇒ \(\mathrm{O}(g)+e \rightarrow \mathrm{O}^{-}(g)+141 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\left(\Delta_i H_1=-v e\right)\)

However, when another electron is added to O^– to form an O^-2-ion, energy is absorbed to overcome the strong electrostatic repulsion between the negatively charged O ion and the second incoming electron. Thus, the second electron-gain enthalpy of oxygen is positive.

⇒ \(\mathrm{O}^{-}(\mathrm{g})+e \rightarrow \mathrm{O}^{2-}(\mathrm{g})-\left(780 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\right)\left(\Delta_i H_2=+v e\right)\)

Question 19. Use the periodic table to answer the given questions. Identify an element with 5 electrons in the outer subshell. Identify an element that would tend to lose 2 electrons. Identify an element that would tend to gain 2 electrons. Identify the group having metal, non-metal, liquid, and gas at room temperature
Answer:

Fluorine. Its configuration is ls²2s²2p⁵

• Magnesium. Its configuration is ls2²s2²p⁶3s². So, Mg loses 2 electrons from its outermost shell to form Mg²⁺ and attains a stable configuration.
• Oxygen. Its configuration is ls²2s²2p⁴ So, O agains 2 electrons to form O^2- and attains stable configuration.
• Group-17. The metallic character of astatine (At) is much greater than its non-metallic character and its melting point is very high (302°C).
• So, astatine is considered as a metal. So in group-17 there is a metal (At), non-metals (F₂, Cl₂, Br₂, I₂), liquid (Br₂) and gas (F₂, Cl₂).

Question 20. The order of reactivity of group-1 LI < Na < K < Rb < Cs whereas that of group-17 elements Is F > Cl > Br >I. Explain.
Answer:

There is only one electron in the valence shell of the elements of group 1.

• Thus, they have a strong tendency to lose this single electron.
• The tendency to lose electrons depends on the ionization enthalpy.
• As ionization enthalpy decreases down the group, the correct order of increasing reactivity of group 1 elements is Li < Na < K < Rb < Cs.
• On the other hand, there are 7 electrons in the valence shell of the elements of group-17.
• Thus, they have a strong tendency to gain a single electron. The tendency to gain electrons depends on the electrode potentials of the elements.
• As the electrode potential of elements decreases down the group, the correct order of activity is F > Cl > Br >I.

Alternate explanation:

In the case of halogens, their reactivity increases with the increase in electron-gain enthalpy.

Order of electron-gain enthalpy:

F < Cl > Br >I. As electron gain enthalpy decreases from Cl to, the order of reactivity also follows this sequence. However, fluorine is the most reactive halogen as its bond dissociation energy is very low.

Question 21. Assign the position of the element having outer electronic configuration:

1. ns²np⁴ for n = 3, 
2. (n-1)d²ns² for n = 4
3. (n-2)f⁷(n-1)d¹ns² for n = 6, in the periodic table.

Answer:

1. As n= 3, the element belongs to the period. Since the last electron enters the p-orbital, the given element is a p-block element.

For p-block elements, group no. of the element = 10+no. of electrons in the valence shell.

• The element is in the (10+6) = 16th period.

2. As n = 4, the element belongs to the fourth period. Since is present in the element, it is a block element. For d-block elements, group no. of the element = no. of ns electrons + no. of(n-1) f electrons = 2+2 = 4. Therefore, the element is in the 4th period.

3. As n – 6, the element belongs to the sixth period. Since the last electron enters the f-orbital, the element is a f-block element. All f-block elements are situated in the third group of the periodic table.

Question 22. The first (Δ_iH₁) and second (Δ_iH₂)) ionization enthalpies (klmol^-1) and the (Δ_cgH)electron gain enthalpy (in kj.mol^-1 ) of a few elements are given below:

Which of the above elements is likely to be:

1. The least reactive clement?
2. The most reactive metal.
3. The most reactive non-metal.
4. The least reactive non-metal.
5. The metal can form a stable binary halide of the formula MX₂(X = halogen).
6. The metal that can form a predominantly stable covalent halide of the formula MX (X = halogen)?

Answer:

⇒ \(\begin{array}{|c|c|c|c|}
\hline \text { Elements } & \left(\Delta H_1\right) & \left(\Delta H_2\right) & \left(\Delta_{c g} H\right) \\
\hline \text { 1 } & 520 & 7300 & -60 \\
\hline \text { 2 } & 419 & 3051 & -48 \\
\hline \text { 3 } & 1681 & 3374 & -328 \\
\hline \text { 4 } & 1008 & 1846 & -295 \\
\hline \text { 5 } & 2372 & 5251 & +48 \\
\hline \text {6 } & 738 & 1451 & -40 \\
\hline
\end{array}\)

1. Element 5: Element 5 is the least reactive metal as it has the highest first ionization enthalpy & positive electron-gain enthalpy.
2.  Element 2: Element 2 is the most reactive metal as it has lowest first ionization enthalpy & low negative electron-gain enthalpy.
3. Element 3: Element 3  is the most reactive non-metal because it has very high first ionization enthalpy and very high negative electron-gain enthalpy.
4. Element 4: Element 4  is the least reactive non-metal because it has a high negative electron-gain enthalpy but not so high first ionization enthalpy.
5. Element 6: Element 6 has low first and second ionization enthalpy. Again, the first ionization enthalpy of this element is higher than those ofthe alkali metals. Thus, the given element is an alkaline earth metal and can form a stable binary halide ofthe formula MX2.
6. The first ionization enthalpy of elements is low but its second ionization enthalpy is high. So, it is an alkali metal and can form a stable covalent halide (MX)

Question 23. Predict the formulas of the stable binary compounds that would be formed by given pairs of elements:

1. Li and O
2. Mg and N
3. Al and I
4. Si and O, P and F
5. Element with atomic numbers 71 and F.

Answer:

Question 24. What will be the name (IUPAC) and symbol if the element with atomic number 119 is discovered? Write its electronic configuration. Also, write the formulas of the stable chloride and oxide of this element.
Answer:

IUPAC name : Ununennium, Symbol: Uue

Atomic number ofthe element =119 = 87 + 32

• It is known that the element with atomic number 87 is francium (Fr). Fr belongs to group 1 in the 7th period of the periodic table.
• So, the element with atomic number 119 will take its position in group 1 and 8th period just below francium(Fr).
• The electronic configuration of this element will be [UuojBs1, (where Uuo = Ununoctium, Z = 118). It will be an alkali metal with valency=1
• If the symbol ofthe element is ‘M’ then the formulas of its stable chloride and oxide will be MCI and M₂O respectively.

Question 25. Elements A, B, and C have atomic numbers (Z- 2), Z, and (Z +1) respectively. Of these, B is an inert gas. Which one of these has the highest electronegativity? Which one of these has the highest ionization potential? What is the formula of the compound formed by the combination of A and C? What is the nature of the bond in this compound?
Answer:

Since element B (atomic no =Z) is an inert gas, the element ‘A’ with atomic number (Z- 2) is included in group 6A.

On the other hand, the element C, having an atomic number (Z + 1) must belong to groups (alkali metal). Hence, the electronegativity of the element A is maximum.

• The element B, being an inert gas, has the highest value of ionization potential.
• The valency of the element A, belonging to group (8- 6) = 2, and that ofthe element C, being an element of group IA, is 1.
• Therefore, the formula of the compound formed by A and C will be C₂A.
• Being a strongly electronegative element and C being a strongly electropositive element complete their octet through gain and loss of electrons respectively.
• So, the nature of the bond formed between C and A in C₂A isionic or electrovalent bond.

Question 26. The atomic radius of ₁₀Ne is more than that of ₉F —why?
Answer:

Fluorine forms diatomic molecules, thus the atomic radius of fluorine is a measure of half of the internuclear distance in its molecule (i.e., half of the covalent bond length of an F₂ molecule) but neon being an inert gas, its atoms are incapable of forming covalent bonds by mutual combination amongst themselves.

• The only force that comes into play between the atoms is the weak van der Waals force.
• So a measure of the atomic radius of Ne is equal to its van der Waals radius but the van der Waals radius is always greater than the covalent radius.
• Thus, the atomic radius of neon is larger than that of fluorine.
• Furthermore, due to an increase in the number of electrons in the outermost 2p -orbital of Ne, there occurs an increase in electron-electron repulsion.
• So 2p-orbital of Ne suffers expansion leading to its increased atomic radius.

Question 27. The first electron affinity of oxygen is negative but the second electron affinity is positive—explain.
Answer:

• When an electron is added to the valence shell of an isolated gaseous O-atom in its ground state to form a negative ion, energy is released.
• Because a neutral oxygen atom tends to complete its octet with electrons. So, the electron affinity of oxygen is an exothermic process and its value is negative.
• When an extra electron is added to an O- ion, that second electron experiences a force of repulsion exerted by the negative charge ofthe anion. So, first, this process requires a supply of energy from an external source.
• This accounts for the endothermic nature of second electron affinity and has a positive value.

Question 28. The electron affinity of sodium is negative but magnesium has a positive value—why?
Answer:

Electronic configuration of ₁₁Na: ls²2s²2p⁶3s¹

Electronic configuration of ₁₂Mg: ls²2s²2p⁶3s²

• The addition of one electron to the 3s -orbital of Na leads to a comparatively stable electronic configuration with a fulfilled orbital.
• So, this process of the addition of electrons to Na is an exothermic, process. So, the electron affinity of Na is negative. On the other hand, Mg has fulfilled 3s orbital and has a stable electronic configuration.
• So the addition of an electron to the 3p -orbital destabilizes the electronic configuration of Mg.
• Additional energy is required for the addition of electrons to the outermost shell of Mg i.e., this process is endothermic and thus the value of electron affinity ofMg is positive.

Question 29. If the electron affinity of chlorine is 350 kJ. moI^-1, then what is the amount of energy liberated to convert 1.775 g of chlorine (existing at atomic state) to chloride ions completely (in a gaseous state)
Answer:

Atomic mass of chlorine = 35.5 g .mol^-1

The energy liberated in the conversion of 35.5 g of Cl to Cl^– ion =350 kj

Energyliberatedin the conversion of1.775 g of Cl to Cl^–

⇒ \(\text { ion }=\frac{350}{35.5} \times 1.775=17.5 \mathrm{~kJ}\)

Question 30. The second ionization enthalpy of Mg is sufficiently high and the second electron-gain enthalpy of O has a positive value. How do you explain the existence of Mg²⁺ O^2- rather than Mg⁺O^– ?
Answer:

The lattice energy of an ionic crystal depends on the force of attraction between the cations and anions

⇒ \(\left(F \propto \frac{q_1 q_2}{r^2}\right)\)

So, the magnitude of lattice energy increases as the charges on the cation and anion increase. Consequently, the lattice energy of Mg²⁺ O^2- is very much greater than that of Mg⁺ O^–.

The lattice energy of Mg²⁺ O^2- is so high that it exceeds the unfavorable effects of the second ionization enthalpy of Mg and the second electron-gain enthalpy of 0.

So, Mg²⁺ O^2- is a stable ionic compound, and its formation is favored over Mg⁺ O^–.

Question 31. The atomic numbers of some elements are given below. Classify them into three groups so that the two elements in each group exhibit identical chemical behavior: 9, 12, 16, 34, 53, 56.
Answer:

Atomic Number →  Electronic configuration

9 → \(1 s^2 2 s^2 2 p^5\)

12 → \(1 s^2 2 s^2 2 p^6 3 s^2\)

16 → \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^4\)

34 → \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^2 4 p^4 \)

53 → \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^2 4 p^6 4 d^{10} 5 s^2 5 p^5\)

56 → \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^2 4 p^6 4 d^{10} 5 s^2 5 p^6 6 s^2\)

Elements with atomic numbers 9 and 53 belong to the -block and they have similar outer electronic configurations (ns²np⁵). So they will exhibit similar chemical properties. Their group number = (10 + 2 + 5) = 17. Elements with atomic numbers 12 and 56 belong to s -block and they have similar outer electronic configurations (ns²).

So they will exhibit similar chemical properties. Their group number = 2. Elements with atomic numbers 16 and 34 belong to the -block and they have similar outer electronic configurations (ns²np²).

So, they will exhibit similar chemical properties. Their group number = (10 + 2 + 4) = 16. So, based on similarity in chemical properties, the given elements are divided into three groups :

Group-2 → 12,56 (Atomic number)

Group-16 → 16,34 (Atomic number)

Group-17→ 9,53 (Atomic number)

Question 32. Though the nuclear charge of sulfur is more than that of phosphorus, yet the ionization potential of phosphorus is relatively high”—why?
Answer:

1. \({15} \mathrm{P}: 1 s^2 2 s^2 2 p^6 3 s^2 3 p_x^1 3 p_y^1 3 p_z^1\)

2. \({16} \mathrm{~S}: 1 s^2 2 s^2 2 p^6 3 s^2 3 p_x^2 3 p_y^1 3 p_z^1\)

3p -orbital of the outermost shell of the P -atom being half filled, this electronic configuration is very stable. So, the removal of one 3p -electron to produce a P+ ion requires a sufficiently high amount of energy.

On the other hand, the amount of energy required for removing one electron from a partially filled 3p -orbital of the S -atom to yield an S⁺ ion is relatively less, since the half-filled 3p -orbital of S+ assumes the extra stability due to the loss of this electron. This accounts for the higher value of ionization potential of phosphorus, relative to sulfur.

Question 33. Mg has relatively higher ionization enthalpy than A1 although the atomic number of the latter is more than the former—explain why.
Answer:

Electronic configuration of ₁₂Mg \(: 1 s^2 2 s^2 2 p^6 3 s^2\)

Electronic Configuration of \({ }_{13} \mathrm{Al}: 1 s^2 2 s^2 2 p^6 3 s^2 3 p^1\)

The penetration effect of the s-electron is greater than that of the electron. So it is easier to remove the 3p-electron from the outermost shell of Al.

Furthermore, the removal of this electron gives Al+, which has filled 3s-orbital (stable electronic configuration) in the outermost shell. On the other hand, Mgatom has filled 3s-orbital (stable electronic configuration) in its ground state.

Removal of one electron from the 3s-orbital of Mg-atom will require a large amount of energy because the resulting Mg⁺ ion will have a less stable electronic configuration (ls²2s²2p⁶3s¹). Furthermore, it is rather difficult to remove an electron from the s-orbital having a greater penetration effect. So ionization enthalpy of Mg is greater than that of A.

Question 34. Why are electron-gain enthalpy of Be and N positive?
Answer:

The fact that Be and N have positive electron-gain enthalpy values can be explained by considering the given electron-gain processes.

It is observed that the stable electronic configuration of both Be and N -atoms is disturbed by the addition of an electron to each of them.

Consequently, such electron addition processes involve the absorption of energy and hence, both Be and N have positive electron-gain enthalpy values.

Question 35. The electron affinity of lithium is negative but the electron affinity of beryllium is positive”—why?
Answer:

Electronic configuration of ₃Li: ls²2s¹

Electronic configuration of ₄Be: ls²2s²

In addition of an electron to Li-atom, the 2s -orbital of Li becomes filled with electrons, and consequently, that electronic configuration attains stability.

• This process is accompanied by the liberation of energy. On the other hand, Be has a stable electronic configuration with a frilly-filled 2s subshell.
• When an electron is added to Be-atomic occupies 2p -subshell causing destabilization ofthe stable electronic configuration.
• This process is accompanied by the absorption of heat. Naturally electron affinity of Li is negative but the electron affinity ofBe is positive.

Question 36. Which of the following statements related to the modern periodic table is incorrect?

• p -p-block has 6 columns because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
• d -block has 8 columns, as a maximum of 8 electrons can occupy all the orbitals in a d -subshell.
• Each block contains many columns equal to the number of electrons that can occupy that subshell.
• Block indicates the value of azimuthal quantum number (l) for the last subshell that received electrons in building up electronic configuration

Answer: The statement is incorrect because d -the block has 10 columns because a maximum of ten electrons can occupy all the orbitals in a d -subshell.

Question 37. Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?

1. Valence principal quantum number (n)
2. Nuclear charge (Z)
3. Nuclear mass
4. Number of core electrons.
5. Nuclear mass does not affect the valence shell electrons (such as, and H have similar chemical properties.
6. The size of isoelectronic species F-, Ne, Na+ is affected by: nuclear charge (Z)
7. valence principal quantum number (n)

Question 38. What do you mean by isoelectronic species? Name a species that will be isoelectronic with each of the given atoms or ions:

1.  F^–
2. Ar,
3. Mg²⁺
4.  Rb⁺

Answer:

Electronic ions are ions of different elements that have the same number of electrons but different magnitudes of nuclear charge.

1. There are (9 + 1) or 10 electrons in F^–.

Isoelectronic species of F^– are :

• Nitride (N^3-) ion [7 + 3]
• Oxide (O^–₂) ion [8 + 2]
• Neon (Ne) atom [10]
• Sodium (Na⁺) ion [11-1],
• Magnesium (Mg²⁺) ion [12-2]
• Aluminum (Al³⁺) ion [13-3].

2. There are =18 electrons in Ar.

Isoelectronic species are:

• Phosphide (P^3-) ion [15 + 3]
• Sulfide (S^2-) ion [16 + 2]
• Chloride (Cl^– ion [17 + 1]
• Potassium (K⁺) ion [19 -1], and
• Calcium (Ca²⁺) ion [20-2 ].

3. There are (12-2) = 10 electrons in Mg²⁺

Isoelectronic species are: 

• Nitride (N^3-) ton [7 + 3]
• Oxide (O^2-) ion [8 + 2]
• Fluoride (F^–)ion [9+1]
• Sodium (Na+) ion [11-1].

4. There are (37-1)= 36 electrons In Kb.

Isoelectronic species are:

• Rb+ is bromide (Br^–) Ion [35 + 1],
• Krypton (Kr) atom [36]
• Strontium (Sr^2-) Ion [30^2-].

Question 39. The atomic numbers of three elements A, B, and C are 9, 13, and 17 respectively.

1. Write their electronic configuration.
2. Ascertain their positions in the periodic table.
3. Which one is most electropositive and which one is most electronegative?

Answer:

1. Electronic configurations of

₉A: ls²2s²2p⁵

₁₃B: ls²2s²2p⁶3s²3p¹

₁₇C: ls²2s²2p⁶3s²3p⁵

2. All three elements are p -p-block elements. Hence, their group and period numbers are as follows

Element → Period number → Group number

A → 2 → 10 + 2 + 5 = 17

B→ 3→ 10 + 2 + 1 = 13

c→ 3→ 10 + 2 + 5 = 17

3. Element B can easily donate 3 electrons from its outermost shell to attain a stable inert gas configuration. So, it is the most electropositive element

Elements A and C are electronegative because they can accept one electron to attain a stable inert gas electronic configuration. These elements (A and C) have similar outer electronic configurations
(ns²np⁵) but the size of A is smaller than that of C. So, the electronegativity of A is greater than that of C. Hence, A is the most electronegative element.

CBSE Class 11 Chemistry Notes For Chapter 4 Chemical Bonding And Molecular Structure Theory Of Valency Kossel-Lewis Theory

Earlier the term ‘valency’ was defined as the combining capacity of an element. In order words, an element can combine with another element.

Theories of valency were a direct consequence of the development of the atomic structure. w. Kossel and G.N. Lewis were the pioneers in this field, who provided logical explanations of valency, based on the internees of noble gases, which was later known as the noctule. The rule was later modified by Langmuir based on the following-

One or more than one electron(s) of the valence shell i.e., the outermost shell (both penultimate and ultimate shell in some cases) of an atom participates in the chemical reaction. Hence they are responsible for the valency of the atom. These electrons are called valence electrons.

Helium (₂He), Neon (₁₀Ne), Argon (₁₈Ar), Krypton (₃₆Kr), Xenon (₅₄Xe), and Radon (₈₆Rn) — these noble gaseous elements possess very high ionization potential but very low electron affinity.

They do not undergo any reaction under normal conditions. Thus they are called inert gases and are placed in the zero group ofthe periodic table which means they are zero-valent. The electronic configuration of He is 1s² and the general electronic configuration of other inert gases is ns²np⁶. Thus, the total no. of electrons in the outermost shell of the inert gases (other than Helium) is 8. As the noble gas elements are reluctant towards chemical bond formation, this type of electronic configuration is assumed to be stable.

Octet And Duplet Rule

The study of noble gases showed that they are chemically inert, as they have very stable electronic configurations. Kossel and Lewis stated that the stability of noble gases is due to the presence of eight electrons in their valence shell (called octet)

or two electrons (called duplet) as in the case of helium. Most of the two electrons (called duplet) as in the case of helium.

Most of the elements take part in chemical reactions or bond formation to complete their respective octet or duplet.

1. Octet rule: Atoms of various elements (except H, Li and Be) combine either by transfer of valence electron(s) from one atom to another (gain or loss) or by sharing of valence electrons so that they have eight electrons (an octet) in their outermost (valence) shells.
2. Duplet rules: the element’s dose to helium (H, Li, Be) in the periodic table to attain the stable electronic configuration of He (Is²) by gaining, losing, or sharing electrons in their valence shells.

Significance and Limitations of Octet Rule:

Significance of Octet Rule:

The reason behind the formation of covalent or ionic bonds by the atoms of different elements can be explained by the octet rule.

Limitations of Octet Rule: A few limitations of octet rule are as follows—

1. The central atom of some molecules, despite having more or less than 8 electrons in the valence shell is quite stable.
2. There are some molecules whose outermost shell contains an odd number of electrons. For example— NO, NO₂, etc.
3. Though the octet rule is based on the inertness of the noble gases, some noble gases form compounds with oxygen and fluorine This rule cannot predict the shape of molecules.
4. The comparative stability of the molecules cannot be predicted from this rule.

Lewis symbols

• All the electrons present in an atom are not involved during chemical combination. It was proposed that the inner shell electrons are well protected and generally do not take part in chemical combinations.
• It is mainly the outer shell electrons that take part in chemical combinations. Hence, these are also called valence shell electrons.
• G. N. Lewis introduced simple notations to represent the valence electrons in an atom. The outer shell electrons are shown as dots surrounding the symbol of an atom.
• These notations showing the symbol of an atom surrounded by an appropriate number of dots are known as Lewis symbols or electron dot symbols.

Lewis symbols ignore the inner shell electrons. For example, the Lewis symbols for the elements of the second period are—

Significance of Lewis symbols:

• The number of dots present in the Lewis symbol of an atom gives the number of electrons present in the outermost (valence) shell of that atom. This number is useful for the calculation of the common valency of an element.
• The common valency of an element is either equal to the number of dots in the Lewis symbol (when these are < 4) or % 8 – the number of dots (when these are > 4).
• For example, the common valencies of Li, Be, B, and C are equal to the number of electrons present in their valence shell i.e., 1, 2, 3, and 4 respectively, while those of N O, F, and Ne are 8 minus number of dots, i.e., 3, 2, 1 and 0 respectively.
• When the monovalent, divalent, trivalent, etc., metal atoms are converted to unipositive, dipositive, tripositive, etc., ions, no electrons are present in their outermost shell.

On the other hand, when the monovalent, divalent, trivalent, etc., non-metal atoms are converted to negative, negative, try negative, etc., ions, the outermost shell of each of them contains 8 electrons. Lewis symbols of some cations and anions are given below.

Chemical bonding

Chemical bonding Definition:

The force of attraction between the atoms participating in a chemical reaction, to attain the stable electronic configuration of the nearest noble gas by gaining, losing, or sharing electrons in their valence shells, is called chemical bonding.

Atoms acquire the stable inert gas configuration in the following ways

By complete transfer of one or more electrons from one atom to another:

This process leads to the formation of a chemical bond termed an electrovalent bond or ionic bond.

By sharing of electrons: This occurs in two ways.

1. When two combining atoms contribute an equal number of electrons for sharing, the bond formed is called a covalent bond. The shared electron pair(s) remains common to both the atoms.
2. When the shared pair of electrons is donated by one of the two atoms involved in the formation of a bond, then the bond.

Types of chemical bonds: Chemical bonds are of 3 types

1. Electrovalent or ionic bond
2. Covalent bond and
3. Coordinate or dative bond.

Electrovalency And Electrovalent Bond

An ionic or electrovalent bond is formed by the complete transfer of one or more electrons from the valence shell of an electropositive (metal) atom to that of an electronegative (nonmetal) atom so that both atoms can achieve the stable electronic configuration of the nearest noble gas.

In this process, the metal atom and non-metal atom result in the formation of a cation and an anion respectively. These two oppositely charged species combine through the electrostatic force of attraction to form an ionic crystal (electrovalent compound.

Elecrovalency:

To achieve the stable electronic configuration, some atoms give up one or more valence electrons completely to form stable cations while some other atoms gain these electrons to form stable anions and ultimately these two types of oppositely charged species combine through electrostatic forces of attraction to form compounds. The capacity of the elements for such a chemical combination Is known as electrovalency.

Ionic Or Electrovalent Bond:

The coulomblc or electrostatic force of attraction which holds the oppositely charged Ions of combining atoms formed by the complete transfer of one or more electrons from the electropositive to the electronegative atom is called an ionic or electrovalent bond.

Ionic Or Electrovalent Compound:

• Compounds Containing inoinc or electrovalent bonds are called electrovalent compounds.
• In the formation of an electrovalent compound, the number of electrons (s) lost or gained by an atom of any participating element gives the measure of its electrovalency.
• For example, in the compound sodium chloride (NaCl), the electrovalent of sodium = 1 and the electrovalency of chlorine = 1. In magnesium chloride (MgCl₂), each Mgatom loses two electrons and each Cl-atom gains one electron, so the electrovalency of magnesium and chlorine are 2 and 1 respectively.

Examples of ionic compound formation: The formation of some ionic compounds is discussed below—

Formation of sodium chloride (NaCl):

The electronic configuration of sodium (11Na) atom: Is²2s²2p⁶3s¹ and that of chlorine (17C1) atom: ls²2s²2p⁶3s²3p⁵. Na has only one electron in its valence shell.

• Being an electropositive element, sodium tends to lose electrons.
• So, it loses its valence electron to acquire the configuration of the nearest noble gas, Ne (ls²2s²2p⁶). On the other hand, Cl^– atom has seven electrons in its valence shell. Being an electronegative element, chlorine tends to gain electrons.
• So, it can acquire the stable electronic configuration of the nearest noble gas. Ar (ls²2s²2p⁶3s²3p⁶) by gaining one electron.
• Therefore, when a Na-atom combines with a Cl^–atom, the former transfers its valence electron to the latter resulting in the formation of sodium ion, Na⁺, and chloride ion, Cl^– respectively.
• These two ions may be considered as two charged spherical particles.
• The charge is distributed throughout the surface of these spheres, and the field of the electrostatic attraction is distributed in all directions.
• Hence, even after the formation of the ion-pair Na+Cl^–, these ions can attract oppositely charged ions towards themselves, the Columbia attractive forces of these two oppositely charged ions are not satisfied.
• Because of this, a large number of Na+Cl^– ion pairs attract each other to form an aggregate and consequently, the energy of the system decreases.

This process is completed when a stable crystal of NaCl having a suitable shape and size is obtained. Since the formation of crystal lattice is a thermodynamically favorable process, therefore, the formation of ionic compounds like NaCl is a result of the formation of a three-dimensional crystal lattice.

Formation of calcium oxide (CaO):

Formation of magnesium nitride (Mg₃N₂):

Formation of aluminium oxide(Al₂O₃):

Electrovalent or ionic bond and periodic table

The tendency of an element to form a cation or an anion depends on its position in the periodic table.

• The elements of group-1(1A), the alkali metals, and group-2(2A), the alkaline earth metals, belonging to s-block are highly electropositive and have very low ionization enthalpy because of larger atomic size.
• Therefore, to achieve the octet, these elements can easily form monovalent or divalent cations by losing one or two valence electrons respectively.
• On the other hand, the elements of groups 15 (5A), 16(6A), and 17(7A) belonging to the p-block are highly electronegative and possess higher values of electron gain enthalpy (electron affinity] owing to smaller atomic size.
• Therefore, to achieve the octet, these elements can easily accept 3, 2, or 1 electron respectively to form the corresponding anions.
• Hence, the metals of groups 1(1A) and 2(2A) react chemically with the non-metals (nitrogen, oxygen, halogen, etc.) of groups- 15(5A), 16(6A),and17(7A) to form ionic compounds.
• Note that nitrides, halides, oxides, sulfides, hydrides, and carbides of alkali metals (Na, K, Rb, Cs) and alkaline earth CaO metals (Mg, Ca, Sr, Ba) are generally ionic compounds.

Factors favoring the formation of ionic compounds

• Number of valence electrons: The atom forming the cation should have 1, 2, or 3 valence electrons, whereas the atom forming the anion should have 5, 6, or 7 electrons in its outermost shell.
• The difference in electronegativity: There should be a large difference in electronegativities of the combining atoms. The greater the difference in electronegativities of the two atoms, the greater the ease of of ionic bond.
• Sizen of the ions: The formation of ionic bonds is favored by large cations and small anions.
• Ionization enthalpy and electron affinity: The lower the Ionisation enthalpy of the electropositive atom, the easier the formation of the cation. The higher the negative electron gain enthalpy of the electronegative atom, the easier the formation of the anion.
• The magnitude of charges: The higher the charge on the ions, the greater the force of attraction. Hence, the larger the amount of energy released, the greater the stability ofthe bond.
• Lattice enthalpy (or energy): The higher the value of of lattice enthalpy (electrostatic attraction between charged ions in a crystal), the greater the tire stability of the ionic bond and hence greater tire ease of formation of the compound.

Lattice Energy

The cations and anions combine to form three-dimensional solid substances known as ionic crystals. During the formation of a crystalline ionic compound, the ions of opposite charges come closer from an infinite distance and pack up three-dimensionally in a definite geometric pattern.

This process involves the liberation of energy because the attractive force between the ions of opposite charges tends to decrease the energy of the system. The energy thus liberated is called lattice energy or lattice enthalpy. Generally, it is denoted by U.

Lattice Energy Definition

Lattice enthalpy or lattice energy may be defined as the amount of energy evolved when one gram-formula mass of an ionic compound is formed by the close packing of the oppositely charged gaseous ions.

M⁺ (g) + X^–(g) → M⁺ X^–(s) + U (lattice energy)

The higher the value of lattice enthalpy, the greater the stability of the ionic compound. It has been observed that

⇒ \(U \propto \frac{\text { Product of ionic charges }}{\text { Distance between cation and anion }}\)

Thus, lattice enthalpy depends on the following factors:

• Charges on ions: The higher the charge on the ions, the greater the forces of attraction, and consequently, a larger amount ofenergy is released. Thus, the lattice enthalpy is high.
• Charges on ions Example: The lattice enthalpy of CaO (3452.7 kJ^–mol^-1 ) is greater than that of NaF(902.9 kJ^–mol-1). This is because the charges on the two ions in CaO (+2 and -2) are greater than those on the two ions in NaF (+1 and -1).
• Size of the ions: The smaller the size of ions, the lesser the internuclear distance. Thus, the interionic attraction is greater resulting in higher lattice enthalpy.
• Size of the ions Example: The lattice enthalpy of KF (802.6 kJ.mol^-1) is higher than that of KI (635.4 kJ.mol^-1). This is because the ionic radius of the F^– ion (1.36 Å) is less than that of the r ion (2.16 Å).

Similarly, the lattice energy of NaCl is greater than that of KCl. However, lattice energy is more dependent on the charge of an ion rather than its size.

Hence, the order of lattice energy is—

1. LiF > NaF > KF > RbF > CsF
2. LiF > LiCl > LiBr >Li

Born-lande equation:

Lattice energy (U) cannot be determined directly. However, its theoretical value can be calculated using the equation given below

⇒ \(U=-\frac{A e^2 Z_{+} Z_{-} N}{r}\left(1-\frac{1}{n}\right)\)

Where

A = Madelung constant

N=Avogadro’s number

n = a constant called Bom exponent (depends on the repulsive force arising from interionic penetration and is generally taken to be 9),

e = charge of an electron, Z+ and Z ¯ are the charges on the cation and the anion respectively

r = interionic distance (minimum distance between the centers of oppositely charged ions in the lattice).

Calculation of lattice enthalpy from the Born-Haber cycle:

The lattice enthalpy of an ionic compound cannot be measured directly by experiment. It can be measured indirectly from the Born-Haber Cycle. In 1919, Max Bom and Fritz Haber proposed a method based on Hess’s law for calculating lattice enthalpy by relating it with other thermochemical data.

It can be illustrated as follows:

Calculation of lattice enthalpy of NaCl:

The ionic compound, NaCl(s) may be formed from its constituent elements by two different paths. It may be formed by the combination of its constituent elements directly i.e., from sodium and chlorine. This is an exothermic process. The heat evolved at 25°C and 1 atm pressure is called standard enthalpy of formation \(\left(\Delta \boldsymbol{H}_f^{\mathbf{0}}\right)\).

⇒ \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_2(s) \rightarrow \mathrm{NaCl}(s) ; \Delta \mathrm{H}_f^0=-411.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The formation of 1 mol of NnCl(s) may also be considered indirectly through the following steps: O Sublimation of solid Nu to gaseous Na:

The energy needed to break down the metal lattice of sodium is called sublimation energy (S) or enthalpy of sublimation \(\left(\Delta H_s^0\right)\)

⇒ \(\mathrm{Na}(s) \rightarrow \mathrm{Na}(\mathrm{g}) ; \Delta H_s^0=+108.4 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Formation of gaseous Na⁺ ion:

The energy required in this process is called the ionization energy (I) or ionization enthalpy \(\left(\Delta H_i^0\right)\)

⇒ \(\mathrm{Na}(\mathrm{g}) \rightarrow \mathrm{Na}^{+}(\mathrm{g})+e^{-} ; I=+495.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Dissociation of Cl₂ molecule into Gl-atoms:

The dissociation energy (D) or enthalpy of dissociation \(\left(\Delta H_d^0\right)\) is the amount of energy needed to convert 1 mol of Cl₂ molecules into 2mol of Cl^– atoms. To produce 1 mol of Cl-atoms, half dissociation energy is required.

⇒ \(\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightarrow \mathrm{Cl}(\mathrm{g}) ; \frac{1}{2} \Delta H_d^0=+121 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Formation of Cl- ions:

It is an exothermic process. The energy evolved is called electron-gain enthalpy (JE).

⇒ \(\mathrm{Cl}(\mathrm{g})+e \rightarrow \mathrm{Cl}^{-}(\mathrm{g}) ; E=-348.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Formation of NaCl(s) from Na⁺(g) and Cl^–(g):

It is an exothermic process that results in the liberation of energy known as lattice energy lattice enthalpy (U).

According to Hess’s law:

⇒ \(\Delta H_f^0=\Delta H_s^0+I+\frac{1}{2} \Delta H_d^0+E+U\)

∴ U =-411.2 -108.4 -495.6 -121 -(-348.6)

= -787.6 kl-mol^-1

Thus, the lattice enthalpy of NaCl(s) has a large negative value. This indicates that the compound is highly stable.

Importance of lattice enthalpy:

1. A negative value of lattice enthalpy indicates that the formation of a crystalline ionic compound from its constituent ions is an exothermic process, i.e., an ionic crystal is more stable compared to its constituent ions.
2. The magnitude of lattice enthalpy gives an Idea about the forces and the stability of the ionic compound. The higher the negative value of lattice enthalpy greater the stability of the ionic compound.
3. A higher value of lattice enthalpy indicates that the ionic crystal is hard, has a high melting point, and is less soluble in water.

The order of lattice enthalpy of some ionic compounds:

1. LiX > NaX > KX > RbX > CsX
2. MgO > CaO > SrO > BaO
3. MgCO₂ > CaCO₃ > SrCO₃ > BaCO₃
4. Mg(OH)₂> Ca(OH)₂ > Sr(OH)₂ > Ba(OH)₂

Most of the ionic compounds are formed by reaction between cations (of metals) and anions (of non-metals). However, ammonium ion (NH), a cation obtained from two non-metallic elements is a very common exception.

Variable electrovalency and exceptions to the octet rule

Several metals like Ga, In, Tl, Sn, Pb, Bi, etc. (belonging to groups 13, 14, and 15 of-block) and Cr, Mn, Fe, Cu, etc. (belonging to d-block) exhibit variable electrovalency by losing different numbers of electrons. The reasons for exhibiting several electrovalency are

1. Unstable electronic configurations of the ions and
2. Inert pair effect. Some exceptions to the octet rule are also observed in the case of these metals.

Variable valency of heavy p-block elements:

Some heavier p -p-block elements having the valence shell electronic configuration: ns²npl^-6 exhibit variable electrovalency.

• However, their primary valency is equal to the number of electrons present in the ultimate and penultimate shells.
• Why is it so? The electronic configuration of these elements has revealed that both d and f-electrons are present in the valence shell of these elements.
• Due to the poor screening effect of the d and orbitals, the s-electrons of the outermost shell are held tightly to the nucleus.
• As a result, a pair of electrons present in the ns orbital are reluctant to take part in the reaction. This is called the inert pair effect. Due to the inert pair effect, heavier p-block elements show variable valency.

Example:

• Lead (Pb) shows a +2 oxidation state predominantly due to the inert pair effect.
• Pb: [Xe]4f¹⁴5d¹⁰6s²6p². The two 6p¯ electrons are easily lost to attain the ⁺2 oxidation state.
• However, due to the very poor shielding effect of the 4f and 5d -electrons, the pair of 6s¯ electrons get closer to the nucleus and hence, are more tightly bound by the nuclear force.
• A large amount of energy must be expended to unpair the electron pair in the 6s -orbital and hence, they tend to exist as an inert pair.

Hence, the common oxidation state of Pb in most of its compounds is ⁺2. It is only in the presence of highly electronegative elements like fluorine and oxygen that the pair of electrons in the 6s¯ orbital can be unpaired and one of the electrons is promoted to the 6p¯ orbital giving rise to the ⁺4 oxidation state of Pb.

Hence, only two compounds of Pb in the ⁺4 oxidation state are known, viz., lead tetrafluoride (PbF₄) and lead dioxide (PbO₂).

Variable valency of d-block (transition) elements:

• The general electronic configuration of d-block elements is (n-1)d^1-10-10ns^1-2
• Here, apart from the s-electrons of the 1st shell, one or more d-electron(s) of (n-1)th shell contribute to the valency and hence to the oxidation state of the elements.
• Hence, the d-block elements exhibit variable valency.

Examples:

• The electrovalency of iron (Fe) may be 2 or 3. The electronic configuration of Fe ls²2s²2p⁶3s²3p⁶3d⁶4s². It forms a Fe2⁺ ion by the loss of two electrons from the 4s orbital. Soin ferrous compounds, the electrovalency of Fe is ⁺2. Fe2⁺ ion has a less stable d6 configuration.
• Therefore, it loses one electron more to form Fe3⁺ ion thereby attaining a relatively more stable configuration. So, the electrovalency of Fe in ferric compounds is 3.
• In the outermost shells of Fe2⁺ and Fe3⁺ ions, there are 14 and 13 electrons respectively. Thus, the octet rule is violated in both cases. The electrovalency of copper (Cu) can either be 1 or 2, i.e., it may either form Cu⁺ or Cu2⁺ ions.
• The electronic configuration of Cu is  ls²2s²2p⁶3s²3p⁶3d¹⁰4s¹. A single Cu-atom loses one 4s -electron and gets converted into a Cu⁺ ion. So in cuprous compounds, the electrovalency of Cu is 1.
• The nuclear charge of Cu is not sufficient enough to hold 18 electrons of Cu⁺ ion present in the outermost shell and hence, to acquire greater stability, Cu⁺ ion loses one more electron from the 3d -orbital to form Cu2⁺ ion.
• So, the electrovalency of copper in cupric compounds is 2. In the outermost shells of Cu⁺ and Cu2⁺, there are 18 and 17 electrons respectively. Thus, the octet rule is violated in both cases.

Exceptions to the octet rule in some elements

Shapes of ionic compounds

For the maximum stability of ionic compounds, cations, and anions form crystals by arranging themselves in regular and definite geometrical patterns so that the coulom forces of repulsion among the ions of similar charge, as well as electron-electron repulsion among the extranuclear electrons, are minimum.

• The shape of the crystals depends on the charges of the ions, their packing arrangements, and the ratio of the cation to anion radius.
• It can be shown by simple geometrical calculation that if the radius ratio is greater than 0.414 but less than 0.732, each cation is surrounded by the six nearest anionic neighbors. Such an array gives rise to the octahedral crystal ofthe compound.

So, during the formation of a NaCl crystal, each Na+ ion is surrounded by six neighboring Cl^–ions, and each such Cl^– ton is similarly surrounded by six neighboring Na⁺ ions, each ion lies at the center of an octahedron and the oppositely charged ions reside at tire corners of that octahedron. This type of arrangement is called 6-6 coordination.

Coordination number:

• In an ionic crystal, the number of oppositely charged adjacent ions that are equidistant from a particular ion (in 3D close packing) is called the coordination number (C.N.) of that ion.
• Positive and negative ions both have the same coordination number when there are equal numbers of both types of ions present (NaCl), but the coordination numbers for positive and negative ions are different when there are different numbers of oppositely charged ions (CaCl₂).

Example:

When the radius ratio (r+/r-) is less than 0.414, the coordination number is less than 6, but when the radius ratio (r+/r-) is more than 0.732, the coordination number is more than 6. In cesium chloride (CsCl), rCs+/rcr = L69A/1.81 Å = 0.933.

• So in a CsCl crystal, each Cs+ ion is surrounded by eight Cl- ions, and each Cl- ion is similarly surrounded by eight Cs+ ions. In this case, the coordination number of both the Cs+ and Cl- ions is 8. On the other hand, in zinc sulfide (ZnS) crystal, rZni+/rS^2- = 0.74A/1.84Å = 0.40.
• In the ZnS crystal, each Zn Ion is surrounded by four S^2- ions, and each S^2- lon is surrounded by four Zn²⁺ ions. So, the coordination number of both ions is 4.
• In the formation of ternary ionic compounds [such as calcium fluoride crystal (CaF₂) to maintain electrical neutrality, the coordination number of calcium ions (Ca²⁺) becomes twice the coordination number of fluoride ions (F^–).
• In CaF, crystal, each Ca²⁺ ion is surrounded by eight F^– ions while each F^– ion is surrounded by four Ca²⁺ ions.

Role of cation and anion in the formation of stable crystal:

The definite position of the anions surrounding a cation in a stable octahedral crystal is shown. Two anions, aligned vertically above and below the central cation have not been shown.

• In this case, the radius ratio (r+/r_) is in the range: of 0.414 – 0.732. If the size of the cation is small, then the value of (r+/r-) will be diminished and in this condition, the anions in contact will repel each other.
• But the cation, not being in contact with tire anions, will not attract them. Consequently, a stable octahedral crystal will not be formed. Instead, the ionic compound assumes a tetrahedral structure with coordination number 4 by disposing of its ions suitably, so that it gains stability.
• Stated differently, if the cation is much smaller in size than the anion, four anions are sufficient to surround the central cation—six anions are not required. When the cationic size is very large, the value of (r +/r-) increases.
• In such a case, the anions touch the cation but do not touch each other. So, a stable octahedral structure will not be formed. Hence, more anions should surround the central cation so that they touch each other to give rise to

A stable cubic structure with coordination number 8.

Radius ratio (r⁺/r^–),C.Nand crystalstructure

The given rules relating radius ratio with crystal structure apply only to those ionic compounds in which the cation and the anion bear the same charge (electrovalency).

For example, in NaCl, the electrovalency of both Na⁺ and Cl^– is 1, and in ZnS, the electrovalency of both Zn²⁺ and S^2- is 2, etc. The radius ratio rule (mentioned above) is not applicable for those ionic compounds in which the electrovalencies of cations and anions are not equal.

Note that a conglomeration of countless cations and anions leads to the formation of crystals of an ionic compound. Hence, there is no existence of a separate molecule and the entire crystal exists as a giant molecule.

Properties and characteristics of ionic compounds

• Physical state: In ionic compounds, there is no existence of separate molecular entities. Oppositely charged ions arrange themselves three-dimensionally, forming a crystal of definite geometrical shape. The compounds are solids at ordinary temperature and pressure.
• Melting and boiling points: In ionic compounds, the oppositely charged ions are held together, tightly by strong electrostatic forces of attraction, and hence a huge amount of energy is required to overcome these forces, i.e., to break the compact and hard crystal lattice. As a result, the melting and boiling points of ionic compounds are generally very high.
• Directional nature: Electrostatic force in an ionic compound extends in all directions. Hence, ionic bonds are non-directional.
• Isomerism: Due to the non-directional nature of ionic bonds, ionic compounds do not exhibit isomerism.
• Electrical conductivity: Ionic compounds do not conduct electricity in the solid state because oppositely charged. Ions are held together strongly with a coulomb force of attraction extending in all directions.
• But in the molten state or solution in a suitable solvent (like water), the ions being free from the crystal lattice, conduct electricity.
• Solubility: Ionic compounds generally dissolve in polar solvents i.e., solvents possessing high dielectric constant, (e.g., water), and insoluble in non-polar solvents (e.g., carbon disulfide, carbon tetrachloride, benzene, etc.).
• Isomorphism: Isoelectronicionic compounds generally exhibit the property of isomorphism (both of the ions have similar electronic configurations).

Two pairs of isomorphous compounds are—

Sodium Fluoride (Naf) and magnesium oxide (MgO):

Potassium sulfide (K₂S) and calcium chloride (CaCl₂)

Example:

⇒ [Na+(2, 8) F^–2, 8)], [Mg²⁺(2, 8) O^2-(2, 8)]; [K⁺(2, 8, 8) S²⁺(2, 8, 8)], [Ca²⁺(2, 8, 8) Cl^–(2, 8, 8)]

Ionic reaction and its rate:

In an aqueous solution, electrovalent compounds exist as ions. In any solution, the chemical reaction of ionic compounds is the chemical reaction of the constituent ions of that compound. As a result, a chemical reaction between ionic compounds in solution is very fast.

For example: On addition of an aqueous solution of AgNO₃ to an aqueous solution of NaCl,

A white precipitate of AgCl is formed immediately:

⇒ \(\mathrm{Na}^{+} \mathrm{Cl}^{-}+\mathrm{Ag}^{+} \mathrm{NO}_3^{-} \rightarrow \mathrm{AgCl} \downarrow+\mathrm{Na}^{+} \mathrm{NO}_3^{-}\)

Solvation of ions and solvation energy or enthalpy

Ionic compounds dissolve in polar solvents (for example water).

• Such solvent molecules strongly attract the ions present in the crystal lattice of solid ionic compounds and detach them from the crystal.
• When any ionic compound dissolves in a polar solvent, the negative pole of the solvent molecule attracts the cation that forms the crystal while its positive pole attracts the anion.
• As a result, the electrostatic force of attraction between the cations and anions decreases.
• If the magnitude of this attractive force of the polar solvent molecules exceeds the lattice energy of the solute, the ions present in the crystal get detached from the crystal lattice and are dispersed in the solvent.

Ions present in the solvent, being surrounded by a suitable number of solvent molecules (i.e., being solvated) are stabilized.

• For example, at the time of dissolution of NaCl in water, each Na⁺ and Cl^– ion being surrounded by six water molecules, becomes solvated to form stable hydrated ions. This process is known as solvation.
• The amount of energy released when one mole (one gram formula mass) of an ionic crystal is solvated in a solvent, is known as the solvation.

Energy evolved in the dissolution of ionic compounds:

The solvation energy is die driving force that brings about die total collapse of the structural framework work of the crystal.

• Higher the dielectric constant the capacity of the solvent to weaken the forces of attraction) and dipole moment of the solvent, the higher the die magnitude of die solvation energy.
• Moreover, solvation energy also depends on the sizes ofthe cations and anions. ΔH _solution = ΔH _solvation lattice where ΔH_solution = energy evolved in the dissolution of the ionic compound, ΔH _solvation = solvation enthalpy, and ΔH_lattice enthalpy of the ionic compound.
• If the solvation energy exceeds the lattice energy, then that ionic compound is soluble in that solvent but if It is much less, then the ionic compound is insoluble in that solvent.

Example: CaF₂ is insoluble in water while CaCl₂ is appreciably soluble. This implies that the lattice energy of CaF₂ is more than the solvation energy of its constituent ions, but the lattice energy of CaCl₂ is less than the solvation energy of its constituent ions.

• It is to be noted that ionic compounds do not dissolve in non-polar solvents (turpentine oil, gasoline, etc.) because solvation of ions by the non-polar solvent is not possible.
• For most of the ionic compounds, ΔH° is +ve, i.e., the dissolution process is endothermic. Hence, the solution solubility of most of the salts in water increases with the temperature rise.

Covalency And Covalent Bond

Formation of ionic bonds is not possible when the atoms of similar or almost similar electronegativities combine. This is because the electron affinity of both atoms is of the same or approximately the same order.

• Therefore, the electron transfer theory’ (as discussed in the case of ionic bond formation) cannot explain the bonding in molecules such as H₂, O₂, N₂, Cl₂, etc.
• To explain the bonding in such molecules, G. N. Lewis (1916) proposed an electronic model, according to which, the chemical bond in a non-ionic compound is covalent.
• He suggested that when both the atoms taking part in a chemical combination are short of electrons than the stable electronic configuration of the nearest noble gas, they can share their electrons to complete their octets (duplet in the case of H).
• This type of bond, formed by mutual sharing of electrons, is called a covalent bond.
• During the formation of a covalent bond, the two combining atoms contribute an equal number of electrons for sharing. The shared electrons are common to both atoms and are responsible for holding the two atoms together.

Since such a combination of atoms does not involve the transfer of electrons from one atom to another, the bonded atoms remain electrically neutral.

Covalency:

To achieve the electronic configuration of the nearest noble gas, an equal number of electron(s) from the outermost shells of two combining atoms remaining in the ground state or excited state, form one or more electron pairs that are evenly shared by the two atoms. The capacity of the elements for this type of chemical combination is called covalency.

Covalent bond: 

The force Of attraction that binds atoms of the same or different elements by the mutual sharing of electrons is called a covalent bond. The atoms involved in covalent bond formation contribute an equal number of electrons for sharing. The shared electron pair(s) are common to both atoms.

Covalent Molecules:

• The molecules that consist of atoms held together by covalent bonds are called covalent molecules.
• The number of valence electrons shared by an atom of an element to form covalent bonds is called the covalency of that element.
• Therefore, the covalency of an element in a covalent molecule is, in fact, equal to the number of covalent bonds formed by its atom with other atoms of the same or different element.

For example:  In a carbon dioxide molecule (0=C=0), the covalency of carbon is 4 and that of oxygen is 2.

Driving force behind covalent bond formation:

Any covalent bond is formed by the combination of two electrons of opposite spin.

The driving forces behind the formation of a covalent bond are the electromagnetic force of attraction developed in the pairing of two electrons of opposite spin and the attainment of stability by forming an inert electron core.

Types of covalent molecules

• Homonuclear covalent molecule: The molecules formed when atoms of the same element are joined together by covalent bonds are called homonuclear covalent molecules, for Example; H₂, O₂, N₂, Cl₂, etc.
• Heteronuclear covalent molecule: The molecules formed when atoms of different elements are joined together by covalent bonds are called heteronuclear covalent molecules, for Example; NH₃, H₂O, HC₁, CH₄, etc.

Types of covalent bonds

• Single bond: The bond formed by the sharing of one electron pair between two atoms is known as a single bond and is represented by ( — ).
• Double bond: The bond formed by sharing two electron pairs between two atoms is known as a double bond and is represented by (=).
• Triple bond: The bond formed by the sharing of three electron pairs between two atoms is known as a triple bond and is represented by (=).

Examples: There exists a single bond between the two hydrogen atoms in a hydrogen molecule (H —H), a double bond between the two oxygen atoms in an oxygen molecule (0=0), and a triple bond between the two nitrogen atoms in a nitrogen molecule (N=N).

Lewis dot structure

The structure of a covalent compound expressed by writing Lewis symbols ofthe participating atoms using one pair of dots between each pair of atoms for each covalent bond where a dot represents an electron is called Lewis dot structure.

Electrons are normally represented by dot or cross (x) signs. Lewis dot structures of fluorine and hydrogen chloride are shown here.

Valence electrons that do not participate in covalent bond formation are simply written as pairs of dots surrounding the symbol of the concerned atom.

The steps involved in writing Lewis dot structure are as follows:

1. The total number of valence electrons ofthe atoms present in a particular molecule, Orion should be calculated.
2. If the species is a cation, the number of electrons equal to the units of +ve charge should be subtracted from the total, and if the species is an anion, the number of electrons equal to the units of -ve charge should be added to the total. This gives the total number of electrons to be distributed.
3. The skeletal structure is written by placing the least electronegative atom in the center (except hydrogen) and more negative atoms in the terminal positions. Note that the monovalent atoms like H and F always occupy the terminal positions.
4. One shared electron pair should be placed between every pair of atoms to represent a single bond between them. The remaining pairs of electrons are used either for multiple bonding or to show them as lone pairs, keeping in mind that the octet of every atom (except) is completed.
5. Remember that oxygen atoms do not bond to each other except in cases of O2, O3, peroxides, and superoxides.

Example: Lewis dot structure of HCN molecule:

1. Total number of valence electrons of the atoms in HCN molecule =I (for H-atom) +4 (for C-atom) +5 (for Natom)=10.
2. The skeletal structure of the molecule is HCN.
3. One shared pair of electrons is placed between H and C and one shared pair is placed between C and N. The remaining electrons are treated as two lone pairs on N and one lone pair on C. H:C: N:O
4. Since the octets of C and N are incomplete, multiple bonds are required between them.
5. To complete their octets, a triple bond (i.e., two more shared pairs of electrons) should be placed between them. Thus, the Lewis dot structure of the hydrogen cyanide molecule is:

Lewis dot structure of some molecule or ions 

Example of covalent bond formation:

Formation of a chlorine molecule (Cl₂): Two Cl-atom combine to form a Cl₂ molecule. Electronic configuration of a Cl -atom (Z = 17) 1s²2s²2p⁶3s²3p⁵ i.e., 2, 8, 7

• Thus, each Cl -atom has seven electrons In Its valence shell and needs one more electron to attain a stable electronic configuration of Ar (2, 8, 8), i.e., to achieve the octet.
• During combination, both the Cl -atoms contribute one electron each to form a common shared pair. In this way, both of them complete their octets.
• As a result, a covalent bond Is formed between the two chlorine atoms to produce a chlorine molecule. The completed octets are generally represented by enclosing the dots around the symbol of the element by a circle or ellipse.

The electron pair (s) shared by the bonding atoms is known as the shared pair or bond pair and the electron pair not involved In sharing is known as the unshared pair or lone pair.

Formation of oxygen molecule (O₂):

Formation of nitrogen molecule (N₂):

Formation of water molecule (H₂O)

Formation of carbon dioxide molecule (CO₂):

Formal Charge

A molecule Is neutral and its constituent atoms do not carry charges. In polyatomic ions, the net charge is possessed as a whole and not by individual atoms. In some cases, charges are assigned to individual atoms. These are called formal charges.

The formal charge of an atom in a polyatomic molecule or ion is defined as the difference between the number of valence electrons of that atom in an isolated atom and the number of electrons assigned to that atom in the Lewis structure. It can be expressed as follows.

If the atom has more electrons in the molecule Orion than in the free or isolated state, then the atom possesses a negative formal charge and if the atom has fewer electrons in the molecule or ion than in the free or isolated state, then the atom possesses a positive formal charge.

Calculation of Formal Charges Of some Molecules And Ions 

Advantages of formal charge calculation:

• The main advantage of the calculation of formal charges is that It helps to select the most stable structure from many possible Lewis structures for a given molecule or Ion.
• Generally, the lowest energy structure (most stable) is the one with the lowest formal charges on the constituent atoms for a particular molecule or ion.

Factors favoring the formation of covalent bonds

• Number of valence electrons:
  • Formation of a covalent bond is favored when each of the combining atoms possesses 4, 5, 6, or 7 (except H) valence electrons.
  • Such atoms can form 4, 3, 2, or 1 electron pair(s) with one or more atoms to achieve the octet by mutual sharing. So, elements of groups 14, 15, 16, and 17 form covalent bonds easily.
• High ionization enthalpy:
  • The atoms having high ionization energy are unable to form electrovalent bonds. They form molecules through the formation of covalent bonds.
  • This behavior is observed in the case of p-block elements.
• Comparable electron-gain enthalpies:
  • The formation of a covalent bond is favored when the participating atoms have equal or nearly equal electron-gain enthalpies, they should have equal or nearly equal electron affinity.
• Comparable electronegativities:
  • The two atoms involved in covalent bond formation should have equal or nearly equal values of electronegativity because in that case no transfer of electrons from one atom to another takes place and thus, the formation of a covalent bond is favored.
• High nuclear charge and small internuclear distance:
  • During the formation of a covalent bond, the electron density is concentrated between the two nuclei of the combining atoms, which is responsible for holding the two nuclei together.
  • The greater the nuclear charge and the smaller the internuclear distance, the greater the tendency for the formation of covalent bonds.

Characteristics of covalent compounds

• Physical state: Covalent compounds are composed of discrete molecules. The intermolecular forces of attraction between them are usually very weak. Hence, covalent compounds exist in a gaseous or liquid state. However, a few covalent compounds such as urea, sugar, glucose, etc. exist as solids because of stronger intermolecular forces.
• Melting and boiling points: The attractive force between the molecules of covalent compounds is usually weak and consequently, a lesser amount of energy is required to overcome these forces. As a result, covalent compounds possess low melting and boiling points compared to ionic compounds.
• Electrical conductivity: Covalent compounds do not possess negatively and positively charged ions so, they usually do not conduct electricity in the fused or dissolved state.

However, in some cases, the covalent compound dissolved in a polar solvent reacts with the solvent molecules to form ions and thus conduct electricity.

For example, being a covalent compound, hydrogen chloride is a non-conductor of electricity in the pure state but when dissolved in water, it reacts with water to form hydronium ions and chloride ions. Hence, an aqueous solution of hydrogen chloride (i.e., hydrochloric acid) is capable of conducting electricity.

⇒ \(\mathrm{HCl}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Cl}^{-}\)

Solubility:

• Covalent compounds are usually soluble in non-polar solvents but insoluble in polar solvents (in conformity with the principle, “like dissolves like”).
• For example, the covalent compound carbon tetrachloride does not dissolve in a polar solvent (like water) but it readily dissolves in the non-polar solvent, benzene. However, some covalent compounds such as alcohol, acetic acid, hydrogen chloride, glucose, etc. dissolve In a die polar solvent, water.
• This Is because they are themselves polar compounds and react with water or participate In the formation of hydrogen bonds with water molecules. For example, HC1 dissolves In water and forms H3O+ and Cl- Ions while glucose (C₆H₁₂O₆) having five hydroxyl (-Oil) groups, dissolves In water by forming hydrogen bonds with water molecules.

Rate of chemical reaction:

• The reactions of covalent compounds Involve the breaking of strong covalent hond(s) present In their molecules.
• Since It requires sufficient energy and time, the chemical reactions of covalent compounds occur at a relatively slower rate. For example, the formation of ethanol from glucose by fermentation takes nearly 3 days.

⇒ \(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \rightarrow{\text { Zymase }} 2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{CO}_2\)

Directional characteristics of covalent bond:

• Since the atomic orbitals have definite spatial orientation and the covalent bonds are formed by overlapping of atomic orbitals, the bonds possess directional properties.
• For example, 4 covalent bonds of a sp³ -hybridized C-atom are directed toward the four corners of a tetrahedron and for days, the shape of the CH₄ molecule is tetrahedral.

Isomerism:

• Since the covalent bonds are rigid and directional, the atoms involved in the formation of a covalent molecule may be oriented differently.
• Two or more structurally different compounds having different chemical and physical properties may be represented by a single molecular formula.
• Such compounds are called structural isomers, In other words, covalent compounds exhibit structural isomerism.

For example, both dimethyl ether and ethyl alcohol have the same molecular formula (C₂H₆O), but different structural formulas i.e., they are isomers.

For elements such as hydrogen and nitrogen, oxygen, and fluorine (elements of the second period of the periodic table), the number of unpaired electrons (s) in their valence shells gives a measure of their covalency. Therefore, the covalencies of H, N, O, and F are 1,3, 2, and 1 respectively.

⇒ \(\begin{gathered}
\mathrm{H}: 1 s^1 ; \mathrm{N}: 1 s^2 2 s^2 2 p_x^1 2 p_y^1 2 p_z^1 ; \mathrm{O}: 1 s^2 2 s^2 2 p_x^2 2 p_y^1 2 p_z^1 \text {; } \\
\mathrm{F}: 1 s^2 2 s^2 2 p_x^2 2 p_y^2 2 p_z^1
\end{gathered}\)

For elements such as Be, B, and C, their covalencies are not determined by the number of unpaired electrons(s) in their valence shells in the ground state.

In the excited state, out of two electrons in the 2s-orbital, one electron gets promoted to the 2p-orbital. Thus, there are 2, 3, and 4 unpaired electrons respectively in the valence shells of Be, B, and C. So, their covalencies are 2, 3, and 4 respectively.

• Elements belonging mainly to the 3rd, 4th, and 5th periods of the periodic table possess variable covalencies.
• These elements possess vacant orbitals in their valence shell and are capable of promoting one of the paired electrons of that shell to the vacant orbitals. The number of electrons to be promoted depends on the energy available for excitation.
• Thus, such an element exhibits more than one covalency depending on the availability of unpaired electrons. This is termed as variable covalency. For example— In PCl₃ and PCl₅, covalencies of P are 3 and 5 respectively;
• In H₂S, SF₄, and SF₆, sulfur exhibits covalencies of 2, 4, and 6 respectively, and in IC1, IC1₃, IF₅, and IF₇ iodine has covalence of 1, 3, 5, and 7 respectively.

Variable covalency of phosphorus: 3,5

1. Trlcovalency of P:

The outermost electronic configuration of phosphorus in its ground state:

In the ground state, phosphorus1 has only three unpaired valence electrons in 3px, 3py, and 3pz orbitals. So, the normal covalency of phosphorus is 3.

Example: In phosphorus trichloride (PCl₃), phosphorus (sp³ hybridized) exhibits a covalency of 3.

2. Pentacovalency of P:

In the excited state, phosphorus possesses five unpaired electrons in its valence shell by promoting one of its 3s -electrons to the vacant 3dorbital. Thus, it exhibits a covalence of 5.

Example: In phosphorus pentachloride (PCl₃), phosphorus (sp³d -hybridized) possesses a covalency of 5.

Variable covalency of sulphur: 2,4,6

1. Bicovalency of S:

The outermost electronic configuration of sulfur in its ground state.

In the ground state, sulfur has only 2 unpaired electrons in 3py and 3pz orbitals. Hence, the covalency of sulfur in the ground state is 2.

Example: In hydrogen sulfide (H₂S), sulfur (sp³ hybridized) exhibits a covalency of 2.

2. Tetracovalency of S:

On excitation, one of the paired electrons in 3px -orbital is promoted to vacant 3d -orbital. This results in 4 unpaired electrons in its valence shell.

So, the covalency of sulfur in the first excited state is 4. Example: in sulfur tetrafluoride(SF₄), sulfur (sp3 d hybridized) exhibits a covalency of 4.

3. Hexacovalency of S:

In the case of the second excitation, one of the 3s -electrons gets promoted to vacant 3d -orbital. This results in 6 unpaired electrons in its valence shell.

Hence, the covalency of S in the second excited state is 6.

Example: In sulfur hexafluoride (SF₆), sulphur {sp³d² hybridised) exhibits a covalency of 6.

Variable covalency of iodine: 1,3, 5, 7

1. Monocovalency of I:

The outermost electronic configuration of iodine in its ground state.

One unpaired electron indicates the monovalency of I.

2. Tricovalency of I:

In its first excited state, one of the paired electrons from the 5p -orbital gets promoted to a vacant 5d -orbital. This leads to the presence of unpaired electrons in its valence shell.

Hence, the covalency of iodine in the first excited state is 3.

Example: In iodine trifluoride (IF₃), iodine (sp²d -hybridized) exhibits a covalency of 3.

3. Pentacovalency of I:

In the second excited state, one of the paired electrons from 5p_x -orbital is promoted to vacant 5d -orbital. This results in the presence of 5 impaired electrons in the valence shell of the I-atom.

Example: In iodine pentafluoride (IF₅), iodine (sp³d³ – hybridized) exhibits a covalency of 5.

4. Heptacovalency of I:

I-atom attains the third excited state by promoting one of the paired electrons from 5s -orbital to vacant 5d -orbital. This leads to the presence of 7 unpaired electrons in the valence shell of i-atoms.

Since the Iodine atom now possesses 7 unpaired electrons the covalency of iodinein in the third excited state is 7.

Example: In iodine heptafluoride (IF), iodine (sp³d³- hybridized) exhibits a covalency of7.

Maximum covalency:

Maximum covalency of an element is the maximum number of unpaired electrons that an atom of an element possesses after promoting electrons from s and the p -orbitals to d -orbitals, i.e., it is the maximum number of covalent bonds that an atom of an element can form. For example, the maximum covalencies of P, S, and I are 5, 6, and 7 respectively.

Limitations of the octet rule

The octet rule, although useful for understanding the structures of most organic compounds, fails in many cases and has several exceptions.

Some important exceptions to the rules are as follows:

Incomplete octet of the central atom:

• Elements of groups, 2, and 13 are not expected to form covalent compounds as they possess less than four electrons in their valence shell and cannot achieve an octet by sharing electrons.
• But, several covalent compounds of these elements are known to exist, which violates the octet rule such as LiCl, BeCl₂, BF₃, AlCl₃, etc.

These compounds with an incomplete octet of the central atom are called hypovalent compounds or electron-deficient compounds.

Expansion of octet of the central atom:

The octet rule is also found to be violated in compounds like PCl₅, SF₆, and IF₇ in which the central atoms possess more than eight electrons in their valence shells, i.e., they possess expanded octets.

These compounds with an expanded octet of the central atom are called hypervalent compounds.

Odd electron molecules:

There are some molecules and ions in which the atoms bonded to each other contain an odd no. of electrons (usually 3).

These bonds formed by three electrons are called odd electron bonds and the corresponding molecules are called odd electron molecules. Octet rule is not satisfied for all the atoms of such molecules. Some common examples are as follows:

Formation of compounds by noble gases:

Noble gases have filled octets and hence are not expected to form compounds.

• However, it has been found that some noble gases like xenon and krypton combine with oxygen and fluorine to form a large number of compounds such as XeF₂, XeOF₂, XeOF₄, XeF₆, KrF₂, etc. In these compounds, Xe or Kr have expanded octets.
• This theory cannot explain the shapes of covalent molecules. It cannot explain the relative stability of the molecules in terms of energy.

The explanation for the deviations from the octet rule

Sidgwick’s concept of maximum covalency:

• According to Sidgwick, it is not always necessary for an atom of an element to achieve the octet combination.
• He postulated his views in the form of an empirical rule called the rule of maximum covalency. According to this rule, the maximum covalency of an element depends on its position in the periodic table.
• For example, the maximum covalency for H belonging to the first period is 2, the second period (Li to F) is 4, the third (Na to Cl) and fourth period (K to Br) is 6, and for elements of higher periods, it is 8. So, the formation of compounds like PCl₅ and SF₆ where P and S exhibit penta covalency and hexa covalency respectively, is not irrelevant. The modem electronic concept supports Sidgwick’s concept.

Explanation: In the formation of covalent bonds, atoms of the elements belonging to the second period use one orbital of 2s -subshell and three orbitals of 2p -subshell of L -shell {n = 2). Therefore, they can share a maximum of 8 electrons to form covalent bonds, i.e., their maximum covalency is 4.

The maximum covalency shown by the elements of the third period and that of the higher periods is more than 4. This is because their atoms can form compounds using s, p, and d -orbitals of their outermost shell [M(n = 3), N(n = 4), O(n = 5), etc], For exhibiting higher valency, one or more electrons are promoted from s and p -orbitals to vacant d -orbitals having slightly higher energy. So these elements, depending on the requirement of valency, can utilize 5, 6, 7, or 8 orbitals.

Sugden’s concept of single electron linkage or singlet linkage:

According to Sugden, the central atoms of molecules like PCl₅, SF₆, etc., attain octets by the formation of one or more one-electron bonds. To explain their structures, he proposed the formation of a new type of bonding called singlet linkage.

Example: In the PCl₅ molecule, the P-atom with the help of its 5 valence electrons, forms 3 shared pairs or normal single covalent bonds with three Cl-atoms, and the remaining 2 electrons are used to bond with two Cl-atoms by singlet linkages.

Similarly, the S-atom in SFg molecule forms 2 normal single covalent bonds with two F-atoms using 2 of its 6 valence electrons and the remaining 4 electrons are used to form singlet linkages with four F-atoms.

Limitations of the concept of singlet linkage:

Singlet electron linkage is weaker than a normal covalent bond.

• Vapors of PCl₅ dissociate into PCl₅ and Cl₂ at 300°C. This indicates that two P —Cl bonds in PClg are comparatively weaker than the remaining three P—Cl bonds.
• However, SFg is a very stable molecule and experimental results show that all the six S —F bonds are similar.
• Hence, it is not possible to distinguish between singlet electron linkage and normal covalent bond in the case of SF₆. There are 5 valence electrons present in the outer shell of both nitrogen and phosphorus.
• However, nitrogen forms only trihalides (NX₃, X = Cl, Br, I), while phosphorus forms both trihalides (PX₃) and pentahalides (PX₅). If singlet electron linkages exist, then nitrogen would also have formed pentahalides.
• Although the concept of singlet electron linkage explains the formation and properties of a few molecules, it fails in most of cases.
• There are 5 valence electrons present in the outer shell of both nitrogen and phosphorus. However, nitrogen forms only trihalides (NX₃, X = Cl, Br, I), while phosphorus forms both trihalides (PX₃) and pentahalides (PX₃).
• If singlet electron linkages exist, then nitrogen would also have formed pentahalides. Although the concept of singlet electron linkage explains the formation and properties of a few molecules, it fails in most of the cases octet occurs for them.
• Group 16 elements like S, Se, etc., belonging to third, and fourth, periods form covalent compounds in which the normal valency of the elements is 2 and higher valencies of the elements are 4 and 6.
• In normal valency, those elements have fulfilled octet. In higher valencies, octets occur. Group 17 elements like Cl, III, I (tie, belonging to third, fourth, fifth….. form covalent compounds In

Ionic distortion and development of covalent character In Ionic compounds:

• Fajan’s rule hikes several covalent compounds possessing ionic characters, many Ionic compounds are also found to carry a partial covalent nature.
• For example, Is an Ionic compound hut due to its significant covalent character, It Is more soluble in organic solvents water.

Development of covalent character in an ionic compound:

When two oppositely charged Ions approach each other, the cation attracts the electron cloud of the anion but repels Its nucleus.

• This results in distortion of (the electron cloud around the anion.
• This Is known as the polarisation of the anion.
• The power of the cation to polarise the anion Is called Its polarising power.
• The tendency of the anion to get polarised by the cation is called Its polarisability. Such polarisation results In the transportation of electron cloud towards the cation to produce an overlapping zone.
• Consequently, the Ionic character of the bond decreases and the covalent character increases. The following picture shows the gradual development of covalent character with an increase in polarisation.

It is to be noted that the reverse polarisation of the cationic charge cloud by the anion will indeed be very small, as the cation has a more compact charge cloud.

The polarising power of the cation is expressed by the term ‘ionic potential: It is the ratio of charge to radius of the corresponding cation and is expressed by the sign, (Phi). Thus,

⇒ \(\text { Ionlc potential, } \phi=\frac{\text { Charge of the cation }}{\text { Radlus of the cation }}\)

On moving front from left to right In a period, the charge of the cation Increases while Its radius decreases. This results In an Increase lu the value of phi.

⇒ \(\text { For example, } \phi\left(\mathrm{Na}^{+}\right)<\phi\left(\mathrm{Mg}^{2+}\right)<\phi\left(\mathrm{Al}^{3+}\right)<\phi\left(\mathrm{Si}^{4+}\right)\)

On the other hand, on moving down the group the cationic charge remains unaltered but the cationic radius increases. Consequently, the value of decreases. For example

⇒\(\phi\left(\mathrm{LI}^{+}\right)>\phi\left(\mathrm{Na}^{+}\right)>\phi\left(\mathrm{K}^{+}\right) ; \phi\left(\mathrm{Be}^{2+}\right)>\phi\left(\mathrm{Mg}^{2+}\right)>\left(\mathrm{Ca}^{2+}\right)\)

In the case of some metals with different oxidation states, the value of phi Increases with an Increase In oxidation number,

⇒ \(\text { For example, } \phi\left(\mathrm{Sn}^{2+}\right)<\phi\left(\mathrm{Sn}^{4+}\right) ; \phi\left(\mathrm{Fe}^{2+}\right)<\phi\left(\mathrm{Fe}^{3+}\right)\)

With an Increase In the value of <p, the polarising power of the cullon increases which ultimately Increases the covalent character of the Ionic compound.

Fajan’s rule:

The polarising power of the cation and the polarisability of the anion (i.e., the extent of polarization causing the development of covalent character in an ionic compound) are governed by certain rules known as Fajan’s rules.

According to these rules, the covalent character of an ionic compound depends on the following factors:

Size of the cation:

For the cations having the same charge, the value Φ increases with a decrease in the size of the cation. Hence, the deformation of the anion increases which in turn enhances covalency.  From the table given below, it is observed that the melting point decreases (i.e., the covalent character of anhydrous chlorides of alkaline earth metals increases) with a decrease in the radii of the cations.

The melting point however decreases from NaCl to KCl to RbCl, due to successive decreases in in the lattice energy.

Size of the anion: In a large-sized anion:

• The outermost electrons are less tightly held by the nucleus and hence, would be more easily distorted by the cation.
• Thus, the larger the anion, the higher its polarisability and the greater the covalent character of the compound formed.

The following table shows that the melting points decrease (i.e., the covalent character of the anhydrous calcium halides increases) with the increase in the size of the anion.

The greater amount of charge on the cation or anion:

The ionic potential Φ of the cations increases with an increase in cationic charge and a decrease in cationic radii. Consequently, the resulting compound is found to possess a more covalent character. It becomes evident.

As the charge on an anion increases, valence electrons become more loosely held by the nucleus and therefore, it gets more easily deformed by the cation. Thus, the greater the charge on an anion, the higher its polarisability, and the greater the covalent character of the compound formed.

Configuration of cation:

• Between the two cations having the same size and charge, the one with 18 electrons in the outermost shell (S²p⁶d¹⁰) i.e., with pseudo noble gas configuration, has greater polarising power than the other with 8 electrons in the outermost shell (S²p⁶), i.e., with noble gas configuration.
• This is because, in the case of cations having 18 valence electrons, there is a poor screening effect due to the presence of d electrons.
• Thus, a more effective nuclear charge polarises the anion to a greater extent causing the development of a more covalent character in the compound formed, the following table shows that the melting points of the anhydrous chlorides of coinage metals are less compared to those of the anhydrous chlorides of alkali metals with noble gas electronic configuration.

The chlorides of Cu, Ag, and Au, therefore, possess a greater covalent character.

The dielectric constant of the medium:

A polar medium possessing a high dielectric constant tends to weaken electrostatic forces of attraction existing between oppositely charged ions. As a result, ions remain separated in a polar medium and effective polarisation does not take place.

However, effective polarisation takes place in a nonpolar medium having a low dielectric constant. Hence, an ionic compound exhibits more covalent character in a non-polar medium than in an apolar medium.

Effect of polarisation on the properties of compounds

• Solubility: As polarisation increases, the covalent character as well as the tendency of ionic compounds to get dissolved in non-polar solvents increases.
• Solubility of silver halides (AgX) in water: Order of polarisability of halide ions: I¯> Br¯ > Cl¯ > F¯ (polarisability increases with increase in size).
• Therefore, the covalent character of silver halides follows the order: Agl > AgBr > AgCl > AgF, i.e., the ionic character of these halides follows the reverse order. Consequently, the solubility of silver halides in the polar solvent, water, follows the order: AgF > AgCl > AgBr > Agl.
• Solubility of KC1 and K1 in alcohol: Since I¯ ion is larger than Cl¯ ion, I¯ gets more easily polarised than Cl¯. So, KI possesses more covalent character compared to KC1 and thus, it is more soluble in alcohol (a less polar solvent having low dielectric constant) compared to KC1.
• Thermal stability of metal carbonates: For the carbonates of Be, Mg, and Ca belonging to group-2 and possessing common anion (CO₃), ionic potential (cf²) of the cations follows the order: Be²⁺ > Mg²⁺ > Ca²⁺.

So, the ionic nature of these compounds runs as follows:

BeCO₃ < MgCO₃ < CaCO₃. Thus, their thermal stability follows the order: of BeCO₃ < MgCO₃ < CaCO₃. Therefore, on moving down a particular group, the thermal stability ofthe metal carbonates gradually increases.

Color of different salts of metal:

• The tendency of the anions to get polarised increases with an increase in size. This facilitates the transition of electrons from the filled orbital of anions to the unfilled orbital of cations.
• The energy required for the electronic transition of an anion having high polarisability is lower than the energy required for that having low polarisability.
• Anions having high polarisability obtain the energy required for the electronic transition from the visible range while those with low polarisability, from the ultraviolet region.
• Thus the compounds having anions with high polarisability, are generally colored, depending on the wavelength absorbed, while those having anions with low polarisability are generally white.

For example, HgCl₂ is white but Hgl₂ is red; AgCl is white but Agl is yellow; PbCl₂ is white, but Pbl₂ is golden yellow.

• Non-existence of compound: PbCl₄ exists but Pbl₄ has no existence. In Pbl₄, the charge on the cation, Pb⁴⁺ is much higher and it strongly polarises the large anion, I-.
• The degree of polarisation is so high that the two I- ions are oxidized to the I₂ molecule by donating two electrons and the Pb⁴⁺ ion is reduced to the Pb²⁺ ion by gaining two electrons.

⇒ \(\stackrel{+4}{\mathrm{PbI}_4} \rightarrow \stackrel{+2}{\mathrm{PbI}_2^{-1}}+\stackrel{0}{\mathrm{I}}_2\)

Hence, Pbl₄ does not exist. On the other hand, PbCl₄ exists as the degree of polarization of relatively small Cl¯ is not very high, and hence no such electron transfer occurs. For the same reason, Fel₃ does not exist but Fel₂ does.

Coordinate Covalency Bond And Coordinate Bond Or Dative Bond

In 1921, Perkins suggested a special type of covalency known as coordinate covalency.

Coordinate Covalency:

Coordinate Covalency is a special type of covalent bond in which the shared pair is contributed by only one of the two combining atoms. This electron pair is shared by both of the combining atoms due to which both of them attain octet and the valency hence generated is called coordinate covalency.

Coordinate Bond:

A coordinate bond is a special type of covalent bond in which the shared pair of electrons is contributed by one of the two combining atoms.

Coordinate Compounds Compounds

• A coordinate bond is formed between two atoms, one of which has completed its octet and the other is short of two electrons to complete its octet.
• The former atom which donates a pair of electrons (lone pair) is known as the donor and the latter atom which accepts the electron pair to complete its octet is known as an acceptor.
• A coordinate bond is represented by an arrow pointing from the donor towards the acceptor Like a covalent bond, a coordinate bond is formed by overlapping of atomic orbitals of two atoms.
• As the atomic orbitals have specific orientations in space, coordinate bonds also have specific orientations in space. Note that a coordinate bond once formed cannot be distinguished from a covalent bond.

Conditions for the formation of coordinate bonds

1. The donor atom must contain at least one lone pair of electrons in its valence shell.
2. The acceptor atom must have at least one vacant orbital in its valence shell where the lone pair of electrons from the donor atom can be accommodated.
3. The lone pair of the donor atom must be equally shared by both the donor and the acceptor atoms.

The electron pairs, which present the valence shell ofthe atoms or ions, which do not participate in the bond formation are termed ‘lone pairs of electrons.

For example, N-atom in NH₃ molecule and O-atom in H₂O molecule The molecules having atoms with lone pair of electrons e.g., ammonia (NH₃), water (H₂6), methyl amine (CH₃NH₂), aniline (CgH₅NH₂), phosphine (PH₃), triphenylphosphine (PH₃P), alcohols (ROH), phenol (C₆H₅OH), diethyl ether (C₂H₅OC₂H₅), etc.] act as a donor in the coordinate bond formation.

On the other hand, hydrogen ions (H+) or molecules having atoms with electron sextet (e.g., BF₃, BH₃, etc.), or metal ions containing vacant orbital in their valence shell act as acceptors in the coordinate bond formation.

Mechanism for the formation of coordinate bond

The donor atom transfers one electron of its lone pair to the acceptor atom and as a result, the donor atom acquires a positive charge and the acceptor atom acquires a negative charge.

The two ions then contribute one electron each and this electron pair is shared by both the atoms to form a single covalent bond between them.

⇒ \(\stackrel{+}{\mathrm{A}}+\stackrel{\rightharpoonup}{\cdot} \longrightarrow \stackrel{+}{\mathrm{A}}: \overline{\mathrm{B}} \text { or } \stackrel{\mathrm{A}}-\overline{\mathrm{B}} \text { or } \mathrm{A} \rightarrow \mathrm{B}\)

Thus, the formation of a coordinate bond involves the transfer of electrons (as in the formation of an electrovalent bond) as well as the sharing of electrons (as in the formation of a covalent bond). Therefore, a coordinate bond may be regarded as a combination of a polar electrovalent bond and a non-polar or less polar covalent bond. For this reason, a coordinate bond is termed a semipolar bond.

Examples of coordinate bond formation:

Formation of an addition compound (complex) involving ammonia and boron trifluoride: In ammonia (NH₃), the nitrogen atom has a lone pair and the boron atom in boron trifluoride (BF₃)is short of two electrons to achieve its octet.

Therefore, when NH₃ is subjected to react with BF3, the N-atom donates its lone pair to the Batom to form a coordinate bond which holds them together forming the addition compound, \(\mathrm{H}_3 \mathrm{~N} \rightarrow \mathrm{BF}_3.\)

An ion or a molecule that can donate an electron pair is called a Lewis base and an ion or a molecule that can accept an electron pair is called a Lewis acid. In the above example, ammonia is a Lewis base while boron trifluoride is a Lewis acid.

Formation of ammonium ion (NH⁺₄ ): [Donor: N-atom of NH₃ molecule, Acceptor: H⁺-ions]

• From experimental observations, it can be said (bat all the four N— 1-1 bonds in ammonium ions are equivalent.
• So the ammonium ion can be represented as shown above.
• This concept is also applicable to those compounds in which coordinate bonds are present.

Formation of fluoroborate ion (BF^–₄): [Donor: F^– -ion, Acceptor: B -atom of BF₃ molecule]

 Formation of hydronium ion or hydroxonium ion (H₃O⁺): [Donor: O-atom of H₂O molecule, Acceptor: H⁺-ion]

Formation of ozone molecule (O₃): [Donor: Central O-atom, Acceptor: Terminal O-atom]

Formation of the sulphuric acid molecule (H₂SO₄): [Donor: S-atom, Acceptor: O-atom]

Formation of the nitric acid molecule (HNO₃): [Donor: N atom, Acceptor: O-atom]

Formation of Al₂Cl₆ (dimer of aluminium chloride): [Donor: Cl-atom, Acceptor: Al-atom]

Formation of an orthophosphoric acid molecule (H₃PO₄) : [Donor: P-atom, Acceptor: O-atom

Formation Of ammonium chloride (coexistence of electrovalent valency, covalency, and coordinate valency):

Ammonia reacts with an aqueous solution of hydrogen chloride to form ammonium chloride. In the HCl molecule, the highly As a result, polarity develops in the H—Cl bond.

In the presence of H₂O molecules, the polar H —Cl bond undergoes dissociation forming the H+ ion and Cl ion. The O-atom of H₂O donates a pair of electrons to H⁺ to produce hydroxonium ion (H₃O⁺) through the formation of a coordinate bond.

In the NH₃ molecule, since the N-atom is less electronegative than the O-atom, it exhibits a greater tendency to donate its unshared pair. So, NH₃ accepts a proton (H⁺) from H₃O⁺ and produces an NH₄ ion by forming a coordinate bond.

The NH₄ ion thus formed combines with Cl- ion through electrostatic force of attraction to produce crystals of NH₄Cl.

Thus, in NH₄Cl, three H -atoms are attached to the N-atom by three covalent bonds, the fourth H-atom is attached to it by a coordinate bond and the two ions (NH₄ and Cl-) are held together by an ionic bond, i.e., in NH₄Cl, there exists electrovalency, covalency and Coordinate covalency. Some other examples of this type of compound are LiAlH₄, NaBH₄, Na₂HPO₄, etc.

Characteristics of coordinate compounds

Coordinate bonds are a special type of covalent bond and coordinate compounds are in fact, covalent compounds. Hence, the characteristics of coordinate compounds are similar to those of the covalent compounds.

Some of their important characteristics are described below:

1. Physical state: Coordinate compounds exist as gases, liquids, and solids under ordinary conditions.
2. Melting and boiling points: Coordinate bonds are semipolar. Due to this, coordinate compounds are more polar than covalent compounds but less polar than ionic compounds.
3. Consequently, the melting and boiling points of these compounds are usually higher than those of covalent compounds but lower than those of ionic compounds.
4. Solubility: Coordinate compounds are usually insoluble or less soluble in polar solvents like water but soluble in non-polar (organic) solvents.
5. Electrical conductivity: Coordinate compounds do not ionize in a fused state or solution and hence, these compounds do not conduct electricity.
6. Isomerism: Since coordinate bonds are rigid and possess directional properties, coordinate compounds exhibit the property of isomerism.
7. Type of relictions: Court in compiles undergoes molecular reactions which are much slower than those of ionic reactions.

Similarities and dissimilarities between covalent and coordinate bonds δ

Some Important Bond Parameters

Covalent bonds are characterized by certain parameters such as bond length, bond dissociation enthalpy or bond enthalpy, and bond angle.

Bond length

• Bond length is defined as the equilibrium distance between the centers of the nuclei of two bonded atoms in a covalent molecule.
• The bond lengths of different covalent bonds are determined by X-ray diffraction electron diffraction or spectroscopic methods. For a covalent bond, it is the sum of the covalent radii of the bonding atom.
• For example, if in a covalent molecule A — B, r_A, and r_B are the covalent radii of the atoms, A and B respectively, and the bond length is d, then d = r_A + r_B.
• It is generally expressed in terms of angstrom (lA = 10-10m) picometer (1 pm = 10‾12m).

Factors affecting bond length: Bond length depends on the following factors.

1. Size of the atoms: Bond length increases with an increase in the size of the atoms. For example, bond lengths of —X
2. Follow the order: H—I > H—Br > H—Cl > H—F.
3. This is because the order of covalent radii of halogen atoms follows the sequence: I > Br > Cl > F.
4. Bond multiplicity: Bond length decreases with an increase in multiplicity.

Bond lengths of different carbon-carbon bonds follow the order:

C=C (120 pm) < C=C(134 pm) < C—C(154 pm)

Types of hybridization (discussed later in article 4.8):

• Any s-orbital is closer to the nucleus than a p -p-orbital. So, electrons in the s -s-orbital are more tightly held by the nucleus than the electrons in the p-orbital.
• For this AATB [covalent molecule] [atoms or free radicals] A + B reason, with an increase in s -the character of the hybrid orbital the attraction on the electron increases and so, the length of the hybrid orbital decreases.
• As a consequence, the length of the bond obtained by overlapping the hybrid orbital with the s -s-orbital of another atom decreases.
• The s -characters of sp³, sp², and sp orbital are 25%, 33.33%, and 50% respectively.

Thus, the lengths of C—H bonds involving C -atoms with different hybridizations follow the order:

Electronic effects:

Bond length also depends on resonance, hyperconjugation, aromaticity, etc. For example, due to resonance, the carbon-carbon bond length in benzene is 1.39A while in the case of ethylene, the carbon-carbon bond length reduces to 1.34Å.

Bond dissociation enthalpy or bond enthalpy

• When a bond is formed between two atoms, some amount of energy is released. The same amount ofenergy is required to break the bond to get the atoms separated.
• This is called bond dissociation enthalpy which is a measure of bond strength and may be defined as the amount of energy required to break be gaseous state to produce neutral gaseous atoms or be gaseous state to produce neutral gaseous atoms or free radicals.

The bond dissociation enthalpy is usually expressed kJ.mol¯¹. It is to be remembered that the greater the bond dissociation enthalpy, the stronger the bond.

When a compound contains two or more same type of bonds, the average of their bond dissociation enthalpies is considered as bond enthalpy or bond energy.

Example: Bond dissociation enthalpies of four C — H bonds of a methane molecule (CH₄) are 435.7, 444.14, 444.14, and 339.4 kj.mol¯¹.

Therefore, the average value of bond dissociation enthalpies of four C-H bonds = (435.7 + 444.14 + 444.14 + 339.4)/4= 415.84 kj-mol-1.

This average value of bond dissociation enthalpy is the bond enthalpy or bond energy of the C — H bond of methane. In the case of diatomic molecules like H₂, Cl₂, O₂, N₂, HC1, etc., the bond dissociation enthalpy and bond enthalpy are the same. For example, the bond dissociation enthalpy and bond enthalpy of chlorine (Cl—Cl) molecule are the same (247 kj.mol¯¹ ).

Factors affecting bond dissociation enthalpy:

• Size of the bonded atoms: The larger the size of the bonded atoms, the greater the bond length and less the bond dissociation enthalpy. Thus, the bond enthalpy decreases on moving down a group in the periodic table. For example, the bond dissociation enthalpy of the H —Cl bond (431 kj.mol¯¹) is larger than the bond dissociation enthalpy of the H—Br bond (368 kj.mol¯¹ ).
• Bond multiplicity: The Greater the bond multiplicity, the greater the bond dissociation enthalpy of the bond between two atoms.
  • For example—C—C < C=C < C=C; N—N <N=N < N=N
• No. of lone pairs of electrons on the bonded atoms: As the number of lone pairs of electrons present on the bonded atoms increases, the electron-electron repulsion between the lone pairs of electrons on the two atoms increases. Thus, bond dissociation enthalpy decreases.

For example: Bond dissociation enthalpies of C—C (with no lone pair), O —O (with 2 lone pairs on each atom), and F—F bond (with 3 lone pairs on each atom) follow the order:

C—C (377 W-mol¯¹) >(213kj- mol¯¹ > F—F (159 kj-mol¯¹ ).

Types of hybridization: Bond enthalpy increases with an increase in s -character but decreases with an increase in p -character of the hybrid orbitals.

For example:  C(sp) —C(sp) (435.1 kj.mol¯¹) > C(sp²)—C(sp²) (384.6 kl-mol-1) > C(sp³)—C(sp³) (347.6 kj.mol¯¹).

Types of bond:

The bond enthalpy of a sigma (cr) bond is greater than that of an api (r) bond.

Bond angle

The bond angle is defined as the angle between two bonds around the central atom in a molecule. For example, the H — C —H bond angle in methane (CH₄) is 109°28′, the H — N —H bond angle in ammonia (NH₃) is 107.3° and the H — O —H bond angle in water (H₂O) is 104.5°.

Factors affecting bond angle:

1. Types of hybridization: The bond angle depends on the type of hybridization of the central atom in a molecule. For example, in the case of sp₃ -hybridization of carbon, the bond angle is 109°28′, for sp₂ -hybridization, it is 120°, and for sp -hybridization, it is 180°.
2. Number of lone pairs of electrons: As the number of lone pairs of electrons present on the central atom increases, the bond angle decreases.
3. Electronegativity of the central atom: As the Electronegativity of the central atom of a molecule of type ABx increases the bond angle increases.
4. Electronegativity of the atoms hooded to the central atom: As electronegativity of the atom bonded to the Shape of the molecule Hood angles central atom of a molecule of AB- type decreases, the bond angle Increases.

[The last three points are discussed later in VSEPR theory.]

Shapes Of Covalent Molecules And Valence Shell Electron Pair Repulsion (VSEPR) Theory

As already mentioned, Lewis’s concept is unable to explain the shapes of molecules. The first simple theory providing the simple procedure to predict the shapes of covalent molecules is known as the Valence Shell Electron Pair Repulsion (VSEPR) theory. The theory was proposed by Sidgwick and Powell in 1940 and was further developed by Nyholm and Gillespie in 1957. VSEPR theory may be expressed in terms of the following five rules:

Rule 1:

The shape of a molecule depends on the total number of valence shell electron pairs i.e., the total number of bond pairs and lone pairs of electrons or steric number (SN) around the central atom. All electron pairs repel each other. To minimize repulsions, the electron pairs tend to occupy a geometrical position such that the angular distance between them is maximum.

If the central atom of the n molecule does not possess a pair of electrons, the geometry of the molecule will be regular ami Is determined only by the bond pairs.

Number of bond pairs and shapes of molecules

Rule 2:

If the central atom Is surrounded by bond pairs as well as lone pairs of electrons, the repulsions among themselves are different. As a result, the molecule possesses an irregular or distorted geometry.

The repulsive interactions of various electron pairs decrease in the order:

Lone pair-lone pair (Ip-lp) > lone pair-bond pair (Ip-bp) > bond pair-bond pair {bp-bp).

When the angle between two electron pairs increases, the extent of repulsion decreases. These repulsions are found to be relatively greater If the electron pairs are to each other. If the angle between them Is 120° the repulsion becomes comparatively weaker and at 180° It Is the minimum.

Effect of lone pairs of electrons:

Nyholm and Gillespie (1957) pointed out that there is an important difference between the lone pair and the bond pair.

• While a lone pair is localized on the central atom (i.e., it is under the influence of only one atom), each bond pair is shared between two atoms.
• As a consequence, a lone pair occupies more space compared to a bond pair. This results in greater repulsive interactions between two lone pairs as compared to lone pair-bond pair and bond pair-bond pair repulsions.
• For example, In a CH₄ molecule, the C -atom contains four electron pairs in its valence shell and they are situated tetrahedrally.
• Thus, CH₄ is a regular tetrahedron with H —C —H bond angles of 109°28′. But, in the case of NH₃, the central N atom contains 3 bond pairs and 1 lone pair of elections, and as die Ip-bp repulsion is greater than that of bp-bp repulsion, the H —N —2 bond angle shrinks from 109°28´ lo 107.3°, In H₂, O-atom possesses 2 lone pairs along will) 2 bond pairs.
• Due to the strong repulsive forces of these two lone pairs acting on each bond pair, the — O —H bond angle significantly reduces to 104°2B´.

Rule 3:

• As the electronegativity of the hooded atoms for the central atom increases, the extent of repulsion between two bond pairs decreases and this is because the electron pairs are shifted away more from the central atom towards the bonded atoms.
• On the contrary, if the electronegativity of the central atom increases the electron pairs move towards the central atom giving rise to an increase in repulsion between the bond pairs.

Example:

The lone pairs and the bond pairs are tetrahedrally arranged in both NH₃ and NF₃ molecules. In the NF₃ molecule, the N —p bond pair is drawn more towards the more electronegative F-atom. But in the NH₃ molecule, the N — 11 bond pair is drawn more towards the more electronegative N -atom.

Therefore, bp-bp repulsion in NH₃ is more than that in NF₃.

Consequently, there is more distortion in NF₃ (F—N—F bond angle:

• 102°29′  when compared to the NH₃ molecule (H —N — H bond angle: 107.3°).
• For the same reason, the bond angle of H₂O (104.5°) is greater than that of F₂O (102°).

Rule 4:

The effect of electrons involved in the formation of a π-bond is not generally considered in determining the geometrical shape of a molecule.

• The electron cloud of a JI -bond is not tightly held by the nuclei of two atoms like that in a rr -bond.
• Therefore, a triple bond (one rr and two r -bonds) and a double bond (one a and one n -bond) occupy more space than a single bond.
• So, a multiple bond causes more repulsion than a single bond and the order is: multiple bond-multiple bonds> multiple bond single bond > single bond-single bond.

Example:

In the ethylene (CH₂=CH₂) molecule, the H—C—H bond angle reduces from 120° to 116° because of greater repulsion between the C —H and C=C bonds. For a similar reason, the Cl—C—Cl bond angle in phosgene (COCI₂) reduces to 112°.

Two exceptional cases:

PH₃(94°) < PF₃(98°):

• Fluorine is more electronegative than hydrogen. So, according to rule III, the order of bond angles is expected to be the reverse [NF₃(102.2°) <NH3 107.3°). This can be explained in terms of bond multiplicity.
• A coordinate covalent pn-dn bond is found to be formed between a filled 2p -orbital of F and an incomplete 3d -orbital of P.
• Due to resonance, each P —F bond in PF₃ assumes a partial double bond character.
• Consequently, the P —F bond order becomes greater than 1, i.e., its multiplicity increases.

With the increase in bond multiplicity, repulsion between bond pairs increases and consequently, the F—P—F bond angle becomes greater than the H —P —H bond angle.

In the PH3 molecule, on the other hand, similar coordinate covalent n-bond formation is not possible because hydrogen has no unshared electron pair.

F₂O(102°) < H₂O(104.5°) < C1₂O(111°):

The increasing order of electronegativity of the terminal atoms of these compounds is H < Cl < F.

Therefore, the order of bond angles should be:

F₂O< Cl₂O< H₂O. In this case, also, the correct order of bond angles can be explained in terms of bond multiplicity.

The coordinate covalent pn-dn bond is formed between the filled 2p -orbital of an oxygen atom and the vacant 3d -orbital chlorine atom and because of resonance, each O—Cl bond possesses a partial double-bond character. Due to an increase in bond multiplicity, the repulsive force operating between bond pairs increases and as a result, the value of the bond angle increases.

As there is no vacant orbital in the valence shells of hydrogen and fluorine, the formation of similar π-bonding is not possible. Therefore, it is the electronegativity of the terminal atoms which decides the bond angles of H₂O and F₂O.

Rule 5:

The unshared electron pair on a central atom having an incomplete valence shell (i.e., with vacant d-orbital) causes greater repulsion towards bond pairs or other lone pairs resulting in a significant contraction in bond angle compared to that on a central atom having a complete valence shell.

Example:

• There is no vacant orbital in the valence shells of the elements such as C, N, and O belonging to the second period of the periodic table.
• These elements can accommodate four electron pairs in their valence shells which are tetrahedrally arranged.
• In such cases, the repulsive interaction caused by the lone pair is less and this results in a small deviation in bond angle. For example, in an NH₃ molecule, the H —N — H bond angle is 107.3° and in H₂O, the H — O — H bond angle is 104.5°.
• On the other hand, the elements such as Si, P, and S belonging to the third period have vacant d -d-orbitals in their valence shells. If four electron pairs in their valence shells are tetrahedrally placed, then as a result of stronger repulsion by the lone pairs, considerable contraction in bond angle occurs.
• In fact, due to the availability of larger space, repulsion between bond pairs decreases and the stronger lone pair-bond pair repulsion compresses the bond angle almost to 90°. For example, in PH₃ and H₂S, the bond angles, instead of being 109°28´, are reduced to 94° and 92° respectively.

Determination of shapes of molecules and ions by valence shell electron pair repulsion (VSEPR) theory

A central atom having 2 electron pairs in its valence shell:

1. Beryllium chloride (BeCl_a) molecule:

In BeCl₂, the total number of electrons in the valence shell of the central Beatom = 2 valence electrons of Be-atom + 2 electrons of two Cl-atoms involved in cr -bond formation = 4 electrons = 2 electron pairs of = 2 cr -bond pairs.

These two bond pairs experience minimum repulsion when they remain at an angle of 180°. Hence, the shape of the BeCl₂ molecule is linear.

2. Carbon dioxide (CO₂) molecule:

The total number of electrons in the valence shell of the central C -atom of CO₂ molecule = 4 valence electrons of C -atom +4 electrons of two doubly-bonded O-atoms = 8 electrons = 4 electron pairs = 2cr -bond pairs + 2n -bond pairs, π-bond pairs have no role in determining the shape of a molecule. Therefore, the shape of the molecule is determined only by the two cr bond pairs.

Repulsion between these two bond pairs is minimal if they exist at an angle of 180°. Hence, the angular distance between two C=0 bonds is 180°, i.e., the shape ofthe CO₂ molecule is linear.

[The shape of the carbon disulfide (S=C=S) molecule is also similar to that of the carbon dioxide molecule.]

3. Hydrogen cyanide (HCN) molecule:

In the HCN molecule, the total number of electrons in the valence shell of the central C -atom = 4 valence electrons of C-atom +3 electrons of one triply-bonded N -atom +1 electron of one singly-bonded H atom= 8 electrons = 4 electron pairs -2a -bond pairs +2 7T -bond pairs, n -bond pairs play no role in determining the shape of a molecule.

The two bond pairs experience minimum repulsion when they remain at an angle of 180°. Hence, the shape of the molecule is linear.

4. Acetylene (HC=CH) molecule:

The number of electrons surrounding each carbon atom of acetylene molecule = 4 valence electrons of carbon atom +3 electrons of one triply-bonded C-atom +1 electron one singly-bonded H -atom=8 electrons =4 electron pairs =2 a -bond pairs +2n bond pairs.

To minimize the force of repulsion between the bond pairs, the shape of the acetylene molecule is linear.

. A central atom having 3 electron pairs in its valence shell: Boron trifluoride (BF₃) molecule:

The total number of electrons in the valence shell ofthe central B-atom of BF₃ electron of b- atom +3 electron of three sin gly bonded f atoms =6 elctron3 electron pair trigonal planar bf3 Molecule= 3 bond pairs.

The three bond pairs experience minimum repulsion if they remain at a 120° angle concerning each other.

1. Therefore, the geometrical shape of BF₃ is trigonal planar Nitrate (NOg) ion:

The total number of electrons surrounding the N -atom of NO₂ ion =5 valence electrons of N-atom + 2 electrons of one doubly-bonded O-atom +1 electron of one singly-bonded O -atom (no electron from the O-atom.

Attached by a coordinate bond because bond oxygon In or deb iron bond n-bond only, loud puli plays no rob In the dimming shape of Ion, bond pahs, To minimize repulsion among hinted towards Trigonal planar of a triangle, nil NO bond minimi are 120-, Honca, the shape of Ion In trigonal planar.

2. Sulfur (NO₂) molecules In HO₂ molecule:

Electrons surrounding Central H -atom valence electrons of 8-atom 2 electrons of one doubly-hooded O-aloin (no electron from the O-atom a coordinate covalent bond) electrons or 4 electron pair n 2 <r -bond 11 lone pair r 1 n -bond pair. The n-bond pair has no role In determining the shape of the molecule.

To minimize the extent of repetition, 2 electron palms are oriented toward the corners of an equilateral triangle. However, because of greater lone pair-bond pair repulsion compared to bond pair-bond pair the— S — 0 bond angle IN reduced from 120° to 119.5°. Thun, SO molecule is angular.

The n-bond pair has no role In determining the shape of the molecule. To minimize the extent of repetition, 2 electron palms are oriented toward the corneum of an equilateral triangle. However, because of greater lone pair-bond pair repulsion compared to lo bond pair-bond pair population, the — S — 0 bond angle IN reduced from 120° to 119.5°. Thun, SO₂ molecule IN angular

3. Ammonium (NH₃) Molecule:

Ion, the total number of electrons surrounding the central N-atom i.e., In Its valence shell 3 valence electrons of N-atom +3 electrons of three singly-bonded H-atoms (no electron from the Hatom attached by a coordinate covalent bond) =8 electrons or 4 electron pairs =4 rr -bond pairs.

To Ammonium to be torn In NH Ion, the total number of electrons surrounding the central N-atom i.e., In Its valence shell 3 valence electrons of N-atom +3 electrons of three singly-bonded H-atoms (no electron from the Hatom attached by a coordinate covalent bond) =8 electrons or 4 electron pairs =4 rr -bond pairs. To 107.3°, i.e., the tetrahedron is somewhat distorted. Excluding the lone pair, the shape of the molecule is trigonal pyramidal.

4. Water (H₂O) molecule:

In an H₂O molecule, the total number of electrons surrounding the central O-atom = 6 valence electrons of O-atom +2 electrons of two singly-bonded Hatoms =8 electrons or 4 electron pairs =2o- -bond pairs +2 lone pairs. To minimize the extent of mutual repulsion, these four electron pairs are oriented towards the four comers of a tetrahedron.

However, the tetrahedron is somewhat distorted due to the strong repulsive forces exerted by the lone pairs on each bond pair of electrons. The H —O —H bond angle is reduced to 104.5° from the normal tetrahedral angle of 109°28′. Excluding the lone pairs, the shape of the molecule is angular or V-shaped.

5. Hydrogen sulfide (H₂S) molecule:

In an H₂S molecule, the total number of electrons surrounding the central S-atom =6 valence electrons of S-atom +2 electrons of two singly bonded H-atoms =8 electrons or 4 electron pairs =2crbond pairs +2 lone pairs.

These 4 electron pairs experience minimum repulsion if they occupy the four comers of a tetrahedron. Since the repulsion between two lone pairs is greater than that between two bond pairs, the tetrahedron is distorted and the H — S — H bond angle is decreased to 92° from the ideal tetrahedral angle (109°28′).

Due to the presence of vacant d -d-orbital in S-atom, the bond angle, in this case, reduces significantly. Therefore, excluding the lone pairs, the shape of the molecule is angular or V-shaped.

A central atom having 5 electron pairs in its valence shell:

1. Phosphorus pentachloride (PCI₅) molecule:

In the PCl₅ molecule, the total number of electrons surrounding the central P-atom =5 valence electrons of P-atom +5 electrons of five Cl-atoms =10 electrons or 5 electron pairs =5<r -bond pairs. These electron pairs experience minimum mutual repulsion if they orient themselves towards the five vertices of a trigonal.

In this geometry, all five P—Cl bonds are not equivalent. The three bonds lying In the trigonal plane are called equatorial bonds. The remaining two bonds, one lying above and the other below the trigonal plane and both making an angle of 90° with the plane, are called axial bonds. The axial bonds arc slightly longer than the equatorial bonds.

It is to be noted that the structure of the PC15 molecule is unsymmetrical. As a result, it is less stable and therefore, is more reactive.

2. An axial bond is longer than an equatorial bond:

It can be explained in terms of repulsive forces between electron pairs due to different bond angles. Let us consider the repulsive interactions experienced by an axial and an equatorial bond pair. An axial bond pair is repelled by three equatorial bond pairs at 90° and one axial bond pair at 180°.

On the other hand, an equatorial bond pair is repelled by two axial bond pairs at 90° and two equatorial bond pairs at 120°. It is known that the repulsion between two electron pairs decreases with an increase in the angle between them and hence, the repulsion between electron pairs at 120° and 180° may be neglected in comparison to hose at 90°.

Thus, considering only the repulsive interactions between electron pairs situated at 90° to each other, we find that each axial bond pair is repelled by three electron pairs while each equatorial bond pair is repelled by two electron pairs.

Therefore, an axial bond pair experiences greater repulsion than an equatorial bond pair and as a consequence, an axial bond becomes slightly longer than an equatorial bond.

3. Sulfur tetrafluoride (SF₄) molecule:

In the SF₄ molecule, the total number of electrons surrounding the central Satom = 6 valence electrons of S-atom +4 electrons of four F-atoms =10 electrons or 5 electron pairs =4 a -bond pairs +1 lone pair. Thus, to minimize the extent of repulsion, the five electron pairs around sulfur. Orient themselves in a trigonal bipyramidal geometry.

The lone pair preferably occupies the equatorial position to stabilize the structure. For such orientation of the lone pair, the trigonal bipyramidal structure is distorted (the bond angles are 89° and 177° instead of 90° and 180° respectively). Thus, the shape of the molecule is described as a distorted tetrahedron or a see-saw.

4. Chlorine trifluoride (CIF₃) molecule:

In the ClF₃ molecule, the total number of electrons surrounding the central Clatom =7 valence electrons of Cl-atom +3 electrons of three singly bonded F-atom = 10 electrons or 5 electron pairs =3 σ -bond pairs +2 lone pairs.

Thus, to minimize mutual repulsion, the five electron pairs orient themselves in a trigonal bipyramidal geometry in which two equatorial positions are occupied by two lone pairs. This is because, in such orientation, the structure acquires maximum stability. But, due to the presence of two lone pairs in the equatorial position, the trigonal bipyramidal structure is distorted (Ffl —Cl —Fe bond angle becomes 87°29′) and the molecule is T-shaped.

5. ICI₂– ion:

In the ICl₂ ion, the total number of electrons surrounding the central I-atom = 7 valence electrons of atom +2 electrons of two cr -bonded Cl-atoms +1 electron for the negative charge = 10 electrons = 5 = 2 electron pairs bond pairs and 3 lone pairs.

These five electron pairs arrange themselves in a trigonal bipyramidal geometry with three equatorial positions occupied by the three lone pairs because such an arrangement ensures maximum stability.

Since the three lone pairs are present at the comers of an equilateral triangle, there is no distortion of the Cl —I — Cl bond angle of 180°. Hence, the ion has a linear shape. Similar examples are XeF2.

A central atom having 6 electron pairs in its valence shell:

1. Sulfur hexafluoride (SF₆) molecule:

In the SF₆ molecule, the number of electrons surrounding the central S -atom = 6 valence electrons of the S -atom + 6 electrons of six σ bonded F-atoms = 12 electrons or 6 electron pairs = 6 cr-bond pairs.

To have the minimum force of repulsion, the six electron pairs are oriented towards the corners of a regular octahedron. Hence, the shape of the SF₆ molecule is octahedral with a bond angle of 90°.

2. Bromine pentafluoride (BrF₅) molecule:

In BrF₅, the total number of electrons surrounding the central Br-atom = 7 valence electrons of Br-atom +5 electrons of five crbondedF-atoms = 12 electrons or 6 electron pairs =5 bond pairs +1 lone pair. To minimize the extent of mutual repulsion, these six electron pairs arrange themselves octahedrally in which any one of the positions (all positions are equivalent) is by the lone pair.

Due to the presence of a lone pair of electrons, the Br-atom slightly deviates from the equatorial plane. So, the BrFg molecule is square pyramidal. A similar example is IF5.

3. Xenon tetrafluoride (XeF₄) molecule:

In the XeF₄ molecule, the total number of electrons surrounding the central Xe-atom = 8 valence electrons of Xeatom + 4 electrons of four cr-bonded F-atoms =12 electrons or 6 electron pairs = 4 cr-bond pairs + 2 lone pairs. To minimize the extent of mutual repulsion, these electron pairs arrange themselves octahedrally in which two opposite axial positions are occupied by the two lone pairs. Therefore, the shape of the XeF4 molecule is square planar.

A central atom having 7 electron pairs In Its valence shell Iodine hepts (1F7) molecule/in IF7 molecule, the total number of electrons surrounding the central Iatom = 7 valence electrons ofI-atom +7 electrons of seven σ -bonded F-atoms = 14 electrons or 7 electron pairs – 7 σ-bond pairs.

To have the minimum force of repulsion, the seven electron pairs are oriented toward the corners of a pentagonal bipyramid. Hence, the shape of the IF7 molecule is pentagonal bipyramidal with bond angles of 72°, 90°, and 180°.

Shapes of different types of molecules or ions according to VSEPB theory

Modern Bond Concept(Vbt) Of Covalency Valence Bond Theory

The valence bond theory was given by W. Heitler and F. London in 1927 and was later improved and developed by L. Pauling and J. C. Slaterin in 1931. It is based on atomic orbitals, electronic configurations of elements, overlap criteria of atomic orbitals, and stabilities of molecules.

Basic characteristics of valence bond theory

1. A covalent bond is formed by overlapping the atomic orbitals of the two combining atoms having unpaired electrons of opposite spin. Opposite spins of the two electrons are mutually neutralized during the formation of the covalent bond.
2. The extent of overlapping of the two half-filled atomic orbitals determines the strength of a covalent bond. The greater the overlapping of atomic orbitals, the stronger the covalent bond formed.
3. The atomic orbitals having only unpaired electrons are involved in overlapping.
4. Multiple bonds are for median atoms possessing more than one atomic orbital containing unpaired electrons.
5. Atoms do not lose their identity in the molecule formed by the combination.
6. During bond formation, only the valence electrons of each bonded atom lose their identity. The other electrons remain unaffected.
7. The formation of a bond is accompanied by the release of some energy. The larger the amount of energy released, the stronger the bond.
8. The orientations of the atomic orbitals involved in the overlapping determine the orientation of the covalent bond formed.

Explanation of the formation of H₂ molecule with the help of valence bond theory

Letus consider the formation of hydrogen molecule which is the simplest of all molecules. Consider two hydrogen atoms, A and B approaching each other having nuclei NA and Nfi respectively and their electrons are represented by eA and eB. When the two atoms are far apart from each other, there is no attractive or repulsive interaction between them, and the potential energy of the system (isolated atoms)is assumed to be zero. When the two atoms come closer to each other, new attractive and repulsive forces start operating. These are:

1. The force of attraction between the nucleus and its electron i.e., N_A– e_A and N_B– e_B
2. The force of attraction between the nucleus of B and electrons of A (N_A-e_B) and the nucleus of A and electrons of B (N_A-e_B), the
3. The force of repulsion between nuclei ofthe two atoms (N_B– e_A ) and
4. The Force repulsion between electrons of the two atoms (e_A-e_B)- The diagrammatic representation of these forces is given in

We know that attractive forces tend to bring the atoms closer while repulsive forces tend to push them apart.

It has been observed experimentally that the magnitude of new attractive forces is greater than the new repulsive forces. As a consequence, the potential energy of the system decreases gradually as the two atoms come closer and closer.

Finally, a stage is reached where the total force of attraction is just balanced by the total force of repulsion. In this situation, the two hydrogen atoms are said to be bonded together to form a stable molecule, and the distance (r0) between the two nuclei is called bond length which is equal to 74 pm.

If the two atoms are brought still closer, the repulsive forces predominate. As a consequence, the potential energy of the system increases and the system becomes unstable. Hence, the two hydrogen atoms cannot be brought closer than 74 pm. The change in potential energy takes place in the formation of a hydrogen molecule.

Since a certain amount of energy is released when a bond is formed between the two H -atoms, the hydrogen molecule is more stable than the isolated hydrogen atoms.

⇒ \(\mathrm{H}(\mathrm{g})+\mathrm{H}(\mathrm{g}) \longrightarrow \mathrm{H}_2(\mathrm{~g})+435.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

The energy so released is known as bond enthalpy or bond energy. The larger the amount of energy released, the stronger the bond formed and vice-versa. Conversely, 435.8 kg of energy is required to break the bond, i.e., to separate the atoms in one mole of H₂ molecules.

⇒ \(\mathrm{H}_2(\mathrm{~g})+435.8 \mathrm{~kJ} \longrightarrow \mathrm{H}(\mathrm{g})+\mathrm{H}(\mathrm{g})\)

It is to be noted that the decrease in energy of the system during the formation of a chemical bond determines the strength of the bond formed and vice-versa.

Non-existence of helium molecule:

• When two helium atoms (HeA and HeB ) approach each other, four new forces of attraction and five new forces of repulsion come into play.
• The old and new attractive as well as repulsive forces, Since the overall repulsive forces are more than the attractive forces, the energy of the system increases.
• Hence, the formation of a chemical bond between two atoms is not possible.

Atomic Orbitals

• The three-dimensional region in space around the nucleus of an atom where the probability of finding an electron is maximum is called an atomic orbital.
• The size and shape of any orbital depend on the energy of the electron present in that orbital, i.e., on the principal energy level and subshell in which the electron resides.
• According to the energy content of electrons, the orbitals are expressed as s, p, d, and f.
• With the increase in principal quantum number, the size of an orbital of the same type (i.e., s, p, d, or) increases.

S-orbital:

• It is the spherical three-dimensional region in space around the nucleus having a fixed radius where the probability of finding the electron is maximum. Electron density on the surface ofthe sphere is maximum.
• Although different s -orbitals (Is, 2s, 3s, etc.) are expressed as spheres of different radii, the density of the electron cloud is not the same throughout the sphere (as 1=0 and m – 0);
• For Example:  For 2s -orbital, electron density increases upto some distance from the nucleus then decreases and again increases at the surface of the sphere.
• The intermediate space where the electron density is minimum is called a spherical node.
• 2s and 3s-orbitals contain one and two nodes respectively but the ls-orbital does not contain any node.

P-Orbital:

In the case of the p -subshell,l = 1 and m =-1,0 and +1.

• So, it consists of three orbitals, designated as px, py, and pz. From the solutions of Schrodinger’s wave equation, it is known that p -p-orbitals have three possible orientations along the x-axis, y-axis, and z-axis, mutually perpendicular to each other.
• Each orbital has two lobes, separated by a plane where the probability of finding the electron is zero.
• This plane is called the nodal plane and the point at which the two lobes meet indicates the position of the nucleus of the atom and is called the node of the orbital. Each orbital is thus dumbbell-shaped.
• The electron density is maximum on the surface of the dumbbell. Being situated along the three axes, they have definite directions. Spatial orientations of px, py, and pz.

D-orbital: In the case of d -subshell,l = 2 and m – -2, -1, 0, +1 and +2. So there are five d -orbitals with equivalent energies.

Overlapping of atomic orbitals (AO): Formation of the covalent bonds

During the formation of a covalent bond, the combining atoms approach each other till they acquire minimum energy. At this stage, the half-filled atomic orbitals are so close to each other that they undergo partial interpenetration or partial merging. Such partial merging of atomic orbitals is called orbital overlap.

• This overlapping results in the pairing of electrons (with opposite spins) and causes a lowering of the energy of the system. The probability of finding electrons in the region of overlap is higher than in other places.
• The overlapping of half-filled atomic orbitals (containing electrons with opposite spins) leads to the formation of a covalent bond. The extent of overlap determines the strength of a covalent bond. The greater the extent of overlapping, the stronger the covalent bond.
• It is to be noted that the overlap between the atomic orbitals may be positive or negative depending on the nature of the overlapping of atomic orbitals. A positive overlap that leads to constructive interactions involves the overlap of the lobes of the same signs.
• A negative overlap which leads to destructive interactions, involves the overlap ofthe lobes of opposite signs. As the atomic orbitals (filled, partly filled, or vacant) have certain orientations, the covalent bonds also have specific orientations in three-dimensional space.
• When the electrons present in these orbitals form covalent bonds by pairing, then the covalent bonds formed have three-dimensional orientations. Hence due to the directional property of orbitals, covalent bonds too exhibit directional properties.
• Coordinate bonds also exhibit directional properties because coordinate bonds form due to the overlapping of filled orbitals of the electron donor atom with the empty orbital of the electron acceptor atom. The shape and orientation of the combining orbitals determine the shape of the compound formed.

Depending on the nature of overlapping, covalent bonds may be classified into two types:

1. Sigma (σ) and
2. pi (π) bond.

Sigma (r) bond: When a bond is formed between two atoms by the end-to-end (head-on) overlap of their atomic orbitals along the internuclear axis, i.e., the line joining the centers of the nuclei of the two atoms, it is called a sigma (σ) bond and the electrons constituting it are called sigma electrons.

Axial overlapping between two pure atomic orbitals:

s-s overlapping:

• This involves an overlap between the orbitals of the two approaching atoms. It is to be noted that s -orbitals are spherically symmetrical and can overlap to the same extent in all directions, along any axis.
• Since s orbitals are non-directional, the bond formed by the overlapping of s -s-orbitals is also non-directional.

For example, in the formation of an H₂ molecule, the Is -orbital of one H-atom overlaps with the Is -orbital of another H-atom, forming a cr -bond.

s-p overlapping:

• This involves overlapping of s -the orbital of one atom with p -the orbital of another atom along their axes.
• The bond thus formed is called an (s-p) σ – bond. Since p -orbitals possess directional characteristics, the σ -bond formed by the overlap of s and p -orbitals, will also possess directional characteristics, which is similar to that of die p -orbital.

For example, in the formation of an HF molecule, the Is -orbital of one H-atom overlaps with the 2pzorbital of one F-atom thus forming a <x -bond. Thus, the z-axis is taken as the internuclear axis.

p-p overlapping:

• This involves the overlapping of p -p-orbitals of two approaching atoms along the internuclear axis.
• The bond thus formed is called a (p-p) σ -bond.

For example, the molecule of fluorine is formed by the overlapping of two 2pz -orbitals of two F -atoms.

Formation of σ-bond due to axial overlapping between a pure and a hybrid orbital (Sp³-s overlapping):

σ bond can be formed by the overlapping of pure s -orbital and a hybrid Sp³ -orbital 4 along their axes.

For example: the C—H <T -bond in CH₄ is formed by the overlap between the Sp³ -orbital of the C-atom and the pure Is -orbital of the H-atom.

Formation of σ -bond due to axial overlapping between two hybrid orbitals (Sp³-Sp³ overlapping):

σ-bond can also be formed by the overlapping of two hybrid orbitals along their axes.

For example, the C—C σ -bond in ethane molecule is formed by the overlapping of two Sp³ -hybrid orbitals of two C-atoms.

Pi(π)bond:

• When a bond is formed by the lateral or sidewise overlap of two p -p-orbitals, it is called a pi (π) bond.
• The overlapping of orbitals occurs in such a way that their axes are parallel to each other and perpendicular to the internuclear axis.
• The resulting orbital is called a π-orbital and the electrons constituting it are called π-electrons.
• For example, in a C₂H₄ molecule, a sideways overlap of two parallel 2px -orbitals of two C-atoms occurs to form a n bond.
• A π-bond consists of two saucer-type charge clouds above and below the plane of the participating atoms.

• It is to be noted that a π-bond can be formed only in addition to a sigma bond, i.e., it is always present in the molecules having multiple bonds (double or triple).
• The spherically symmetrical s -s-atomic orbitals do not take part in sideways overlap to form a π-bond.

Strength of sigma and pi bond

• The strength of a covalent bond depends on the extent of overlapping between die-combining orbitals. The greater the overlapping, the greater the energy released and the stronger the bond.
• Axial overlapping of atomic orbitals leads to the formation of a σ-bond and sideways overlapping of orbitals leads to the formation of a σ-bond. Since the extent of axial overlapping is greater as compared to sideways overlapping, a π-bond is stronger than an n-bond.
• The π-bonds obtained by s-s, s-p, or p-p overlapping are not equal strengths.
• Two s -s-orbitals cannot overlap effectively with their spherical charge.
• On the other hand, the charge cloud of the σ-orbitals is concentrated along either the x, y, or z-axis.
• Their projected lobes can overlap effectively with s -s-orbitals and more effectively with other p -orbitals.
• Consequently, a p-p σ -bond is stronger than an s-p σ-bond which in turn is stronger than an s-s bond, i.e.,
• The order of decreasing bond strength is p-p > s-p > s-s.

Hybridization

According to the valence bond theory, a covalent bond is formed by the mutual overlap of two pure atomic orbitals. The formation of molecules like H₂, Cl₂, HCl, etc., can be easily, explained with the help of valence bond theory but, it fails to explain the formation of molecules like H₂O, CH₄, NH₃, etc.

For example: 

The H—O—H bond angle in a water molecule should be 90° as it involves the overlapping of pure 2p orbitals of O-atom but, the actual bond angle is 104.5°. Similar observations are found to occur in the case of NH₃ molecules.

• To explain these experimental observations, a new concept called hybridization was introduced by L. Pauling.
• According to this concept, the atomic orbitals possessing slightly different energies mix up to form a new set of orbitals each having an equivalent amount of energy.
• These orbitals are called hybrid orbitals and the mixing is called hybridisation.

Hybridization Definition:

Hybridization may be defined as the intermixing of atomic orbitals of the same atom having the same or slightly different energies to redistribute their energies and form new orbitals of equal energies and identical shapes.

Salient features of hybridization:

1. The number of hybrid orbitals formed is equal to the number of atomic orbitals taking part in hybridization.
2. The energy and shapes of the hybrid orbitals are always equivalent.
3. Hybrid orbitals are more effective in forming stable bonds as compared to pure atomic orbitals. This is because they can undergo more effective overlapping.
4. The hybrid orbitals are oriented space in some preferred directions to have a stable arrangement with minimum repulsions among themselves. So, types of hybridization govern the geometrical shapes ofthe molecule.

Important conditions for hybridization:

1. Only the orbitals present in the valence shell of the atom are involved in hybridization.
2. The orbitals participating in hybridization should have comparable energies.
3. Promotion of electrons is not an essential condition for hybridization.
4. Not all the half-filled orbitals need to participate in hybridization.
5. Only the half-filled orbitals don’t need to participate in hybridization. Filled orbitals of the valence shell can also participate in hybridization.
6. Hybridization never takes place in isolated atoms. It occurs only when the atom takes part in bond formation.

Explanation of the tetravalency of carbon atoms:

• The electronic configuration of the carbon atom in its ground state is \(1 s^2 2 s^2 2 p_x^1 2 p_y^1 2 p_z^0\), i.e., one odd electron is present in each of 2px and 2py orbitals of carbon atom. Number of odd electrons present in the valence shell of odd electrons present in the valence shell of odd electrons present in the valence shell of an atom generally gives the measure of the covalency of that atom.
• So, the valency of carbon should be two. However, the valency of carbon in almost all organic compounds is 4, except for a few extremely unstable compounds, where the valency of C is 2, like methylene (: CH₂), dichloromethylene (: CCl₂), etc.
• During a chemical reaction, the 2 electrons in 2s-orbital become unpaired by absorbing energy, and one electron is promoted to 2pz -orbital. This is an excited state of a carbon atom and the electronic configuration of the carbon atom in this state is,\(1 s^2 2 s^1 2 p_x^1 2 p_y^1 2 p_z^1\). Thus in the excited state, 4 odd electrons are present in the outermost shell of a carbon atom. The presence of these four unpaired electrons accounts for the tetravalency of carbon atoms.

Sp³-hybridisation

When one s and three p -orbitals ofthe valence shell of an atom merge to form four new equivalent orbitals having the same energy and shape, it results in tetrahedral or Sp³-hybridization. The resulting orbitals are called Sp³ – hybrid orbitals.

Sp³-hybridisation Process

The four Sp³-hybrid orbitals each containing one electron, are directed towards the four comers of a regular tetrahedron making an angle of 109°28′ with one another and the atom lies generally given at the center of the tetrahedron.

Orbitals are oriented in such positions in space that minimum repulsions occur between them. The formation of Sp³-hybrid orbitals by the combination of s, px, py, and pz-atomic orbitals in

Sp³ -hybridization can be illustrated by the following examples:

Formation of methane (CH₄) molecule:

• During the formation of a methane molecule, one 2s -orbital and three 2porbitals of excited carbon atom undergo hybridization to form four equivalent Sp³ -hybrid orbitals.
• The hybrid orbitals are directed toward the four comers of a regular tetrahedron with the C-atom at the center of the tetrahedron.
• Each hybrid orbital containing an unpaired electron overlaps with the Is -orbital of a hydrogen atom resulting in the formation of a total of four C —H bonds.
• Thus, the methane molecule has a highly stable tetrahedral geometry with each H —C —H bond angle equal to 109°28′.

Note that if the four atoms linked covalently to a carbon atom are not the same the geometry of the molecule wooly still be tetrahedral but it may not be regular in shape, Example methyl chloride (CH₃Cl), chloroform (CHCI₃), etc.

In these discs, the bond angles differ slightly from the normal value of 109°28′. Formation of ethane (CH₆) molecule:

During the formation of the ethane molecule, each of the two C -atoms in their excited states undergo Sp³ -hybridization. Two hybrid orbitals, one from each C -atom, overlap axially to form a C — C σ -bond.

The remaining three hybrid orbitals of each carbon atom overlap with the half-filled Is -orbital of hydrogen atoms forming a total of six C — H σ-bonds, Each C —H bond in ethane is (Sp³-s) σ-bond having a bond length of 109 pm. The C —C bond is an (Sp³-Sp³) σ -bond having a bond length of 154 pm.

sp2-hybridisation

When one s -orbital and two p -orbitals of the valence shell of an atom merge and redistribute their energies to form three equivalent new orbitals of equal energy and identical shape, the type of hybridization involved is called sp² -hybridization. The new orbitals thus formed are called sp² -hybrid orbitals.

All three hybrid orbitals each containing one electron lie in one plane making an angle of 120° are directed towards three corners of an equilateral triangle with the carbon atom in the center of the triangle. The unhybridized 2pz -orbital (containing one electron) remains perpendicular to the plane of the triangle with its two lobes above and below that plane.

Therefore, a molecule in which the central atom is sp² -sp²-hybridized has a triangular planar shape and the hybridization is called planar trigonal hybridization. Formation of sp² -hybrid orbitals by the combination of s, px, and py -atomic orbitals.

sp² -hybridization can be illustrated by the following examples:

Formation of ethylene C₂H₄ molecule: in the formation of the ethylene molecule, each ofthe two carbon atoms undergo sp² -hybridization, leaving the 2pz -orbital unhybridised. The three sp²-hybrid orbitals of each carbon atom are planar and oriented at an angle of 120° to each other.

• The unhybridized 2pz -orbital is perpendicular to the plane of sp² -hybrid orbitals. One sp² -hybrid orbital of one carbon atom overlaps axially with one sp² -hybrid orbital of the other carbon atom to form a C — C cr bond.
• The remaining two sp” -hybrid orbitals of each carbon atom overlap with the half-filled Is -orbitals of two hydrogen atoms resulting in the formation ofa total of four C —H σ -bonds. The unhybridized 2pz -orbital of one carbon atom overlaps with that of the other carbon atom in a sideways fashion to form an n-bond between the two carbon atoms.
• The n-bond consists of two equal electron clouds distributed above and below the plane of carbon and hydrogen atoms. All the six atoms in the molecule lie in one plane. Thus, ethylene is a planar molecule.
• In this molecule, the C=C bond (one sp²-sp² σ -bond and one 7σ-bond) length is 134 pm, the C —H bond (sp²-s bond) length is 108 pm and each C—C—H or H —C —H bond angle is nearly equal to 120°.

Formation of boron trifluoride (BF₃) molecule:

The excited state electronic configuration of boron is \(1 s^2 2 s^1 2 p_x^1 2 p_y^1\). One 2s -and two 2p -orbitals of boron undergo hybridization to form three equivalent sp² -hybrid orbitals (each containing one electron). The sp² -hybrid orbitals are directed towards the comers of an equilateral triangle and lie in one plane making an angle of 120° with one another.

Each of the sp² -hybrid orbitals overlaps axially with the half-filled 2p -orbital of fluorine to form three B —F σ- bonds. Due to the sp²-hybridization of boron, the boron trifluoride molecule has a trigonal planar shape. The formation of boron trifluoride (BF₃) molecule.

sp- hybridization

When one s and one p -orbital of the valence shell of an atom merge and redistribute their energies to form two equivalent hybrid orbitals of equal energy and identical shape, tin; hybridization Involved is called ip-hybridization or diagonal hybridization, The orbitals lints obtained are called sp-hybrid orbitals.

The other two p -orbitals remain unhybridized. These sp hybrid orbitals are Inclined to each other at an angle of 180°. Therefore, a molecule in which the central atom is sphybridised is linear in shape and the hybridisation is also known as diagonal hybridisation. The formation of sp-hybrid orbitals by the combination of 2s and 2px -atomic orbitals.

Sp-hybridization can be illustrated by the following examples:

Formation molecules of bothacety lengthy C-atoms(HCsHC)are sp molecule:

Hybridized in Thereacetylene two Orbital overlap in BeF₂ unhybridized orbitals (2py & 2pz) on each C-atom.

• Two sp-hybrid orbitals are linear and are directed at an angle of 180°, Urihybridised p -orbitals are perpendicular to the sp-hybrid orbitals and also perpendicular to each other. One sp -sp-hybridized orbital of one C-atom overlaps axially with a similar orbital of the other C-atom to form a C σ -bond.
• The remaining hybrid orbital of each C-atom overlaps with the half-filled s -s-orbital of the H-atom to form a total of two C—H σ -bonds. Thus, the acetylene molecule is linear, (fn hybridized py -orbitals of two carbons and the unhybridized pz -orbitals of two carbons overlap sideways separately to form two different n -bonds.
• Electron clouds of one n-bond lie above and below the internuclear axis representing the σ-bond whereas the electron cloud of the other π -bond liein the front and back of the internuclear axis. These two sets of n -n-electron clouds merge into one another to form a cylindrical electron cloud around the internuclear axis surrounding the C —C σ -bond.
• In the acetylene molecule, the C-C bond length is equal to 120 pm, the C—H bond length is equal to 108 pm and the C —C —H bond angle is equal to 180°.

Formation of beryllium fluoride (BeF2) molecule:

• The excited state electronic configuration of Be is \(1 s^2 2 s^1 2 p_x^1\). The 2s -orbital and the 2px -orbital containing one electron each undergoes hybridization to form two sp -hybridized orbitals.
• The two sp-hybrid orbitals are linear and oriented in opposite directions at an angle of 180°
• . These two hybrid orbitals overlap axially with the half-filled 2p orbitals of the two fluorine atoms to form two Be—F σ bonds.
• Thus, the BeF2 molecule is linear The formation of beryllium fluoride molecule.

Hybridizations involving s, p, and d-orbitals

Sp³ d-hybridization:

This type of hybridization involves the mixing of one s, three p, and one d -orbitals to form five equivalent Sp³d-hybrid orbitals. These hybrid orbitals are directed towards the five corners of a trigonal bipyramid and hence such hybridization is called trigonal bipyramidal hybridization.

Example: In PC15 molecule, P-atom using one 3s, three 3p, and one 3d -orbitals form five Sp³ d -hybrid orbitals which are directed towards the five comers of a trigonal bipyramid.

The five Sp³d -hybrid orbitals overlap axially with five half-filled 3p -orbital of five Clatoms to form five P bonds. Thus, PCl₅ has a trigonal 120° bipyramidal shape. Each axial (a) P —Cl bond is involved in repulsive Interaction with the equatorial (e) P—Cl bond pair, whereas each equatorial P—Cl bond is Involved in repulsive interaction with two equatorial P—Cl bond pairs.

As a result of this, the length of the P—F axial bond somewhat increases compared to that of the P —F equatorial bond (P—F axial = 158 pm, P—F equatorial = 153 pm).

Sp³d² -hybridization:

In this type of hybridization, one s, three p, and two d -orbitals (dx²-y² and dz²) intermix to form six equivalent Sp³d² -hybrid orbitals. The molecules, in which these orbitals of the central atom are involved have octahedral geometry.

Example:

In the SF₆ molecule, she orbitals (one s, three p, and two d -orbitals) of S-atom hybridize to form six new Sp³d² – hybrid orbitals which are projected towards the six comers of a regular octahedron.

These six equivalent hybrid orbitals overlap with the 2porbitals of six fluorine atoms to form six S—F σ-bonds. SF₇ molecule has a regular octahedral geometry. Since each S—P bond Is involved in similar repulsive interaction with five orthogonal S— F bond pairs, all the six S —F bonds are equal In length.

5p³d³-hybridization:

This type of hybridization Involves the intermixing of one s, three p, and three d -( dxy,dyz, da.) orbitals to form seven Sp³d² -hybrid orbitals that adopt pentagonal bipyramidal geometry.

Example: In IF₇ one 5s, three 5p, and three 5d -orbitals of Iatom hybridize to give seven Sp³d² -hybrid orbitals.

These seven hybrid orbitals overlap axially with the half-filled 2p -orbitals of the atom forming seven I— F or bonds one to Sp³d² -hybridization of the central molecule lists pentagonal bipyramidal geometry.

Five I—F bonds are directed towards 1 the vertices of a regular pentagon making an angle of 72° with each other. The remaining two I—F bonds are directed at an angle of 90- above and below the plane of the pentagon.

• Orbitals of an atom with the same energy level take part in hybridization.
• Before hybridization, the transfer of electrons is not mandatory.
• No. of hybrid orbitals is equal to no. of participating orbitals.
• Not only half-filled orbitals but also filled orbitals can take part in hybridization.
• Hybridization does not occur in discrete atoms.
• Hybridization occurs due to the formation of bonds.
• An idea about the geometrical shape of a molecule can be obtained from the nature of hybridization.
• Hybrid orbitals form more stable bonds than pure atomic orbitals.
• d -orbitals involved in various types of hybridization
• ⇒ \(\begin{aligned}
  & s p^3 d: d_{z^2} ; s p^3 d^2: d_{x^2-y^2}, d_{z^2} ; s p^3 d^3: d_{x y}, d_{y z}, d_{z x} \\
  & d s p^2: d_{x^2-y^2}
  \end{aligned}\)

Determination of hybridization of the central atom

From the following general formula, the number of orbitals of the central atom of a molecule or ion undergoing hybridization (H) can be determined and hence the state of hybridization of that atom can be known.

From the knowledge of the type of hybridization of the central atom, the shape ofthe molecule orion may be ascertained.

⇒ \(H=\frac{1}{2}[V+X-C+A]\)

Where,

H = Number of orbitals involved in hybridization

V = Number of electrons in the valence shell of the central atom,

X = Number of monovalent atoms surrounding the central atom,

C = Charge on the cation, and A = Charge on the anion.

State of hybridization of the central atom of different molecules or Ions

Resonance

Sometimes, all the properties of some molecules or ions, cannot be predicted from a single electronic (Lewis) structure. In such cases, the molecule or ion is represented by two or more Lewis structures which differ in the arrangement of valence electrons keeping the basic structure involving the sigma skeleton the same.

Resonance Definition:

The Various Lewis Structure, That Differ In the Positions Of Non-bonding or Electrons But not in the relative position of the atoms are known as resonance structures or canonical forms, and the concept is called resonance. Resonance structures are imaginary and are considered to explain the physical and chemical properties of the molecules or ions.

The actual molecule or ion is a resonance hybrid of all the canonical forms that are involved in resonance. Resonance is also known as mesomerism. Various resonance structures are connected by double-headed arrows \((\longleftrightarrow)\).

Rules for writing meaningful resonance structures

The rules for writing meaningful resonance structures are:

1. The various resonance structures should differ only in the position of electrons. The basic structure involving bonds between atoms should remain undisturbed.
2. The number of unpaired electrons, if any, in the resonance structures must be the same.
3. Only those atoms should be involved in the resonance which are coplanar (or nearly coplanar).
4. The resonance structures should have nearly the same energy.
5. Each structure must be a proper Lewis structure.

Examples: Carbonate ion \(\left(\mathrm{CO}_3^{2-}\right)\) is represented as a resonance hybrid of the following three resonance structures:1,2 and 3.

1. In all three structures, there are two carbon-oxygen single bonds (1.43A) and one carbon-oxygen double bond (1.20Å). Experimental results have however revealed that all the carbon-oxygen bonds are of equal length (1.28Å) in carbonate ions.
2. Thus, all the three carbon-oxygen bonds are equivalent. This shows that the actual structure of carbonate ion is intermediate between the structures 1, 2, and 3. So carbonate ion is often represented by the non-Lewis structure 4, which is also considered as the hybrid structure.

Nitrate ion (NO-3) is represented as a resonance hybrid ofthe following three equivalent resonating structures:

If we consider any one of the nitrogen-oxygen bonds, then we see that in one structure it is a double bond while in the other two structures is a single bond. So, every N-O bond is equivalent i.e., every N-O bond length is the same.

Resonance energy

Resonance energy Definition:

Resonance energy is defined as the difference in internal energy between the actual or observed value and that of the resonance structure with the lowest internal energy i.e., the structure with the highest stability.

Unit:

Resonance energy is expressed in terms of kcal-mol-1 or kj. mol¯1. With the increase in the value of resonance energy internal energy of the compound decreases and hence stability ofthe compound increases.

Calculation of resonance energy:

Resonance energy is not a measurable quantity. It can only be obtained from thermochemical data.

• Cyclohexene on hydrogenation forms cyclohexane. In this reaction, 28.6 kcal-mol^-1 heat is generated. Thus, on hydrogenation of a double bond, 28.6 kcal-mol^-1 of heat is obtained.
• Accordingly, on hydrogenation of three double bonds of 1,3,5-cyclohexatriene (benzene), 3 × 28.6= 85.8 kcal-mol^-1 heat should be obtained.
• But, in reality, the heat of hydrogenation of benzene is 49.8 kcal-mol^-1
• Thus, the internal energy of benzene is (85.8-49.8) = 36 kcal-mol^-1 less than the predicted value. This 36 kcal-mol^-1 is the resonance energy of benzene.
• The lowering of internal energy by 36 kcal-mol^-1 is responsible for the extra stability of benzene.

Examples of resonance in some molecules or ions:

Electronegativity

Electronegativity Definition:

The electronegativity of an element is the tendency or ability of its atom to attract the bonding or shared pair of electrons towards itself in a covalent bond. It is represented by x.

Example:

In the HCl molecule, Cl is more electronegative than H. As a result, the Cl -atom draws the bonding electron pair towards itself. Thus Cl – atom acquires a partial negative charge and the H-atom acquires a partial positive charge 5+ <5- \([\stackrel{\delta}{\mathrm{H}}-\stackrel{\delta-}{\mathrm{C}}]\). So, the partial ionic character is developed in the covalent molecule HCl.

The Pauling scale of electronegativity

According to Pauling, the bond energy of any compound, A—B is greater than the geometrical jerage of the bond energies of its constituent molecules, A₂ B₂. i.e., \(E_{\mathrm{A}-\mathrm{B}}>\left(E_{\mathrm{A}-\mathrm{A}} \cdot E_{\mathrm{B}-\mathrm{B}}\right)^{1 / 2}\) terms indicate bond energy

The difference between \(E_{\mathrm{A}-\mathrm{B}}>\left(E_{\mathrm{A}-\mathrm{A}} \cdot E_{\mathrm{B}-\mathrm{B}}\right)^{1 / 2}\) is called the Ionic resonance energy of the A—B bond and it is expressed by AA _ \(E_{\mathrm{A}-\mathrm{B}}>\left(E_{\mathrm{A}-\mathrm{A}} \cdot E_{\mathrm{B}-\mathrm{B}}\right)^{1 / 2}\)

Here, EA_B bond energy includes both the ionic and covalent bond energies but, the source of both the bond energies, Δ A_A and EB_B is only the covalent bond. So, the value of AAB determines the extent of the ionic character of the A—B bond. In the opinion of Pauling, the value of AA B is related to the electronegativities of the two elements.

If XA and XB are the electronegativities of A and B respectively, then

⇒ \(\chi_{\mathrm{A}}-\chi_{\mathrm{B}} \propto\left(\Delta_{\mathrm{A}-\mathrm{B}}\right)^{1 / 2}\) or, \(\chi_{\mathrm{A}}-\chi_{\mathrm{B}}=K\left(\Delta_{\mathrm{A}-\mathrm{B}}\right)^{1 / 2}\) [where K constant]

If the die value of AA_B is expressed in the unit of k.J mol-1, then K = 0.208.

Thus, the equation reduces to

⇒ \(\chi_A-\chi_B=0.208 \times\left(\Delta_{A-B}\right)^{1 / 2}\)

As the values of EA_B, EA_ A, and EB_B can be determined by experiments, so for any compound, A—B, the value of AA_B can easily be estimated. Once the value of ΔA_B is determined, then from the knowledge of the electronegativity of either element A or B, the electronegativity of the other element can easily be calculated.

Based on this theory and calculations, Pauling determined the electronegativities of different elements (in normal oxidation state). The values of electronegativities of elements which are determined for electronegativity of 2 (2.1) is called the Pauling scale of electronegativity.

Mulliken’s scale of electronegativity

Mulliken regarded electronegativity as the average value of ionization potential (IE) and electron affinity (EA) of an atom of any element.

Therefore, Electronegativity (xA) = \(\frac{I E+E A}{2}\) [IE and EA are expressed in the union eV-atom-1]

Factors affecting the electronegativity of elements

Atomic size or radius:

If the atomic size or volume is small, the distance of the outermost shell from the nucleus becomes less. Consequently, the positive charge of the nucleus attracts the electron(s) to a greater extent. Hence, the smaller the die atomic size, the higher be electronegativity.

Example:

The electronegativity of elements belonging to group-1A of the periodic table gradually decreases from H to Cs.

Number of shells in an atom:

As the number of shells between the nucleus and the outermost shell increases, because of the shielding effect of the electrons In the inner shells, the electronegativity of elements decreases.

Example: Electronegativity of elements belonging to group A gradually decreases from F to I.

Atomic Electro negativity number or nuclear charge:

If the number of shells m remains the same, Then with an increase in atomic number (i.e., nuclear positive charge), electronegativity increases.

• Oxidation state: For an element that exhibits several oxidation states, electronegativity at a higher oxidation state is more than that in its lower oxidation state
• Example: Electronegativities of Fe, Fe²⁺, and Fe³⁺ are 1.8, 1.84, and 1.95 respectively. The electronegativity of sulfur in SFf) is higher than that in SCI₂.
• Electron affinity: Generally, the higher the electron affinity of any element, the greater its electronegativity.
• Example: Halogens of the group- 7A are strong electronegative elements because of their very high electron affinity values. Alkali metals of group IA have low electron affinity. So, their electronegativity values are very small.
• Hybridization: Hybridisation of the atom of an element has a significant effect on the electronegativity of that atom. The greater the s -s-character of the hybrid orbital, the higher its electronegativity.

Example:

Nature of the substituents attached to the atom:

The electronegativity of an atom increases with an increase in the electronegativity ofthe atom bonded to it.
Example: C-atomic CF₃I is more electronegative than CH₃I.

Bond Polarity And Dipole Moment

Non-polar covalent bond :

• If two similar atoms (atoms having the same electronegativities) form a bond by sharing a pair of electrons, the shared electron parties just in the middle of the two nuclei.
• In other words, the electron cloud constituting the covalent bond is symmetrically distributed between the two bonded atoms.
• As a result, no poles are developed and in such a case, the covalent bond becomes almost pure. This type of covalent bond is called a non-polar covalent bond.
• The corresponding molecule is known as a non-polar covalent molecule.

Some examples of non-polar molecules are H₂(H —H), C₁₂(C1 —Cl), O₂(0=0), N₂(N = N) etc.

Polar covalent bond:

• When a covalent bond is formed between two atoms having different electronegativities, the shared pair of electrons gets displaced more towards the more electronegative atom. As a result, the more electronegative atom acquires a partial negative charge and the less electronegative atom acquires a partial positive charge.
• This type of bond is called polar covalent bond (i.e., covalent bond with partial ionic character), and the molecules having such bonds are called polar covalent molecules or simply polar molecules. HF, for example, is a polar molecule.
• The extent of the percentage of ionic character in a covalent bond depends on the difference in electronegativities of the two bonding atoms. The greater the difference in electronegativities, greater is the greater the percentage of ionic character in the bond (i.e., greater is the polarity of the bond).

For example, the order of polarity of bonds in halogen hydrides is:

H —F(60%) > H —Cl(19%) > H —Br(ll%) > H —1(4%). This is because the electronegativities of halogens decrease in the order F > Cl > Br >I.

If the difference of electronegativities between two combining atoms is 1.7, then the bond is said to possess 50% ionic character.

An approximate relation between the difference in electronegativities of the bonded atom and the resulting ionic character is given in the following table.

Dipole moment

The product of the magnitude of positive or negative charge (q) and the distance (d) between the centers of positive and negative charges in a polar molecule is called dipole moment.

Dipole moment is usually represented by the Greek letter, ‘μ’ (mu). It can be expressed mathematically as μ = q x d.

Unit: The CGS unit of dipole moment is D (Debye). ID = 10-18 esu. cm. The SI unit of dipole moment is Coulomb-metre (C-m) and 1 C.m = 2.9962 × 10²⁹ D.

General characteristics: Some important characteristics of dipole moment are mentioned below—

1. The value of μ for a non-polar molecule is zero. For example, in the case of H₂, O₂, and N₂ molecules, μ= 0 D . For polar molecules,μ has a definite value, Example for HF molecule, μ = 1.91 D. Polarities of molecules increase with an increase in the value of
2. Since dipole moment is a vector quantity, it has magnitude as well as direction. Its direction is usually indicated by a crossed arrow \((\longmapsto)\) pointing from the positive to negative pole in a polar bond or polar molecule as a whole.
3. Molecules formed by two different elements (A —B) are always polar i.e., they possess dipole moment. Due to the differences in electronegativities of the two atoms, the bond between them is polar. For example HC1 is a polar molecule \(\mathrm{H}  \mathrm{Cl}\) (μ= 1.03D).

Polyatomic molecules may be non-polar although they possess polar bonds:

The dipole moment of a polyatomic lu molecule depends <not only on the polarity of the bonds present in it but also on the spatial arrangement of various bonds on the geometrical shape of the molecule because in such cases, the dipole moment of a molecule is considered to be the vector sum of the dipole moments of various bonds

Examples:

1. If a triatomic molecule of the type, AB2 is linear (B—A—B) shaped, two equal but opposite \(A  B\)(B Is more electronegative than A) bond moments cancel out each other, and hence, resultant dipole moment becomes zero. For example, C02, CS2, BeCl2, etc., have no net dipole moment.
2. On the other hand, if the molecule is angular, two A \(A  B\)(B is more electronegative than A or A \(A  B\) (A is more electronegative than B) bond moments will not cancel out each other.
3. So, there will be a resultant dipole moment \(A  B\) and the molecule will be polar. For example, H₂O, H₂S, etc. possess a net dipole moment In fact, symmetrical molecules have no resultant dipole moment although their bonds are polar.

Applications of dipole moment

• To distinguish between polar and non-polar molecules: Molecules having specific dipole moment are polar while those having zero dipole moment are polar. Thus, BeCl₂ (y = 0D) is a non-polar molecule while H₂O (y = 1.85D) is a polar one.
• Comparison of relative polarities of molecules: The relative polarities of molecules can be compared from their dipole moment values because the greater the magnitude of the dipole moment, the higher the polarity of the molecule. Thus, H₂O (1.85D) is more polar than H₂S (1.10D).
• Determination of shapes of molecules: The dipole values help predict the general shapes of molecules containing three or more atoms. If a molecule is symmetrical, its dipole moment is found to be zero.

Example:

Molecules like CO₂, CCl₄, BF₃, etc., have no net dipole moment. Therefore, these molecules have a perfectly symmetrical shape (CO₂ is linear, CCl₄ is tetrahedral and BF3 is triangular planar).

Percentage of ionic character of a bond:

From the knowledge of dipole moment and bond length, it is possible to determine the percentage of ionic character of a covalent bond.

Example:

The extent of the ionic character of the H—Cl bond in a hydrogen chloride molecule can be estimated from the value of its dipole moment (1.03D) and bond length (1.275Å). For 100% ionic character, the charge developed on H and Cl-atoms would be equal to the charge of an electron (4.8 × 10^-10 esu). Therefore, the dipole moment of HCl when its bond is 100% ionic would be,

⇒ \(\mu_{\text {ionic }}=q \times d=4.8 \times 10^{-10} \mathrm{esu} \times 1.275 \times 10^{-8} \mathrm{~cm}\)

= \(6.12 \times 10^{-18} \mathrm{esu} \cdot \mathrm{cm}=6.12 \mathrm{D}\)

∴ Percentageionic character

= \(\frac{\mu_{\text {observed }}}{\mu_{\text {ionic }}} \times 100\)

The distinction between stereoisomers (cis-and trans-isomers):

In a disubstituted ethene, the os-isomer usually has a higher dipole moment than the fracas-isomer. This is because, in the cis-isomer, the bond moments are not canceled out while in the trans-isomer, the bond moments are either totally or partially canceled out.

Example:

The cis-isomer of 1,2-dichloroethene has a definite dipole moment whereas the dipole moment of the transisomeris found to be zero.

The distinction between structural isomers (ortho, meta, and poro-isomers):

• The dipole moment of para-disubstituted benzene (with identical and symmetrical substituents) is zero and that of the ortho-isomer is greater than that of the mete-isomer (for electron-withdrawing substituents).
• From this knowledge, it is possible to distinguish between any two of these isomers or to identify all three isomers.

Prediction of relative boiling points of compounds:

If the value of the dipole moment is much higher, then the compound is highly polar indicating that, the dipole-dipole attractive forces among the molecules are sufficiently strong. Hence, a large amount of energy is required to separate the molecules from each other, i.e., the boiling point of such a compound is much higher.

Dipole moments (µ) of some molecules

Group moment:

The resultant moment of a group obtained as the vector sum of all the individual bond dipoles or bond moments present in it is called the group moment of that group. Benzene, for example, has no net dipole moment because oppositely oriented C —H bond moments cancel out each other.

When a -NO₂ (nitro) group is substituted in place of any H-atom, the value of the dipole moment of nitrobenzene is found to be 3.95D. Therefore, it may be concluded that the group moment of the -NO₂ group when attached to a benzene ring is 3.95D.

Prediction of dipole moments of some molecules from their geometrical shapes

Carbon dioxide (CO₂) molecule:

In a CO₂ molecule, the central carbon atom is sp -hybridized so CO₂ is a linear molecule. For this type of structure, the two oppositely oriented C=O bond moments cancel out each other.

Because of this, CO₂ is a non-polar molecule, i.e., it has no net dipole moment. Viewing alternatively, the center of positive charge and the center of negative charge coincide in this linear molecule; i.e., d = 0 and so, p =qxd = qxO =OD.

Beryllium hydride (8eH₂) molecule:

The structure of the BeH₂ molecule is similar to that of CO₂ and hence, the linear BeH₂ molecule is non-polar (µ = 0).

Carbon tetrachloride (CCI₄) molecule:

c-atom in CCl₄ molecule is sp³ -hybridized. So, it has a symmetrical tetrahedral structure. In such a geometrical shape, the resultant of the three C—Cl bond moments cancels the fourth C—Cl bond moment or the resultant of two C —Cl bond moments cancels the resultant of the other two C —Cl bond moments. Because of this, the dipole moment of the CCl₄ molecule becomes zero, i.e., it is non-polar. Alternatively, the center of the positive charge and the center of the negative charge coincide in this tetrahedral molecule, i.e., d = 0 and = qxd = qxO = OD.

Boron trifluoride (BF₃) molecule:

Since the B-atom in the BF₃ molecule is sp² – hybridized, its geometrical shape is trigonal planar and the bond angles are 120°. Because of such a geometrical shape, the resultant of any two B —F bond moments cancels the third B —F bond moment. Hence, BF3 has no net dipole moment, i.e., it is non-polar.

Alternatively, the center of positive charge and center of negative charge coincide in this trigonal planar molecule, i.e., d = 0 and = qxd = qx0 = 0D.

Benzene (C₆H₆) molecule:

Due to the sp²-hybridization of the atoms involved, the benzene molecule is planar and all bond angles are 120°. Hence the shape of the molecule is hexagonal planar. In such a geometrical shape, each C —H bond moment is canceled out by another C —H bond moment lying opposite to it.

Hence, the molecule has no net dipole moment, i.e., it is non-polar. Alternatively, the centers of positive and negative charges coincide, i.e., d = 0 and µ= qxd=qx0 = 0D.

Chloroform (CHCl₃) molecule:

Chloroform is a tetrahedral molecule because the central carbon atom is sp3 -sp3-hybridized. Because of such a geometrical shape, the resultant of the three C —Cl bond moments and the moment of the C —H bond (although very small) act in the same direction. Hence, the molecule possesses a net dipole moment, i.e., the molecule is polar.

Water (H₂O) molecule:

The O-atom in H₂O is sp³- hybridized. So, in this molecule, the two unshared electron pairs and the two 0 —H bonds are arranged tetrahedrally, and the actual shape of the molecule is angular. The h—O—H bond angle is 104.5°.

Because of such a geometrical shape, the resultant of two O —H bond moments and the resultant moment of the two unshared electron pairs act in the same direction. Thus, a water molecule possesses considerable dipole moment, i.e. it is a polar molecule.

Sulfur dioxide (SO₃) molecule:

• The central sulfur atom in the SO₂ molecule is sp² -hybridized. So, the two S = O bonds and the lone electron pair, are oriented towards the comers of an equilateral triangle, i.e., they lie mutually at an angle of about 120° and the molecule is angular or V-shaped.
• Because of such a shape, the resultant moment of the two polar S=O bonds is partly canceled by the moment contributed by an unshared electron pair acting in opposite directions. Hence, the SO₂ molecule has a net dipole moment.

Nitrous oxide (N₂O) molecule:

• Although N₂O is a linear molecule, it possesses polarity because its structure is unsymmetrical (O -atom exists at one end). The (N → O) bond moment and the moment due to the unshared electron pair act in the opposite directions and the latter moment partly neutralizes the former one.
• Thus, the molecule has a net dipole moment of very small value.

Nitrogen trifluoride (NF₃) molecule:

• The central nitrogen atom of the NF₃ molecule is sp³ – hybridized. Thus, the three N—F bonds and one lone pair are arranged tetrahedrally, and excluding the lone pair, the shape of the molecule is pyramidal.
• Because of such a geometrical shape and higher electronegativity of fluorine, the resultant of the three N—F bond moments and moments contributed by the lone pair of electrons act in opposite directions.
• Thus, the molecule possesses a small but net dipole moment, Le„ the molecule is slightly polar.

Ammonia (NH₃ ) molecule:

The geometrical shape of the ammonia molecule is similar to that of the NF₃  molecule. However, because of the higher electronegativity of nitrogen compared to that of hydrogen, the resultant of the three N—H bond moments and the moment contributed by the lone pair act in the same direction. Consequently, it possesses a net dipole moment much higher than NF₃.

Carbon monoxide (CO) molecule:

⇒\(\text { In }: C=\ddot{O}:\)

The molecule, a <xmoment (due to electron withdrawal through cr-bond) acts from carbon to the more electronegative oxygen atom. Also, a very weak n -moment operates from carbon to oxygen atom. Although the octet of O-atom is completed in this molecule, the carbon atom has a sextet.

So, oxygen donates an unshared pair of electrons to carbon and helps it to complete its octet by forming a dative σ- bond with it. As a result, a much stronger -moment acts from oxygen to carbon atom and since this moment is almost canceled by σ-moment and the weak r -moment acting in the opposite direction, the molecule possesses a very small net dipole moment.

Isomeric 1,2-dichloroethenes:

• Cis-i,2-dichloroethene is more polar than its trans-isomer. In the cis-isomer, the two C —Cl bond moments make an angle of 60° with each other.
• Consequently, a resultant moment acts in the molecule, i.e., the molecule possesses a net dipole moment.
• Conversely, in the trans-isomer, the two C —Cl bond moments acting in opposite directions neutralize each other.

Because of this, the molecule possesses no net dipole moment, i.e., the isomeris non-polar.

Isomeric 1,2- dichlorobenzenes:

Dipole moments of isomeric dichlorobenzenes follow the order:

• Ortho-or 1,2- dichlorobenzene > meta -or 1,3-dichiorobenzene > paraor 1, 4-dichlorobenzene.
• In para-isomer, the two C—Cl bond moments acting in opposite directions neutralize each other So, the isomer possesses no net dipole moment, i.e., the isomer is non-polarIn ortho-and meta isomers, the C- Cl bond moments act at an angle of 60° and 120° respectively.
• Since the resultant moment increases with a decrease in bond angle, the dipole moment of the ortho-isomer is higher than that of the meta-isomer, i.e., ortho-isomerism is more polar than meta-isomer.

The dipole moment of disubstituted benzenes (C₆H₄XY) can be determined by the equation:

⇒ \(\mu=\sqrt{\mu_1^2+\mu_2^2+2 \mu_1 \mu_2 \cos \theta}\)

Where px and p₂ are group moments of the group, X and Y respectively and 9 is the angle between two group moments. In ortho, meta, and para-isomers, the values of 0 are 60°, 120°, and 180 ° respectively.

If any of the two groups is an electron with-drawing one, its group moment is to be represented with a negative sign. For example, the dipole moment of me to-nitrotoluene can be calculated as follows:

⇒ \(\mu_{\text {toluene }}=0.4 \mathrm{D}\)

⇒ \(\mu_{\text {nitrobenzene }}=0.35 \mathrm{D}\)

∴ \(\mu_{\mathrm{CH}_3}=0.4 \mathrm{D}\)

∴ \(\mu_{\mathrm{NO}_2}=-3.95 \mathrm{D}\)

∴ \(\mu_{m-\mathrm{CH}_3 \mathrm{C}_6 \mathrm{H}_4 \mathrm{NO}_2}=\mu\)

⇒\(\sqrt{\mu_{\mathrm{CH}_3}^2+\mu_{\mathrm{NO}_2}^2+2 \mu_{\mathrm{CH}_3} \mu_{\mathrm{NO}_2} \cos 120^{\circ}}\)

= \(\sqrt{(0.4)^2+(-3.95)^2+2 \times 0.4 \times(-3.95) \times-\frac{1}{2}}=4.16 \mathrm{D}\)

Numerical Examples

Example 1. Calculate the percentage of ionic character of HF. Given that the dipole moment of HF is 1.91 D and its bond length is 0.92A.
Answer:

If HF is 100% ionic, each atom would carry a charge equal to one unit, i.e., 4.8 x 10-10 esu. As the bond length of HF is 0.92A, its dipole moment for 100% ionic character would be

⇒ \(\mu_{\text {ionic }}=q \times d=4.8 \times 10^{-10} \mathrm{esu} \times 0.92 \times 10^{-8} \mathrm{~cm}\)

= 4.416 x 10-18 esu.cm = 4.416D

[l0-18 esu cm = ID]

∴ % ionic character \(=\frac{\mu_{\text {observed }}}{\mu_{\text {ionic }}} \times 100=\frac{1.91 \times 100}{4.416}=43.25\)

Example 2. The percentage of ionic character of LiH is 76.81% and the bond length is 1.596A. What is the value of the dipole moment of a molecule? [ID = 3.335 X 10-30Cm]
Answer:

If the molecule is 100% ionic, then

⇒ \(\mu_{\text {ionic }}=q \times d=1.602 \times 10^{-19} \mathrm{C} \times 1.596 \times 10^{-10} \mathrm{~m}\)

= \(2.557 \times 10^{-29} \mathrm{C} \cdot \mathrm{m}\)

⇒ \(\% \text { ionic character }=\frac{\mu_{\text {observed }}}{\mu_{\text {ionic }}} \times 100\)

∴ \(\mu_{\text {observed }}=\frac{\% \text { ionic character } \times \mu_{\text {ionic }}}{100}\)

⇒ \(\frac{76.81 \times 2.557 \times 10^{-29}}{100}=1.96 \times 10^{-29} \mathrm{C} \cdot \mathrm{m}\)

⇒ \(\frac{1.96 \times 10^{-29}}{3.335 \times 10^{-30}} \mathrm{D}=5.87 \mathrm{D}\)

Question 3. Calculate the fractional charge on each atom of HBr. The dipole moment of HBr is 0.78D and its bond length is 1.41A. [Electronic charge, e = 4.8 X 10_1° esu, ID = 10_18esu.cm].
Answer:

Dipole moment, μ = q× d or,

Or, \(=\frac{\mu}{d}=\frac{0.78 \times 10^{-18} \mathrm{esu} \cdot \mathrm{cm}}{1.41 \times 10^{-8} \mathrm{~cm}}\)

=\(0.55 \times 10^{-10}\)

⇒ \(\text { Fraction of charge }(\delta)=\frac{\text { Charge present }(q)}{\text { Electronic charge }(e)}\)

= \(\frac{0.55 \times 10^{-10} \text { esu }}{4.8 \times 10^{-10} \text { esu }}\)

= 0.11

∴ The fractional charge on hydrogen \(\delta_{\mathrm{H}^{+}}=0.11\) and the fractional charge on bromine,δ_Br = -0.11

Example 4. The dipole moment of NaCl is 8.5 D. Interatomic distance between Na+ and Cl- is 2.36 A. Calculate the percentage ionic character of the NaCl molecule.
Answer:

If NaCl is 100% ionic then,

⇒ \(\mu_{\mathrm{Na}^{+} \mathrm{Cl}^{-}}=e \times d=4.8 \times 10^{-10} \mathrm{esu} \times 2.36\)

= \(4.8 \times 10^{-10} \mathrm{esu} \times 2.36 \times 10^{-8} \mathrm{~cm}=11.328 \mathrm{D}\)

∴ % ionic Charchter \(=\frac{\mu_{\text {observed }}}{\mu_{\text {ionic }}} \times 100=\frac{8.5 \times 100}{11.328}=75.035\)

Hydrogen Bond

When a hydrogen atom Is covalently bonded to a small and largely electronegative atom, X (F, 0 or N), the electro¬ negative atom pulls the shared pair of electrons towards itself, resulting in the development ofa partial negative charge on the electronegative atom and a partial positive charge on the hydrogen atom.

This partially positively charged hydrogen atom behaves like a bare proton as Its nucleus gets exposed due to the displacement of the solitary electron.

Such a hydrogen atom of one HX molecule attracts a partially negatively charged Xatom of an adjacent HX molecule or a molecule of HY (where Y is an atom of another highly electronegative element) and the molecules are held together by a strong electrostatic force of attraction. The electrostatic force of attraction that exists between the hydrogen atom and the highly electronegative atom is known as the hydrogen bond.

Hydrogen Bond Definition:

The electrostatic force of attraction existing between the H-atom covalently bonded to an electronegative atom (F, O, or N) in a molecule and the electronegative atom of another molecule (of a similar or different type) is known as a hydrogen bond, Generally, a hydrogen bond Is represented as X —H Y, in which the X—H bond is covalent and H-Y bond is the hydrogen bond. Here, Y is the hydrogen bond acceptor and X —H is the hydrogen bond donor. The hydrogen bond acceptor may or may not contain H-atom.

Conditions for hydrogen bond formation

1. Hydrogen atoms should be bonded to a highly electronegative atom such as F, O, or N. The common examples are HP, H₂O, and NH₃.
2. The size of the electronegative atom should be small. This Is because the smaller the size of the electronegative atom, the higher its charge density and hence greater the electrostatic attraction. Thus NIL, molecules are involved in hydrogen bonding but HC1 molecules are not, although both N and Cl have the same electronegativity.
3. An electronegative atom that participates in H-bonding must have at least one unshared pair of electrons.

Only the firm of elements such as F, O, and N can form effective hydrogen bonds. Although the electronegativities of chlorine and nitrogen are the same (3,0), chlorine cannot form a stable hydrogen bond but nitrogen can do so.

The reason Is that the atomic size of the chlorine atom (atomic radius of Cl-atom – 0.99A) is greater than that of the nitrogen atom (atomic radius of N-atom = 0.70A ).

Exception:

Intramolecular H-bond is found in compounds like chloral hydrate and art to-chlorophenol and rmolecular H-bond is found in para-chlorophenol.

Characteristics of hydrogen bonding

• H-bond is much weaker than a covalent bond but stronger than van der Waals forces of attraction. The strength of the H-bond is only 12.6-41.8 kj.mol-1 while that of a covalent bond is ofthe order of 400 kj. mol-1 .
• The strength of the hydrogen bond depends on the electronegativity of the atom to which the H-atom is attached. The larger the electronegativity of the atom, the greater the strength of the H-bond. For example, electronegativities of F, O, and N decrease in the order: of F(4.0) >0(3.5)>N(3.0).

Consequently, the strength of the H-bond decreases in the order

F—H—F (41.8kj.mo^-1 )> O—H—O (29.3 kj-mol^-1 )>N—H-N (12.6kJ.mol^-1 )

• The position of the H-bond depends on the orientation of the lone pair of the electronegative atom, i.e., like a covalent bond, the hydrogen bond also has a definite direction of bonding.
• Because of the smaller size of hydrogen atoms, the valence shell electrons of the two atoms linked by hydrogen bonds experience considerable repulsion. To minimize the repulsion, all three atoms lie in a straight line [X—H- -X (X = F, O, N) ].
• The magnitude of H-bonding depends on the physical state of the compound. It is maximum in the solid state and minimum in the gaseous state. Thus, hydrogen bonding strongly influences the structure and properties of compounds.
• A large number of molecules remain associated with hydrogen bonding.

Types of hydrogen bonding 

Intermolecular hydrogen Bonding:

When 11-bonding occurs between different molecules of the same or different compounds, it Is called Intermolecular hydrogen bonding.

Examples Hydrogen fluoride( HF):

In an HF molecule, (lie highly electronegative fluorine atom acquires a partial negative charge and the hydrogen atom acquires a partial positive charge. The negatively charged P -atom of one IIP molecule attracts the positively charged hydrogen atom of the other molecule to form a strong hydrogen bond between them.

In this way, a large number of molecules gel associated through hydrogen bonding and exist in a giant molecule which can be represented as (HF)_n. In the solid (crystalline) slate, hydrogen fluoride consists of long zig-zag chains of molecules associated together through hydrogen bonds. In liquid and gaseous state, the length of the chain somewhat shortens.

Water (H₂O):

• A water molecule contains one highly electronegative oxygen atom bonded to two H -atoms. Thus, the oxygen atom becomes partially negatively charged and the two hydrogen atoms become partially positively charged.
• The negative end of one molecule attracts the positive end of the other to form a hydrogen bond. In this way, a large number of water molecules become associated through hydrogen bonding to form a macromolecule which can be represented as (H₂O)n.

Ammonia (NH₃):

In an ammonia molecule, a moderately electronegative nitrogen atom is bonded to three hydrogen atoms and as a consequence, the nitrogen atom acquires a partial negative charge and each hydrogen mam acquires a partial positive charge. For this reason, the molecules of ammonia remain associated through hydrogen bonding to form a polymeric species which may be represented as (NH₃)n.

Acetic acid (CH₃COOH):

In the gaseous state, two acetic acid molecules get associated through hydrogen bonding to form an eight-membered cyclic dimer. However, in a liquid state, a large number of molecules remain associated through hydrogen bonding to form a polymeric species [(CH₃COOH)n].

Intramolecular hydrogen bonding:

When hydrogen bonding takes place within the same molecule, it is called intramolecular hydrogen bonding. This type of hydrogen bonding is also known as chelation as it results in the formation of a ring. It is found to occur frequently in organic compounds and is favored when a six or five-membered ring is formed.

Examples:

It is normally found in disubstituted benzene compounds in which the substituents are attached to adjacent carbon atoms, i.e., they are ortho to each other. For example, ortho-nitrophenol, ortho hydroxybenzaldehyde, salicylaldehyde, and ort/zo-chlorophenol possess intramolecular hydrogen bonding.

It is found in some aliphatic compounds in which the groups involved in hydrogen bonding are situated at adjacent carbon atoms. For example, chloral hydrate [CCl₃CH(OH)₂] and ethylene glycol (HOCH₂CH₂OH) possess intramolecular hydrogen bonding.

Effect of H-bond on different properties of compounds

1. Intermolecular hydrogen bonding causing the association of molecules of any compound results in an increase in the effective molecular mass of that compound. Consequently, the properties such as physical state, melting and boiling points, latent heat of vaporization, viscosity, concentration, surface tension, etc., which depend on molecular mass, are found to increase.
2. Again, as the number of molecules decreases due to hydrogen bonding, the values of colligative properties of compounds such as osmotic pressure, elevation of boiling point, depression of freezing point, etc. decrease.
3. Also, solubilities and acidities of compounds are often markedly influenced by hydrogen bonding. It is to be noted that the effect of intramolecular hydrogen bonding on the physical properties of compounds is negligible.

Question 1. Although oxygen and sulfur belong to the same group (VIA) of the periodic table, at ordinary temperature, the hydride of oxygen (HzO) is a liquid whereas the hydride of sulfur (H₂S) is a gas.
Answer:

Since the electronegativity of oxygen is higher than that of sulfur, the O —H bond in water is more polar than the S —H bond in H₂S. Also, the size of the O-atom is smaller than that of the sulfur atom. So, the molecules of water are associated through intermolecular hydrogen bonding. Hence, water exists in the liquid state but hydrogen sulfide exists in the gaseous state at ordinary temperature.

Question 2. Among the hydrides of halogens (P, C, Br, and I) belonging to group-VIIA of the periodic table, only HF is a liquid at 19.5°C, while each of IICl, IiBr, and HI exists as a gas at this temperature.
Answer:

Because of the highest electronegativity and smaller atomic size of fluorine, HF molecules remain extensively associated through strong JutermoleeuJar hydrogen bonding. So, a large amount of energy Is required to separate the molecules. Hence, HF exists in the liquid state at 19,5°0, On the other hand, Cl, fir, and 1 atom are larger and less electronegative compared to F-atom, So, the formation of intermolecular hydrogen bonding leading to the association of the molecules Is not possible in case of HCI, HUr, and HI, Thus, each of them exists In the gaseous state at this temperature.

Question 3. KHF₂ exists But KHCI₂ has no existence
Answer:

F-atom In HF can form strong hydrogen bonds due to higher electronegativity and smaller size of F-atom. However, Cl does not form a hydrogen bond due to its comparatively low electronegativity and larger atomic size. In an aqueous solution, HF ionizes partially giving only one H+ ion per molecule.

Consequently, HF behaves as a weak monobasic acid. \(\mathrm{H} \rightleftharpoons \mathrm{H}^{+}+\mathrm{F}^{-}\) The fluoride Ion thus produced forms hydrogen bond with undissociated HF molecule to form fluoride anion HF₂-.

⇒ \(\mathrm{F}^{-}+\mathrm{HF} \rightleftharpoons\left(\mathrm{F} \cdots-\mathrm{H}-\mathrm{F}^{-} \text {or } \mathrm{HF}_2^{-}\right.\)

This explains the existence of the salt KHF₂. As Cl-atom cannot form an H-bond, so anion like HCl₂ is not formed. Hence the salt, KHCl₂has no existence.

Question 4. Although the molecular mass of PH₃(34) (the hydride of the second element of group 15) is twice that of NH₃ (17) (the hydride of the first element of the same group), the boiling point of PIf₃ (—87.4°C) is much less than that of NH₃ (-33.4°C).
Answer:

The electronegativity of nitrogen (3.0) is higher than that of phosphorus (2.2) and the atomic size of N is smaller than that of P. So, the N— H bonds in NH₃ are relatively more polar than the P —H bonds In PH₃.

Hence, Nil, molecules remain associated through intermolecular hydrogen bonding, but no H-hond is formed among PH₂ molecules. Therefore the boiling point of NH₃ is much higher than that of PH3 even though the molecular mass of PH₃ is twice that of NH₃.

Question 5. The boiling point of ethyl alcohol (78.5°C) is much higher than that of isomeric dimethyl ether (-24.9°C).
Answer:

Due to the much higher electronegativity and smaller size of oxygen, the O—H bond in ethyl alcohol (C₂H₅OH) is much polar and so, its molecules remain Associated through Intermolecular hydrogen bonding. A greater amount of energy IN is required To separate the molecules Hence, ethyl alcohol volatile unit tin boiling point IN is relatively much higher.

On tin other hand, the O-uto in In dimethyl ether (CH₂OCH₃) IN not bonded to any hydrogen atom. TIUIH, there IN no Possibility of formation of hydrogen bond. Consequently, the ether CXINIH is a discrete molecule. So, It IN more volatile, and the UN boiling point IN much lower than that of ethyl alcohol.

Question 6. The boiling point of formic acid (100.5°C) is higher than that of ethyl alcohol (78.5°C) even though their molecular masses are the same (46) 
Answer:

Explanation:

Both formic acid and ethyl alcohol contain polar O—H bonds and remain associated through hydrogen bonding. Since the polarity of the O —H bond present in the —COOH group of formic acid Is much higher than that of the O—H bond in ethyl alcohol, the strength of the hydrogen bond in formic acid is much stronger than in ethyl alcohol.

(In formic acid, the polar bond of the carboxyl group is involved in the formation of hydrogen bonds). In the vapor state, the formic acid molecules are also involved in forming an eight-membered ring of the diner by intermolecular H-bond.

Because of the stronger hydrogen bond, the boiling point of formic acid is higher than that of ethyl alcohol even though both compounds have the same molecular mass.

In formic acid, the strength of the H-bond increases enormously, due to resonance.

Question 7. n -propylamine (CH₃CH₂C1I2NH₂) boils at a much higher temperature (49″C) than the isomeric trimethylamine (Me., N, 3°C).
Answer:

In n -n-propylamine, the electronegative N -atom is bonded to twoIl-atoms. So Its molecules get associated through Intermolcculur hydrogen bonding. Because of this, it boils at a much higher temperature than the Isomeric trimethylamine in which, the formation of hydrogen bonds is not possible since there is no hydrogen atom directly attached to the electronegative N -atom.

Question 8. The melting and boiling points of p -p-nitrophenol are much higher than those of o-nitrophenol.
Answer:

In o-nitrophenol, -OH and -NO₂ groups are situated at the two adjacent carbon atoms so that they are involved in intramolecular hydrogen bonding. Thus, these molecules exist as discrete molecules and consequently, the compound possesses lower melting and boiling points. On the other hand, in p-nitrophenol, the -OH and -NO₂ groups are situated far apart from each other so intramolecular H-bonding is not possible. The molecules remain associated through intermolecular H-bonding.

Hence, the melting and boiling points of p-nitrophenol are much higher than those of nitrophenol. Due to its low boiling point, o-nitrophenol is steam volatile whereas p-nitrophenol is not.

Intramolecular H-bonding in ortho-nitrophenol molecule

Intermolecular H-bonding in para-nitrophenol molecule For the same reason, the volatility of o -hydroxybenzoic acid is higher than that of its meta- and para-isomers.

Question 9. Ethyl alcohol (C₂H₅OH) dissolves in water in all proportions but dimethyl ether (CH₃OCH₃) is sparingly soluble in water.
Answer:

Bach ethyl alcohol molecule, containing a polar O—H bond, forms 2 hydrogen bonds with two water molecules. So it dissolves in water molecules in all proportions. On die other hand, each dimethyl ether molecule forms one H-bond with a single water molecule through its electronegative O-atom. Hence, dimethyl ether is sparingly soluble in water.

Question 10. Alcohols having lower molecular masses are soluble in water but those with higher molecular masses are insoluble in water.
Answer:

An alcohol molecule (R—OH) is made of a polar -OH group (the hydrophilic part) and a non-polar alkyl group, —R (the hydrophobic part).In lower alcohols (i.e., alcohols having lower molecular masses), the hydrophilic -OH group overcomes the effect of the hydrophobic hydrocarbon group ( —R). Thus, these alcohol molecules dissolve in water by forming an H-bond with water molecules

On the other hand, in higher alcohols (i.e., alcohols having higher molecular masses) the hydrophobic character of the hydrocarbon group predominates over the effect of the polar hydrophilic group. Because of this, the higher alcohols show a negligible tendency to form H-bonds with water molecules and consequently, they are insoluble in water.

Question 11. Ammonium chloride (NH₄Cl ) is more soluble than sodium chloride (NaCl) in water.
Answer:

Na+ ion, obtained by dissociation of NaCl is Stabilized by solvation, involving ion-dipole interactions (water acts as a dipole). But NH₄ ion, obtained by dissociation of NH₄Cl, becomes more stabilized by solvation involving the formation of stronger intermolecular hydrogen bonds with H₂O molecules. Thus, NH₄Cl is more soluble than NaClin water.

Question 12. Despite being organic compounds, glucose (C₆H₁₂O₂) and sugar (C₁₂H₂O₂) are soluble in the polar solvent, water.
Answer:

In glucose and sugar molecules, there are six and eight —OH groups respectively. Due to the presence of a large number of — OH groups, the molecules of these compounds form extensive H-bonds with water molecules. This accounts for the high solubility of these two organic compounds in the polar solvent, water.

Question 13. Although boric acid [H₃BOs or B(OH)₃] contains 3 —OH groups, its solubility in water is very poor.
Answer:

Boric acid molecules form a cyclic two-dimensional network, leading to a giant molecule by forming much stronger intermolecular H-bonds. Boric acid molecules can’t form hydrogen bonds with water molecules by breaking these strong hydrogen bonds. Hence, the solubility of boric acid in water is very poor.

Question 14. Grillo-nitrophenol is less soluble in water than its meta- or para-isomer.
Answer:

In o-nitrophenol, -OH and -NO₂ Groups are involved in intramolecular H-bonding. So, the molecules of o-nitrophenol cannot form H -bonds with water molecules. Hence, o-nitrophenol is less soluble in water. On the other hand, both m-andp-nitrophenol are incapable of forming intramolecular H-bonds, and their —OH and —NO₂ groups are free to form H-bonds with water molecules. So, these are more water-soluble.

o-nitrophottol being loss soluble In water Is steam-volatile. On the other hand, being more soluble In the water is not steam-volatile. Steam distillation, therefore, is an easy method of separation of o-nitrophenol from the other two Isomers.

Question 15. Although H₂ SO₄ And H₃ PO₄ Have the Same Molecular mass H3PO4 HAs a Higher Boiling Point And viscosity.
Answer:

H₃PO₄ molecule contains three —OH groups whereas each H₂SO₄ molecule contains only two —OH groups. So the extent of association through Intennolecular hydrogen bonding in H₃PO₄ is greater than that in H₂SO₄. Because of this, the boiling point and viscosity of H₃PO₄ are higher than that of H₂SO₄, even though their molecular masses are the same.

Question 16. Glycerol (CH₂OHCHOHCH₂OH) is a highly viscous liquid whereas ethyl alcohol is not.
Answer:

Glycerol molecules with three —OH groups, remain extensively associated through intermolecular hydrogen bonding and exist as long chains with many interlocking cross-linkages. For this reason, glycerol is a highly viscous liquid.

On the other hand, ethyl alcohol molecule with a single —OH group, remains associated through intermolecular hydrogen bonding. Though it exists as an associated molecule it does not possess interlocking cross linkages like glycerol. Therefore, ethyl alcohol is not a viscous liquid.

Question 17. ortho-hydroxybenzoic acid Is more acidic than parahy droxybcnzolc acid.
Answer:

Be conjugate base of o-hydroxybenzoic acid gets additional stability through intramolecular hydrogen bonding because the —OH and groups are attached to two adjacent ring carbons. On the other hand, the conjugate base of p-hydroxybenzoic acid is not stabilized by intramolecular hydrogen bonding because the groups are far apart from each other. Because of greater stability ofthe conjugate base, o-hydroxybenzoic acid releases H+ ion more easily, i.e., it exhibits more acidic properties than its para-isomer.

Question 18. 2,6-dihydroxybenzoic acid is much more acidic (nearly 800 times) than benzoic acid.
Answer:

In The Conjugate Base Of 2,6 dihydroxybenzoic acid the -coo- group undergoes intramolecular hydrogen bonding six-membered rings. The acid, PhCOO-, on the other hand, is not stabilized by H-bonding. Hence, 2,6- dihydroxybenzoic acid is more acidic than benzoic acid.

Question 19. The volume of ice decreases on melting or the density of ice is less than that of water.
Answer:

In the crystal of ice, each oxygen atom is surrounded tetrahedrally by four H-atoms, two H-atoms by usual covalent bonds, and the other two by H-bonds. Because of such tetrahedral arrangement and cage-like structure, there remains enough space among the water molecules.

When the ice melts, these tetrahedral arrangements of water molecules are largely destroyed. Water molecules come closer to each other and the intermolecular space is reduced. So, the same number of water molecules are now present in a smaller volume. Hence, the density of ice is less than that of water.

Question 20. The density of water is maximum at 4°C and decreases with an increase in temperature.
Answer:

When water is heated, the breaking of the tetrahedral structure having an ordered arrangement of H-bonds continues up to 4°C and the contraction caused by the proximity of water molecules is greater than thermal expansion (molecular agitation due to vibrational and kinetic energy).

Consequently, the volume goes on decreasing up to 4°C. Normal expansion of volume on heating starts only after 4°C. Therefore, water has its highest density at 4°C and it decreases with an increase in temperature.

Question 21.  On heating a crystal of blue vitriol or cupric sulfate pentahydrate (CuSO₄.5H₂O ), four molecules of water of crystallization are easily removed. However, the removal of the last molecule of the water of crystallization requires thermal energy.
Answer: 

Explanation:

Among the five molecules of water crystalized in the structure of blue vitriol, four molecules are linked to the Cu2⁺ ion through coordinate covalent bonds, and the remaining one molecule of water is joined to the sulfate ion (SO_2-⁴). Two out of four molecules of water are linked also through H-bonds with the fifth water molecule which is further hydrogen bonded with SO_2-⁴ ion.

Dehydration of blue vitriol, i.e., removal of water of crystallization from blue vitriol occurs in three steps.

This is due to the presence of different types of bonds between CuSO₄ and molecules of water of crystallization. Two water molecules are bonded to the Cu²⁺ ion by coordinate bonds. Thus they are loosely held and on heating, they are removed at a comparatively low temperature (30°C).

The other two water molecules are bonded to the Cu²⁺ ion by coordinate bonds as well as to the fifth water molecule by H -bonds. Hence, they are tightly held and are removed at a comparatively high temperature (100°C).

The fifth water molecule is bonded to the third and fourth water molecules as well as to the two O-atoms of the sulfate ion through H-bonds. Thus, it is more tightly held and its removal requires even more energy i.e., much higher temperature (250°C). This water molecule is called anion water.

Participation C—H band in hydrogen hand formation:

Since there is negligible difference in electronegativity between carbon and hydrogen, so H-atom of the C —H bond is not expected to participate in hydrogen bond formation. But if an atom of any strong electronegative atom or electron-attracting group is attached to a carbon atom, then in some cases C—2 bond does take part in the formation of hydrogen bond, Example molecules of acetonitrile (CH₃CN) get associated through intramolecular hydrogen bond; chloroform (CHCl₃) molecules form hydrogen bonds with the molecules of acetone (CH₃COCH₃).

Difference between Hydrogen bond and covalent bond:

Molecular Orbital Theory (MOT)

Although the concept of hybridization could explain the structure and shapes of most of the covalent compounds quite satisfactorily, some of the characteristics of molecules such as their relative bond strengths, paramagnetic and diamagnetic nature, etc., including the formation of odd electron molecules or ions could not be explained successfully by this concept.

To explain these, the molecular orbital theory was put forward by F. Hund and R.S. Mulliken in 1932. As this theory is based on the linear combination of atomic orbitals, it is also called LCAO-MO theory.

Salient features of molecular orbital theory:

When two atomic orbitals overlap, they lose their identity and form new orbitals called molecular orbitals.

• Molecular orbitals are formed only by those atomic orbitals which are of comparable energy and proper symmetry. One s -s-s-orbital, for example, can combine only with another s -s-orbital but not with any 2s-orbital.
• An electron in an atomic orbital is influenced by only one nucleus, but in a molecular orbital, it is under the influence of two or more nuclei depending on the number of atoms in the molecule. Therefore, an atomic orbital is monocentric while a molecular orbital is polycentric.
• The number of molecular orbitals formed is the same as the number of combining atomic orbitals. Half of the new molecular orbitals have lower energy and hence have greater stability than the participating atomic orbitals.
• These orbitals with lower energy are called bonding molecular orbitals. The other half of the new molecular orbitals having higher energy than the participating atomic orbitals are called antibonding molecular orbitals.

Thus two atomic orbitals combine to form two molecular orbitals one of which is a bonding and the other one antibonding. The shapes of the resultant molecular orbitals depend on the shapes of atomic orbitals undergoing combination.

• The molecular orbitals are filled with electrons according to the same rules as applied to the filling of atomic orbitals.
• The molecular orbitals are filled in the order of their increasing energies (Aubau principle).
• A molecular orbital can accommodate a maximum of two electrons having opposite spin (According to Pauli’s exclusion principle).
• The pairing of electrons in degenerate molecular orbitals (molecular orbitals of equal energy) does not take place until each of them is singly occupied (Hund’s rule).

Formation of molecular orbitals: Linear Combination of Atomic Orbitals (LCAO)

In wave mechanics, atomic orbitals are expressed by wave functions which represent the amplitude of the electron waves. These are obtained from the solution of the Schrodinger wave equation.

• A similar wave equation can be formed to describe the behavior of electrons in molecules. However, because of the complex nature of the equation, it is very difficult to determine wave functions of molecular orbitals by solving the equation.
• To obtain die wave functions of molecular orbitals, an approximate method, a linear combination of atomic orbitals (LCAO) method has been adopted.
• Let us apply this method to the homonuclear diatomic H₂ molecule. Let us consider that the hydrogen molecule is composed of two hydrogen atoms HA and HB. In the ground state, each hydrogen atom has one electron 1s -orbital.
• These atomic orbitals may be represented by the wave functions iff and i/fg which represent the amplitudes of the electron waves of the atomic orbitals of the two atoms HA and HB respectively.

According to the LCAO method, the linear combination of atomic orbitals to form molecular orbitals can take place by addition and subtraction of wave functions of the individual atomic orbitals as shown below:

⇒ \(\psi_{\mathrm{b}}=\psi_{\mathrm{A}}+\psi_{\mathrm{B}} \text { and } \psi_{\mathrm{a}}=\psi_{\mathrm{A}}-\psi_{\mathrm{B}}\)

These two types of combinations give rise to two molecular orbitals. The molecular orbital if formed by the addition of wave functions of the two atomic orbitals is called bonding molecular orbital (BMO) and the molecular orbital formed by the subtraction of wave functions of the two atomic orbitals is called antibonding molecular orbital (ABMO).

• The combination of Is -orbitals of the two H-atoms to form molecular orbitals is shown in the given figure. It becomes clear from the figure that in the bonding molecular orbital, the electron density is mainly concentrated between the two positive nuclei.
• This high electron density between the two nuclei decreases their mutual repulsion and holds the nuclei together. Therefore, the hooding molecular orchid always possesses lower energy than either of the combining atomic orbitals. Electrons placed In it bond molecular orbital
• Formation of molecular orbital by a combination of li-atomic orbital stabilizes a molecule. On the other hand, in the antibonding molecular orbital, most of the electron density is mainly located away from the space between the nuclei, creating a node In between them.
• As a consequence, the two nuclei repel each other strongly. Therefore, the antibonding molecular orbital always possesses higher energy than either of the combining atomic orbital and the electrons placed in the antibonding molecular orbital destabilize the molecule.
• The formation of bonding and antibonding molecular orbitals can also be explained In terms of constructive interference (or additive effect) and destructive interference (or subtractive effect) of the electron waves of the combining atoms.

Bonding molecular orbital:

During constructive Interference, the two electron waves are in phase and the amplitudes of the waves get added up and reinforce each other. As a consequence, there is high electron density between the two nuclei, favoring bond formation. The resulting wave corresponds to a bonding molecular orbital.

Antibonding molecular orbital:

When the two electron waves of two combining atoms arc out of phase, there is destructive interference, and the amplitudes of the waves get subtracted. Thus, the resulting wave is weakened and consequently, there is low electron density between the nuclei.

Note that the crests of the electron wave are usually given by the’ +’ sign and the troughs by the ‘- ‘ sign. The bonding molecular orbital is formed by a combination of ‘+’ with’ +’ or with part of the electron waves whereas the antibonding molecular orbital is formed by the combination of ‘ + ’ with the ‘- ‘ part. For a homo-dinuclear molecule, since, if =, the amplitude of ABMO is zero.

Nodal plane:

In the antibonding molecular orbital, the plane perpendicular to the line joining the two nuclei where the probability of finding the electron (i.e., electron density) is zero, is called the nodal plane.

Energy of bonding and antibonding molecular orbitals

Linear combination of two atomic orbitals of hydrogen, for example, gives rise to a bonding molecular orbital of lower energy (higher stability) and an antibonding molecular orbital of higher energy (lower stability).

The energy difference between the bonding molecular orbital and the combining atomic orbitals is known as stabilization energy and the energy difference between the antibonding molecular orbital and the combining atomic orbitals is known as destabilization energy. Further, it may be noted that bonding MO is stabilized to the same extent (A) as the antibonding MO is destabilized.

An electron occupying the bonding molecular orbital lowers the energy ofthe system while an electron occupying the antibonding molecular orbital raises the energy of the system by an equal amount.

Different types of molecular orbitals

The molecular orbitals of diatomic molecules are designated as cr (sigma), 7r(pi), d(delta), etc.

σ-molecular orbitals:

The molecular orbitals formed by overlapping of atomic orbitals along the internuclear axis are called sigma (σ) molecular orbitals. Such molecular orbitals are symmetrical about the internuclear axis or bond axis.

Antibonding sigma molecular orbitals are expressed by the symbol cr. To denote the constituent orbital from which any molecular orbital is formed, the atomic orbital is mentioned beside the symbol of the molecular orbital or as its subscript.

Example: Combination of two σ∗1s -orbitals results in the formation of curls and cr s molecular orbitals.

Formation Of σ∗1s and one antibonding MO of higher energy \(\sigma_{2 s}^*\)

In the same way, additive and subtractive combinations of two Is orbitals result in the formation of curls and crl5 molecular orbitals respectively.

By the combination of two 2pz-atomic orbitals:

According to modem convention, the z-axis is taken as the internuclear axis. Hence, the 2pz -orbital is assumed to lie along the internuclear axis.

Therefore, the molecular orbital formed by the addition of wave functions of two 2pz -orbitals is a bonding σ-MO designated as cr2p the molecular orbital formed by the subtraction of wave functions of two 2pz -orbitals is an antibonding σ-MO designated as

⇒  \(\sigma_{2 p_z}^*\).

x-molecular orbitals:

The molecular orbitals that are formed by sideways or lateral overlap of two atomic orbitals are called pi (π) molecular orbitals. Such molecular orbitals are not symmetrical about the internuclear axis.

Formation of π -molecular orbitals:

By the combination of two 2px -atomic orbitals:

The axis of px -atomic orbital, according to modem convention, is perpendicular to the internuclear axis.

The Molecular orbital formed by the lateral overlap of two 2px- atomic orbitals involving the addition of electron waves is a bonding π-MO designated as 7r2p and the molecular orbital formed by the lateral overlap of two 2px -atomic orbitals involving the subtraction of electron waves is an antibonding π-MO designated as \(\pi_{2 p_x}\)

By a combination of two 2py-atomic orbitals:

The axis of py -the atomic orbital, according to the modem convention, is perpendicular to the internuclear axis.

Therefore, the lateral additive combination of two parallel 2py -orbitals leads to the formation of a bonding π-MO designated as π 2py, and the lateral subtractive combination of two parallel 2py orbital leads to the formation of an antibonding n -MO designated as \(\pi_{2 p_y}^*\).

Differences between σ and π- molecular orbitals

Conditions For the combination of atomic orbitals from molecular orbitals

The following conditions must be satisfied for the combination of atomic orbitals to form molecular orbitals;

The energies of the combining atomic orbitals must be the same or nearly the same.

Example: In the case of the formation of homonuclear species like H₂, N₂, or O₂, the ls-orbital of one atom can combine with the Is-orbital of the other. But there is no possibility of the ls-orbital of any atom to combine with the 2s-orbital of another.

The extent of overlap between the combined atomic orbitals must be large.

The combining atomic orbitals must possess the same symmetry about the molecular axis, i.e., they must have proper orientations.

Examples: 2px -orbital of one atom can combine with 2px orbital of another atom but not with 2 py or 2pz orbital.

0 2s -orbital of an atom can overlap with the 2pz -orbital of another but not with 2 px or 2py, -orbital (which remain perpendicular to the internuclear axis). The reason is that in the second case, the overlapping of V with ‘+’ cancels the overlapping of V with ‘-‘.

The situation is shown below:

According to the present convention, the z-axis is taken as the internuclear axis.

• A bonding molecular orbital is formed by the addition of wave functions of two atomic orbitals.
• Antibonding molecular orbital is formed by subtraction of wave functions of two atomic orbitals.
• Combining atomic orbitals, and forming the molecular orbitals, must be of comparable energy.
• The axis of the pz -atomic orbitals of the combining atoms lie along the internuclear axis. So, pz -orbital is capable of forming cr -molecular orbital by overlapping along this axis.
• The axes of px and -atomic orbitals are perpendicular to the internuclear axis and hence they can form n-molecular orbitals with bilateral overlapping.

Energy level diagram of molecular orbitals:

Order of energy of homonuclear diatomic molecular orbitals

Two Is -orbitals of two atoms combine to form two molecular orbitals— crls and cr*s. In the same manner, two 2s-atomic orbitals give rise to two molecular orbitals (cr2i and or2S) and six 2p-atomic orbitals on combination give six molecular orbitals \(\left(\sigma_{2 p_z}, \sigma_{2 p_z}^*, \pi_{2 p_x}, \pi_{2 p_x}^*, \pi_{2 p_y}, \pi_{2 p_y}^*\right)\).

The energy sequence of these ten molecular orbitals can be determined by the following information:

The energies of molecular orbitals depend on the energies of the corresponding atomic orbitals involved in their formation. For example, since 2s-atomic orbitals have higher energy than tire Is -atomic orbitals, the pair of molecular orbitals obtained from 2s-atomic orbitals have higher energy than that obtained from Is -atomic orbitals.

Within the same pair of molecular orbitals, the bonding MO has lower energy than the antibonding MO.

For Example: \(\sigma_{1 s}<\sigma_{1 s}^*, \pi_{2 p_x}<\pi_{2 p_x}^* .\)

⇒ \(\pi_{2 p_x}\) and \(\pi_{2 p_y}\) molecular orbitals have equal and molecular orbitals also have equal energies i.e., they are also degenerate orbitals than the sideways or lateral overlapping.

This is quite true for O₂ and molecules such as F₂ and Ne₂(hypothetical) where the energy difference between 2s and 2p -atomic orbitals are quite large. However, in molecules up to N₂, the energy difference between these atomic orbitals is very small and the molecular orbitals <x2s and are so close that they experience mixing interactions.

(repulsion between the electrons present in them). As a result, the energy of cr2p orbital increases and becomes greater than N₂p and N₂p, which do not experience such intermixing interactions.

The energy of a2p molecular orbital is greater than that of N₂p or N₂p molecular orbital. From the above information and spectroscopic studies, it has been established that the energy of the first ten molecular orbitals of the homonuclear diatomic molecules or ions such as Li₂, Be₂ (hypothetical), B₂, C₂, N₂ follow the order

⇒ \(\sigma_{1 s}<\sigma_{1 s}^*<\sigma_{2 s}<\sigma_{2 s}^*<\left[\pi_{2 p_x}=\pi_{2 p_y}\right]<\sigma_{2 p_z}\) \(<\left[\pi_{2 p_x}^*=\pi_{2 p_y}^*\right]<\sigma_{2 p_z}^*\)

The energy of the first ten molecular orbitals of the homonuclear diatomic molecules such as O₂, F₂, and Ne₂ (hypothetical) follow the order:

⇒ \(\sigma_{1 s}<\sigma_{1 s}^*<\sigma_{2 s}<\sigma_{2 s}^*<\sigma_{2 p_z} <\left[\pi_{2 p_x}=\pi_{2 p_y}\right]\)\(<\left[\pi_{2 p_x}^*=\pi_{2 p_y}^*\right]<\sigma_{2 p_z}^*\)

The above two energy sequences ofthe molecular orbitals are shown in the following figures:

The difference in MO energy diagrams can be explained as follows. Since cr -bonds are generally stronger, therefore, it would be expected that cr{2p) orbital would be of lower energy than a comparable n{2p) orbited.

However, for most ofthe elements, the σ(2p) orbital lies at a higher energy than n{2p) orbitals because the difference between 2s and 2p atomic energy levels is small (except in the case of O₂, F_2, and Ne₂ (hypothetical) where the difference in energy levels of 2s and 2p is large).

As a consequence, σ(2s), and σ(2p) are so close together that the repulsive forces between the electrons presenting them raise the energy of cr(2p) above that of n(2p) molecular orbitals. In the words, in the case of O₂, F₂, etc., there is no 2s-2p mixing, while for B₂, C₂, and N₂, there is 2s-2p mixing.

Rules for filling up molecular orbitals

1. Molecular orbitals get filled up with electrons according to the Aufbau principle. At first, the electrons enter the molecular orbital of the lowest energy and gradually that higher energy.
2. Molecular orbitals can accommodate a maximum of 2 electrons which must have opposite spin (Pauli’s exclusion principle).
3. The pairing of electrons in degenerate orbitals does not occur until each degenerate orbital is singly filled with electrons having parallel spin (Hund’s rule).
4. Electrons presenting bonding molecular orbitals are called bonding electrons (Nb) and those in antibonding molecular orbitals are called antibonding electrons (Na).

Molecular Behavior and Electronic Configuration The important rules relating molecular behavior with electronic configuration are as follows:

• Stability of molecules in terms of bonding and antibonding electrons: From electronic configuration, we get the number of Nb and Na.
• If Nb > Ng, the molecule is stable. Because a greater number of bonding orbitals are occupied by electrons than antibonding orbitals, resulting in a net force of attraction (stronger bonding).
• IfNa < Nb, the molecule is unstable. Because a greater number of antibonding orbitals are occupied by electrons than bonding orbitals, resulting in a net force of repulsion.
• If Nb = Na, the molecule is unstable. Because the destabilization caused by electrons in antibonding molecular orbitals is greater than the stabilization gained by bonding electrons.

Bond order: The number of covalent bonds in a molecule is expressed by bond order. Bond order is defined as half the difference between the number of electrons present in bonding and antibonding orbitals,

⇒ \(\text { Bond Order (B.O.) }=\frac{1}{2}\left(N_b-N_a\right)\)

Important points regarding bond order

A molecule with B.0. 1, 2, or 3 indicates that it has a single bond, double bond, or triple bond respectively.

The relative stability of molecules in terms of bond order:

The stability (bond dissociation enthalpy) of the diatomic molecules is directly proportional to the bond order.

Thus, a molecule with bond order 3 (for example nitrogen, N =N, having bond dissociation enthalpy 945 kj mol‾1) is more stable than a molecule with a bond order 2 [e.g., oxygen, 0=0, having bond dissociation enthalpy 498 kj. mol¯1 ) which in turn is more stable than a molecule with a bond order 1 (Example fluorine, F—F, having bond dissociation enthalpy 158 kj. mol-1 ).

Bond length in terms of bond order:

Bond order is inversely proportional to bond length. For example, the bond length of the N₂ molecule having bond order 3 is 1.10A whereas that of the O₂ molecule having bond order 2 is 1.21A.

Magnetic nature of molecules:

• It can be predicted successfully by MO theory. If all the electrons in a molecule are paired, it is diamagnetic and if it contains one or more unpaired electrons, it is paramagnetic.
• The greater the number of unpaired electrons in a molecule or ion, the greater its paramagnetic nature. H₂, for example, is diamagnetic while 02 is paramagnetic. The fractional value of bond order indicates that the molecule is paramagnetic.

Electronic configuration & nature of bonding of some homonuclear diatomic molecules & their ions

1. Hydrogen molecule (H₂):

• An H₂ molecule is formed by the combination of two H-atoms with one electron present in each -orbital. The two Is -orbitals combine mutually to form two molecular orbitals (trail and curls).
• The two electrons will occupy the lower energy curls MO and according to Pauli’s exclusion principle, the electrons have opposite spins. So, according to MO theory, the electronic configuration of the hydrogen molecule is 2
• . The MO energy level diagram for the H₂ -H2-molecule is shown below.

Stability: A positive value of B.O. indicates that the H₂ is – 1} stable. A much higher value of bond-dissociation energy (438kj. mol^-1 ) of H₂ indicates that it is a very stable molecule.

Magnetic nature: As the two electrons present in this molecule are paired, the molecule is diamagnetic.

2. Positive hydrogen ion (H₂):

This is formed by the combination of the H -atom containing one electron and one H+ion having no electrons. Therefore, the H₂ ion has only one electron which is occupied in the lower energy bonding MO i.e., curls orbital.

The electronic configuration of H²⁺ is 1. The following figure shows the MO energy level diagram of this ion.

Bond order of H²⁺ ion \(=\frac{N_b-N_a}{2}=\frac{1-0}{2}=\frac{1}{2}\)

Stability: Bond order positive. This shows that the Ion In somehow stable but it In less stable than, the molecule (bond dissociation enthalpy: H²⁺=  280 kj – mol^-1 ). The existence of H₂⁺ has been established spectroscopically.

Magnetic nature:  From the electronic configuration, it is evident that the fl2 Ion contains one odd electron, Therefore, the ion Is paramagnetic.

3. Negative hydrogen ion (H₂^–):

When one hydrogen atom, 11 having one electron in its )s-orbital combines with a hydride Ion (H^–) having two electrons in 1 s-orbital, the anion, (H₂) is formed. Thus, the H₂ ion contains three electrons.

Thus the electronic configuration of H₂^– ion: \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}\right)^1\)

Stability: A small positive bond order suggests that the ion is somewhat stable. Note that the H₂ ion is slightly less stable than the H+ ion even though both have the same bond order (0.5). This is because the antibonding MO of H₂ ion contains no electron whereas the antibonding MO of H₂ contains one electron which causes destabilization.

Magnetic nature: Since the ion contains one unpaired electron, it is paramagnetic.

4. Helium molecule (He₂):

Each He-atom has two electrons in Is -orbital. Therefore, the total number of electrons in the He₂ molecule is four. Two of them occupy orbital while the other two occupy \(\sigma_{1 s}^*\) orbital.

Thus, the electronic brv configuration of He₂ molecule is: \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}\right)^2\).

The MO diagram for this molecule is given as follows:

Bond order of He₂ molecule ,\(=\frac{N_b-N_a}{2}=\frac{2-2}{2}=0\)

Energy diagram of molecular orbital of He₂ molecule As bond order is zero, the He₂ molecule has no real existence. In reality, helium gas consists of monoatomic molecules.

5. Helium molecule-cation (He_2⁺):

This ion contains 3 electrons, two from one He -atom and one from one He⁺ ion. Two electrons occupy the bonding MO\(\left(\sigma_{1 s}\right)\) while the third electron occupies the antibonding MO (curls).

Thus the electronic configuration of He ion is: \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}\right)^1\). The MO energy level diagram for this ion is given below.

Bond order of \(\mathrm{He}_2^{+} \text {ion }=\frac{N_b-N_a}{2}=\frac{2-1}{2}=\frac{1}{2}\)

Stability: The positive value of bond order indicates that the formation of He₂ ion is possible. Its bond dissociation enthalpy is 242.67kj. mol^-1 .

Magnetic nature: Since the ion contains one unpaired electron, it is paramagnetic. 2 Note that the He. J Ion has the same bond order as the H₂ ion and both have the same number of electrons in the antibonding orbitals. Therefore, they should have similar stability (bond dissociation enthalpy), bond length, and paramagnetic nature.

6. Lithium molecule (Li₂):

The electronic configuration of lithium (Z = 3 ) is Is²2s¹. Thus, there are six electrons in the lithium molecule.

The electronic configuration of the Li₂ molecule is written as:

⇒ \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2 \text { or } K K\left(\sigma_{2 s}\right)^2\)

Where KK represents the closed K -K-shell structure:

⇒  \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\)

Bond order of Li₂ molecule \(=\frac{N_b-N_a}{2}=\frac{4-2}{2}=1\)

Stability: From the value of bond order it is clear that there is a single bond between the two Li -atoms and therefore, the Li₂ molecule is stable.

Magnetic nature: Since it has no impaired electron, it is diamagnetic.

7. Beryllium molecule (Be₂):

The electronic configuration of beryllium (Z = 4) is ls²2s². So, the Be²molecule contains a total of eight electrons. The electronic configuration of the Be2 molecule, is, therefore, written as:

⇒  \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\)

Bond order of Be₂ molecule= \(=\frac{N_b-N_a}{2}=\frac{2-2}{2}=0\).

As the bond order of the Be₂ molecule is zero, it has no real existence. These predictions from MO theory are in agreement with the fact that diamagnetic Be molecules are found to exist in the vapor phase as monoatomic molecules

8. Boron molecule (B₂):

The electronic configuration of boron(Z = 5) is 1s²2s²2p¹. So, there are ten electrons in the B₂ molecule.

The electronic configuration of B₂ is, therefore, written as:

KK\(\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^1\left(\pi_{2 p_y}\right)^1\)

Bond order of B₂ molecule = \(\frac{N_b-N_a}{2}=\frac{4-2}{2}\)

= 1

Stability: The molecule has bond order, 1. So it is somewhat stable. The bond dissociation enthalpy of the B₂ molecule is 290 kl-mol^-1

Magnetic nature: Since the molecule contains two unpaired electrons (one each in π2p_x and π2p_y MO ), it is paramagnetic

9. Carbon molecule (C₂):

The electronic configuration of carbon (Z = 6) is ls²2s²2p². There are twelve electrons in the C₂ molecule. The electronic configuration of the C₂ molecule Is written as

KK \(\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\)

Bond order of C₂ molecule \(=\frac{N_b-N_a}{2}=\frac{6-2}{2}\)

= 2

Stability: Since the bond order Is 2, the molecule is stable. The bond dissociation enthalpy of C₂ is 620 kj-mot^-1. The two bonds present in the C₂ molecule are both K bonds because of the presence of four electrons in two n -n-molecular orbitals. This is different from the usual double bonds that are made up of σ -bond and an π -bond.

Magnetic nature: Since a die molecule contains no unpaired electrons, it is diamagnetic. Diamagnetic C₂ molecules have been detected in the vapor phase.

10. Nitrogen molecule (N₂): The electronic configuration of nitrogen (Z = 7 ) is ls²2s²2p³.

Thus the total number of electrons present in the N₂ -molecule is 14. The electronic configuration of N₂ can, therefore, be written as:

KK \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\sigma_{2 p_z}\right)^2\)

Bond order of N₂ molecule = \(=\frac{N_b-N_a}{2}=\frac{8-2}{2}\)

= 3

Generally, the completely filled \(\sigma_{1 s} \text { and } \sigma_{1 s}^*\)  are not consideredin the calculation ofbond order.

Stability:

The bond order value of 3 suggests that the two nitrogen atoms are linked to each other by a triple bond (N = N). This in turn suggests that the molecule must have n much higher value of bond dissociation enthalpy greater stability (compared to all other diatomic molecules) and a shorter bond length. This is in agreement with the experimentally determined values of bond dissociation enthalpy (945kj mol^-1 ) and bond length ( 1. 10 Å). Molecular nitrogen is, therefore, quite inert to chemical combinations.

Magnetic nature: The presence of no unpaired electron indicates that the molecules diamagnetic in nature

11. Oxygen molecule (O₂):

The electronic configuration of oxygen (Z = 8 ) is ls²2s²2p⁴ Thus O₂  molecule contains a total of 16 electrons.

The electronic configuration of O₂ molecules, therefore, is written as:

KK \(\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\left(\pi_{2 p_y}^*\right)^1\)

In the case of elements having atomic numbers greater than nitrogen, the energy of σ 2p_z orbital is lower than that of π2p_x or π2p_y orbital.

Bond order of O molecule = \(=\frac{N_b-N_a}{2}=\frac{8-4}{2}\)

= 2

Stability: The bond order value of 2 suggests the presence of a double bond in the O₂ -molecule. Thus the molecule is considerably stable (bond dissociation enthalpy: 501 kj . mol^-1.).

Magnetic nature: The presence of two unpaired electrons in \(\pi_{2 p_x}^* \text { and } \pi_{2 p_y}^*\)  -molecular orbitals accounts for the paramagnetic nature ofthe molecule. The molecule is weakly attracted by a magnetic field.

12. Superoxide ion (O₂^– ):

O₂^– Ion Is formed by the addition of one electron to O -molecule \(\mathrm{O}_2+e \rightarrow \mathrm{O}_2^{-}\). This electron enters into \(\pi_{2 p_x} \text { or } \pi_{2 p_y}\) molecular orbital. The electronic configuration of 02 ion, is, therefore, written as:

⇒  K K\(\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^1\)

Bond order of \(\mathrm{O}_2^{-} \text {ion }=\frac{N_b-N_a}{2}=\frac{8-5}{2}=1.5\)

Stability: Since its bond order is lower than that of O₂, its bond dissociation enthalpy is lower and its bond length is longer than that of O₂.

Magnetic nature: Owing to the presence of one unpaired electron, it is paramagnetic.

13. Peroxide ion (O₂^2- ) ⇒  \(\mathrm{O}_2^{2-}\) ion is formed when O₂ accepts two electrons: \(\mathrm{O}_2+2 e \rightarrow \mathrm{O}_2^{2-}\)

These two electrons enter into molecular orbitals

Therefore, the electronic configuration of O2- ion can be written as:

KK \(\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^2\)

Bond order of \(2_2^{2-} \text { ion }=\frac{N_b-N_a}{2}=\frac{8-6}{2}=1\)

Stability: Since its bond order is lower than that of O₂. its bond dissociation enthalpy is lower and the bond length is longer than that of O₂.

Magnetic nature: Since the ion contains no unpaired electron, it is diamagnetic

Relative bond dissociates lot enthalpies, liabilities, and bond lengths of

⇒  \(0_2(B .0,=2), 0_2^*(8,0,-2.5), 0_2^{-}(B .0,=1.5)\)

1. Bond dissociation energy: Bond dissociation enthalpy is directly proportional to bond order. Hence, bond dissociation enthalpies follow the order \(\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)
2. Stability: The greater the bond order, the greater the stability The stability order Is: \(\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)
3. Bond length: As bond length Ls is inversely proportional to bond order, the bond lengths of these species follow the order:  \(\mathrm{O}_2^{2-}>\mathrm{O}_2^{-}>\mathrm{O}_2>\mathrm{O}_2^{+}\)

14. Fluorine molecule (F2):

The electronic configuration of fluorine (Z = 9) is 1s²2s²2p⁶. Thus molecule contains a total of 18 electrons.

The electronic configuration of the F₂ -molecule is, therefore, written as:

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^2\)

Bond order of F₂ molecule \(=\frac{N_b-N_a}{2}=\frac{8-6}{2}=1\)

Stability: Since its bond order is 1, its bond dissociation enthalpy (stability) is not too high (129 kj mol‾1 ).

Magnetic nature: The absence of unpaired electrons indicates the diamagnetic nature of the F2 molecule.

15. Neon molecule (Ne₂):

The electronic configuration of (Z = 10) is ls²2s²2p⁶. So, the total number of electrons present in Ne₂ -molecule =10+10 = 20.

The electronic configuration of the Ne2 molecule is, therefore, written as:

KKI \(\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^2\left(\sigma_{2 p_z}^*\right)^2\)

Bond order of Ne₂ molecule \(=\frac{N_b-N_a}{2}=\frac{8-8}{2}=0\)

Since the bond order is zero, no bonds are formed between the two Ne -atoms, or in other words, Ne is a monoatomic gas and the Ne₂ molecule does not exist.  Thus, both He (discussed before) and Ne are monoatomic gases. Hence, it can be concluded that the noble gas elements are monoatomic in nature.

 Molecular orbital population along with molecular properties of diatomic molecules:

NCERT Solutions For Class 11 Chemistry Chapter 5 States Of Matter Gases And Liquids Short Question And Answers

Question 1. How much time would It take to distribute one Avogadro number of wheat grains If 10¹⁰grains are distributed each second?
Answer:

1010 grains are distributed in 1 sec. So, the time required to distribute 6.022 × 10²³grains would be

= \(\frac{1 \times 6.022 \times 10^{23}}{10^{10}}=6.022 \times 10^{13} \sec\)

= \(696.990 \times 10^6 \text { days }\)

= \(1.909563 \times 10^6 \text { years }\)

Question 2. Calculate the total pressure In a mixture of 8 g of dioxygen and 4 g of dihydrogen confined In a vessel of 1 dm³ at 27°C, R = 0.083 bar .dm³. K^-1mol^-1
Answer:

⇒ \(\left(\frac{8}{32}+\frac{4}{2}\right)=2.25 \mathrm{~mol}\) Molar mass of O₂and H₂ are 32 and 2 g mol-1 respectively]

As given: V = 1dm3 and T = (273 + 27)K = 300K

∴ \(P=\frac{n R T}{V}=\frac{2.25 \times 0.083 \times 300}{1}=56.025 \text { bar. }\)

Question 3. What will the nature of the PV vs P graph be for a real gas at Boyle temperature?
Answer:

AOS, At Boyle temperature the value of PV, particularly in the low-pressure region, becomes constant. Thus, the gas shows ideal behaviour. So, at this temperature, the plot pv of PV against P will give a straight line parallel to the P-axis

Question 4. Of the following types of velocity, which one has the highest value and which one has the lowest value at a given temperature?
Answer:

At a given temperature; the average velocity, the root mean square velocity, and the most probable velocity of the molecules of gas with molar mass M are given by average velocity

⇒  \((\bar{c})=\sqrt{\frac{8 R T}{\pi M}}\) and most Probable velocity

⇒ \(\left(c_m\right)=\sqrt{\frac{2 R T}{M}}\)

These expressions indicate that at a given temperature a given gas has the highest value and cm has the lowest value.

Question 5. Why are the deviations from the ideal behaviour of CO₂ and CH₄ greater than those of H₂ and He?
Answer:

The Molar masses of CO₂ and CH₄ are greater than those of H₂ and He. Hence, the intermolecular attractive forces in CO₂ and CH₄ are also greater in magnitude than those in H₂ and He. This results in a greater deviation from ideal behaviour for CO₂ and CH₄ than that for H₂ and He.

Question 6. A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen
Answer:

Let the weight of the mixture be w g. So, in the mixture, the weight of H₂ gas = 0.2 w g and that of O₂ gas =(w-0.2w) =0.8 w g

⇒ \(0.2 w g \text { of } \mathrm{H}_2=\frac{0.2 w}{2}=0.1 \times w \mathrm{~mol} \text { of } \mathrm{H}_2\)

⇒ \(0.8 \mathrm{wg} \text { of } \mathrm{O}_2=\frac{0.8 w}{32}=0.025 \times w \mathrm{~mol} \text { of } \mathrm{O}_2\)

So, the partial pressure of H₂ in the mixture = JCH x total pressure of the mixture =0.8 × 1 bar = 0.8 bar

Question 7. What would be the SI unit for the quantity \(\frac{P V^2 T^2}{n}\)
Answer:

In the SI system, the units of P, V, T and n are N-m~2, m3, K and mol respectively So, the SI unit for the quantity

⇒  \(\frac{V^2 T^2}{n}\) will be \(\frac{\mathrm{N} \cdot \mathrm{m}^{-2} \times\left(\mathrm{m}^3\right)^2 \cdot \mathrm{K}^2}{m o l} \text { i.e., } \mathrm{N} \cdot \mathrm{m}^4 \cdot \mathrm{K}^2 \cdot \mathrm{mol}^{-1}\)

Question 8. The critical temperature for carbon dioxide and methane arc is 31.1 °C and -81.0°C respectively. Which of these has stronger Interiuolecuhir forces and why?
Answer:

The critical temperature of gas Is a measure of the intermolecular forces of attraction In the gas. A gas with stronger intermolecular forces of attraction has a higher critical temperature. Therefore, CO₂., gas possesses stronger intermolecular forces of attraction than CH₄ gas because CO₂ has a higher critical temperature than CH₄.

Question 9. A 2L flask contains 0.4 g O₂ and 0.6 g H₂ at 100°C. Calculate the total pressure of the gas mixture in the flask. 4
Answer:

Total no. of moles of O₂ & H₂ in flask \(=\frac{0.4}{32}+\frac{0.6}{2}=0.3125\)

∴ \(P=\frac{n R T}{V}=\frac{0.3125 \times 0.082 \times 373}{2}=4.78 \mathrm{~atm}\)

Question 10. Arrange CO₂, SO₂ and NO₂ gases in increasing order of their rates of diffusion under the same condition of temperature and pressure with reason.
Answer:

At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. Hence increasing order of rates of diffusion is:

⇒ \(r_{\mathrm{SO}_2}<r_{\mathrm{NO}_2}<r_{\mathrm{CO}_2} .\)

Question 11. Explain the nature of the graphs of log P versus log V and logy versus log T. What are the units of the van der Waals constants ‘a’ and ‘b’?
Answer:

The plot of log P versus log V indicates a straight line with a negative slope (-1). The plot of log V versus log indicates a straight line with positive slope (+1)

Unit of is atm. L² mol¯² a unit of ‘b’ is L-mol^-1

Question 12. Prove that c = [E= total kinetic energy molecules of 1 mol of a gas, M=molar mass of the gas, arms = root mean square velocity of gas molecule]
Answer:

We known \(c_{r m s}=\sqrt{\frac{3 R T}{M}} \quad \text { or, } c_{r m s}=\sqrt{\frac{2}{M} \times \frac{3}{2} R T}\)

The total kinetic energy of the molecules of lmol gas, \(E=\frac{3}{2} R T\)

∴ \(c_{r m s}=\sqrt{\frac{2 E}{M}} \quad \text { (Proved) }\)

Question 13. It is easier to liquefy a gas with a higher critical temperature—explain.
Answer:

The critical temperature of a gas reflects the strength of intermolecular attractive forces in the gas. A gas with a higher value of critical temperature possesses a stronger intermolecular force of attraction. Now, the stronger the intermolecular forces of attraction in a gas, the easier it is to liquefy the gas. Therefore, a gas with a higher critical temperature can be liquified easily.

Question 14. If P is plotted against 1/ V for 1 mol of an ideal gas at 0°C, then a straight line passing through the origin is obtained. What is the slope ofthe straight line?
Answer:

For 1 mol of an ideal gas, PV = RT or, P = RT/V At 0°C or 273K, P = 273R/V.

Hence, the slope of the P vsl/V plot is =273 R

Question 15. The number of molecules in an ideal gas with a volume of V at pressure P and temperature T is ‘n Write down the equation of state for this gas
Answer:

No. of mole of the gas

⇒ \(\frac{\text { Total no. of molecules of the gas }}{\text { Avogadro’s no. }}=\frac{n}{N}\)

∴ The equation of state \(P V=\frac{n}{N} R T\)

Question 16. On what factors does the value of the total kinetic energy of the molecules in a gas depend?
Answer:

The total kinetic energy of the molecules of 1 mol gas \(=\frac{3}{2} R T\)

The total kinetic energy of n mole gas \(=n \times \frac{3}{2} R T\)

Therefore, the total kinetic energy of any gas depends upon the temperature and the quantity ofthe gas.

Question 17. At a definite temperature, the total pressure of a gas mixture consisting of three gases A, B and C is P. If the number of moles of A, B and C are 2, 4 and 6, respectively, then arrange these gases in increasing order of their partial pressures
Answer:

The mole fractions of A, B and C in the gas mixture

⇒ \(x_{\mathrm{A}}=\frac{2}{12}=\frac{1}{6}, x_B=\frac{4}{12}=\frac{1}{3} \text { and } x_C=\frac{6}{12}=\frac{1}{2}\)

Therefore The Partial Pressure Of Gas A ⇒  \(A, p_A=\frac{P}{6}, p_B=\frac{P}{3}\)

⇒  \(p_C=\frac{P}{2}\).

Hence PA<PB<PC.

Question 18. Under identical conditions of temperature and pressure, it takes time t1 for the effusion of VmL of N₂ gas through a porous wall and time t₂ for the effusion of the same volume of O₂ gas through the same porous wall. Which one is greater, t₁ or t₂?
Answer:

At constant temperature and pressure, the rates of effusion of different gases are inversely proportional to the square roots of their molar masses. As the molar mass of N₂ is smaller than that of CO₂, at the given conditions the time required for the effusion of VmL of N₂ gas will be less than that required for V mL CO₂ of gas. Hence, t₁<t₂.

Question 19. Arrange the following gases in the increasing order of their densities at STP: H₂, air, CO₂.
Answer:

The density of a gas at constant temperature and pressure \(d=\frac{P M}{R T}.\)

Thus, at a particular temperature and pressure, the density of the gas is directly proportional to the molar mass of the gas. The order of the molar mass of the given gases is: M_H2< M_air < MCO₂.

Therefore, the order of densities of these gases at STP would be, d_H2 < d_air < dCO₂.

Question 20. At a constant temperature, the pressures of four gases A, B, C, and D are 0.2 atm, 250torr, 26.23k Pa, and 14.2 bar, respectively. Arrange them according to their increasing pressure.
Answer:

⇒  \(A: 0.2 \mathrm{~atm}, B: 250 \mathrm{torr}=\frac{250}{760} \mathrm{~atm}=0.33 \mathrm{~atm}\)

⇒ \(A: 0.2 \mathrm{~atm}, B: 250 \mathrm{torr}=\frac{250}{760} \mathrm{~atm}=0.33 \mathrm{~atm}\)

⇒ \(D: 14.2 \mathrm{bar}=\frac{14.2}{1.013}=14.02 \mathrm{~atm}\)

∴ A<C<B<D

Question 21. Among the four quantities—mass, pressure, temperature, and volume, which are taken to be constant in the following gas laws?

• Boyle’slaw
• Charles’law
• Gay-Lussac’slaw
• Avogadro’s law.

Answer:

• Boyle’s law: Mass and temperature ofthe gas
• Charles’ law: Mass and pressure of the gas
• Gay-Lussac’s law: Mass and volume of the gas
• Avogadro’s law: Pressure and temperature of the gas

Question 22. Why does the volume of a given mass of gas increase by decreasing its pressure at constant temperature?
Answer:

According to Boyle’s law, PV= constant for a fixed mass of gas at a constant temperature. That is, the product of P and V for a fixed mass of gas at a constant temperature is always constant. Suppose, P and V are the pressure and volume of a given mass of gas at a constant temperature.

Keeping the temperature constant, if the pressure is made to \(\frac{P}{x}\)(where x > 1 ), then the volume ofthe gas will be V × x because PV = constant

Question 23. Why does the volume of a gas increase by increasing its number of moles at a given temperature and pressure?
Answer:

According to Avogadro’s law, at a given temperature and pressure the volume (V) of a gas is directly proportional to its number of moles (n), i.e., V ∝ n. Therefore, if the number of moles of a gas is increased at a constant temperature and pressure, its volume will increase

Question 24. Will the nature of the following graphical presentations for a given mass of gas be the same?
Answer: 

P vs V atm constant temperature

For a fixed mass of a gas at a given temperature PV = constant. This relation expresses an equation of a rectangular hyperbola. Therefore, the P vs V plot for a fixed mass of gas at a given temperature will produce a curve of a rectangular hyperbola.

V vs T at constant pressure

For a fixed amount of a gas at a constant pressure, V = KxT (K = constant). This relation expresses an equation of a straight line passing through the origin. Therefore, the V vs T plot for a fixed mass of gas at constant pressure will give a straight line passing through the origin.

Question 25. Using the equation of state PV = nBT, show that at a given temperature, the density of a gas is proportional to the gas pressure P.
Answer:

P V = n R T  or, \( P V=\frac{g}{M} R T\)

M= Molar mass of gas in g.mol^-1

g= mass of gas in g]

Or \(P M=\frac{g}{V} R T[latex]

= d R T

d= g/v= density of the gas

Since, B is constant dec P, when T is constant

Question 26. At a low pressure, the van der Waals equation reduces What is the value of [latex]\left(P+\frac{a}{V^2}\right) V=R T\) compressibility factor (Z) for this case at this condition?
Answer:

⇒ \(\left(P+\frac{a}{V^2}\right) V=R T\)

or, \(P V+\frac{a}{V}=R T \quad \text { or, } \frac{P V}{R T}+\frac{a}{R T V}=1 \quad \text { or, } Z=1-\frac{a}{R T V}\)

Question 27. Rank the gases N₂, CO₂, and CH₅ in order of their increasing densities given temperature and pressure.
Answer:

We know \(d=\frac{P M}{R T}\) Thus, docM at a certain temperature and pressure.

Since, \(M_{\mathrm{CH}_4}<M_{\mathrm{N}_2}<M_{\mathrm{CO}_2}\) at a fixed temperature and pressure, \(d_{\mathrm{CH}_4}<d_{\mathrm{N}_2}<d_{\mathrm{CO}_2}\)

Question 28. Under different conditions, the following graph is obtained for an ideal gas. Mentioning A and B,- identify the conditions.
Answer:

For a fixed amount of gas at a given temperature, PV is constant. This means the value of PV will always be constant for a fixed amount of a gas at a given temperature no matter what the pressure of the gas is. Hence PV vs P or PV vs V plot will give a straight line parallel to the P-axis or V-axis, respectively. Therefore, A will be equal to PV and B will be equal to P or V.

For a fixed amount of a gas at a given pressure V – K (constant) x T, i.e., V/T = K. This relation tells us that the value of V/ T will always be constant for a fixed amount of gas at a given pressure irrespective of the value of temperature of the gas. Hence, the V/T vs T plot will give a straight line parallel to the T axis. Hence, A will be equal to V/ T and B will be equal to T.

Question 29. If a substance were to be in a gaseous state at absolute zero temperature, what would be the theoretical value of its pressure?
Answer:

⇒ \(V_t=V_0\left(1+\frac{t}{273}\right)\)

Pressure and mass of gas being constant] At absolute zero temperature, t = -273°C.

⇒ \(\text { Hence, } V_{-273^{\circ} \mathrm{C}}=V_0\left(1-\frac{273}{273}\right)=0 \text {. }\)

Since the volume of the gas is zero (0), the theoretical value of pressure will be zero (0).

Question 30. Under similar conditions of temperature and pressure, the times it takes for the effusion of the same volume of H₂, N₂, and O₂ gases through the same porous wall are t₁ t₂ and t₃, respectively. Arrange t₁, t₂, and t₃ in order of their increasing values.
Answer:

Let the volume of effused gas be the rates of effusion ofthe three gases will be as follows—

⇒ \(\text { For } \mathrm{H}_2: \frac{V}{t_1} \propto \frac{1}{\sqrt{M_{\mathrm{H}_2}}} \cdots[1] ; \text { For } \mathrm{N}_2: \frac{V}{t_2} \propto \frac{1}{\sqrt{M_{\mathrm{N}_2}}}\)

For \(\mathrm{O}_2: \frac{V}{t_3} \propto \frac{1}{\sqrt{M_{\mathrm{O}_2}}}\)

From [1] and [2] we get, t2/ty = \(\sqrt{M_{\mathrm{N}_2} / M_{\mathrm{H}_2}}\) and form and we get \(t_3 / t_2=\sqrt{M_{\mathrm{O}_2} / M_{\mathrm{N}_2}}.\)

Question 31. What will happen if the collisions ofthe gas molecules with each other are not perfectly elastic?
Answer:

In the case of inelastic collisions, the total kinetic energy of gas molecules decreases, leading to a decrease in molecular speeds. As a result, the gas molecules will gradually settle down at the bottom of the container, thereby causing the pressure ofthe gas to go on decreasing gradually. A time will come when the pressure ofthe gas will come duce to zero.

Question 32. At a given temperature, the root mean square velocities of the molecules of gases A, and B are x and ycm. s^-1, respectively. If x is greater than y, then which gas has a larger molar mass?
Answer:

⇒ \(\text { For gas } A: c_{r m s}=\sqrt{\frac{3 R T}{M_A}}=x \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

⇒ \(\text { For gas } B: c_{r m s}=\sqrt{\frac{3 R T}{M_B}}=y \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

∴ \(\frac{x}{y}=\sqrt{\frac{M_B}{M_A}} ;\) Since x> y, MB will be greater than MA.

Question 33. According to the kinetic theory of gases, the average kinetic energies of O₂ and N₂ molecules are the same at a particular temperature. State whether the velocities of the molecules ofthe two gases at a given temperature will be the same or not.
Answer:

The average kinetic energy of the molecules of a gas depends only on the absolute temperature of the gas. It does not depend on the mass of gas molecules. At a given temperature, a lighter gas molecule has the same average kinetic energy as that of a heavier gas molecule.

On the contrary, the root mean square velocity of the molecules of a gas at a given temperature is inversely proportional to the molar mass ofthe gas. Hence, at a given temperature, the root mean square velocities of the molecules of N2 and O₂ gas will not be the same.

Question 34. At a given temperature, the most probable velocity of the molecules of gas A is the same as the average velocity of the molecules of gas B. Which has a larger molar mass
Answer:

Suppose, the molar masses of gases A and B are MA And MB, and the temperature of body gases is 7′ K. Therefore, at this temperature, the most probable velocity of the molecules of gas A

⇒ \(c_m=\sqrt{\frac{2 R T}{M_A}}\) and the average velocity ofthe molecules of gas B, c

⇒  \(\bar{c}=\sqrt{\frac{8 R T}{\pi M_{\mathrm{B}}}}\) As given, cm = c.

∴ \(\sqrt{\frac{2 R T}{M_A}}=\sqrt{\frac{8 R T}{\pi M_{\mathrm{B}}}} \quad \text { or, } \frac{1}{M_A}=\frac{4}{\pi M_{\mathrm{B}}}\)

∴ \(M_B=1.27 M_A\)

Therefore, gas B has a higher molar mass

Question 35. A real gas follows the equation P(V- nb) = nRT under all conditions of temperature and pressure. Show that the compressibility factor of this gas is always greater than one.
Answer:

The equation of state for the gas is: P(V-nb) = nRT The compressibility factor of the gas can be expressed as

⇒ \(P V-P n b=n R T^{\prime} \text { or, } P V=n R T+P n b\)

Or \(\frac{P V}{n R T}=1+\frac{P b}{R T} \text { or, } Z=1+\frac{P b}{R T}\)

As \(\frac{P b}{R T}\) is always positive the value of z will be greater than 1.

Question 36. For H₂ gas: a = = 0.024 L₂ atm.mol^-2 b = 0.026 Lmol^-1 a = 2.28 L₂-atm .mol^-2, b = 0.042 L-mol^-1. and for CH₄ gas:

1. At ordinary temperature and pressure, which one ofthe two gases will behave more like an ideal gas?
2. Which one of the two gases has a larger molecular size?

Answer:

1. At ordinary temperature and pressure, a real gas behaves more like an ideal gas if the values of ‘a’ and ‘b’ are very small. The values of ‘a’ and ‘b ‘ for H₂ gas are smaller than those for CH₄ gas. Obviously, at ordinary temperature and pressure, H₂ gas will behave more like an ideal gas.
2. The value of ‘b’ for a real gas reflects the sizes of molecules of the gas. A gas whose molecules are large has a high value of ‘b’. As the value of CH₄ gas is greater than that of H₂ gas, the size of the CH₄ molecule will be larger than that of the H₂ molecule.

Question 37. When does the effect of molecular volume dominate over
Answer:

The compressibility factor of a real gas becomes Z>1 when the effect of molecular volume dominates over the effect of intermolecular forces of attraction. Again, Z will be greater than one for a real gas if the pressure ofthe gas is very high. Therefore, the effect of molecular volume becomes greater than the effect of intermolecular forces of attraction if the pressure ofthe gas is very high.

Question 38. At ordinary temperature, why can CO₂ but not O₂ gas be liquefied by applying pressure? Give reason.
Answer:

A gas can be liquefied by applying the necessary pressure if its temperature is equal to or below its critical temperature. The critical temperature of CO₂ is above the ordinary temperature (usually 25°C), while that of O₂ gas is well below one ordinary temperature. Hence, CO₂ can be liquefied by applying pressure at ordinary temperature.

Question 39. The critical temperature and the critical pressure of gas are Tc and Pc, respectively. If the gas exists at a temperature of T and a pressure of P, then under which of the following conditions will the gas not be liquefied?

• T> Tc; P>PC
• r=rc; P>PC
• T = Tc; P<PC
• T<TC-P = PC

Answer: Under the conditions of T> Tc and P> Pc, the gas cannot be liquefied because its temperature is above under the conditions of T = Tc and P>PC, it is possible to liquefy the gas. Because the gas Is at its critical temperature and its pressure is above critical pressure.

Under the conditions of T = Tc and P<PC, it is not possible to liquefy the gas as the minimum pressure needed to liquefy a gas at its critical temperature must be equal to Pc or greater than Pc.

Under the conditions of T< Tc and P = Pc, the gas can be liquefied. Because the gas is below its critical temperature and the pressure of the gas is equal to its critical pressure, the minimum pressure required to liquefy a gas at its critical temperature.

Question 40. The van dar Waab constant ‘a’ for CO₂ and CH₄ gases are 3.6 and 2.3 L²-atm-mol^-2 Which one of these two gases can easily be liquefied?
Answer:

The van der Waals constant for a real gas is a measure ofthe strength of intermolecular forces of attraction in the gas. The stronger the intermolecular forces of attraction in a gas, the greater the value. Again, a gas with stronger intermolecular forces of attraction can easily be liquefied. As the value of CO₂ is greater than that of CH₄, it will be easier to liquefy CO₂ gas.

Question 41. The values of van der Waals constants ‘a’ and ‘b’ for X, Y, and Z gases are 6, 6, 20, and 0.025, 0.15, and 0.11, respectively. Which one has the highest critical temperature?
Answer:

The greater the magnitude of intermolecular forces of attraction of a gas, the higher the critical temperature it will have. Among the given gases, the value of a is maximum for gas Z. Thus, the magnitude of effective intermolecular forces of attraction is also maximum for Z, resulting in its higher value of critical temperature.

Question 42. At 20°C, the surface tension of water is three times that of CCI₄ —give reason.
Answer:

The surface tension of a liquid depends on the magnitude of intermolecular forces of attraction. It decreases or increases, respectively, with a decrease or increase in the magnitude of intermolecular forces of attraction. The only attractive forces that exist in carbon tetrachloride are to London fores exists. As the H-bond is stronger than the London force, the surface tension of water is 3 times that of CCl₄ at 20°C

Question 43. At constant pressure, the value of V/T for different quantities of an ideal gas will be different. Is this statement true or false?
Answer:

The equation of state for n moles of ideal gas is: PV=nRT

Therefore \(\frac{V}{T}=\frac{n R}{P}\)

⇒ \(\frac{V}{T}=\frac{n R}{P}\)

When the pressure remains constant, V/ T ∝ n [since R= constant]

Thus, at constant pressure, V/T for different quantities of an ideal gas will be different.

Question 44. A closed container holds a mixture of H₂, SO₂ and CH₄ gases, each with an amount of 0.5 mol. If these gases effuse through a fine orifice in the container, arrange them in the increasing order of their partial pressures once the effusion begins.
Answer:

According to Graham’s law of diffusion, at constant temperature and pressure, the rate of effusion of a gas is inversely proportional to the square root of its molar mass, i.e \(r \propto \frac{1}{\sqrt{M}}.\)

The increasing order of molar masses ofthe given gases is H₂ < MCH₄ < MSO₂– SO₂, and the order of the rate of effusion of these gases at a particular temperature and pressure will be rH₂>rCH₄>rSO₂– Once effusion begins, the order of their number of moles will be nu2 < raCH4< WSO₄ Therefore, the order of their partial pressures will be pH <pCH <PSO₂.

Question 45. The value of the compressibility factor (Z) for a gas at STP is less than 1. What is the molar volume of this gas at STP?
Answer:

We know, \(Z=\frac{V}{V_i}\) = molar volume of a real gas at a certain temperature arid pressure, V₂ – molar volume of an ideal gas at the same temperature and pressure.]

According to the question, Z <1

∴ \(\frac{V}{V_i}<1 \text { or, } V<V_i \text { or, } V<22.4 \mathrm{~L}\)

Since the molar volume of an ideal gas at STP = 22.4 litres] The volume ofthe real gas at STP will be less than 22.4 litres.

Question 46. At 0°C, plots of PV vs P for three real gases A, B and C are given below.

1. Which gas is present above its Boyle temperature? 
2. Which gas can be liquefied more easily?

Answer:

1. The PV vs P plot for gas above its critical temperature does not possess any minimum, and the value of PV increases continuously with pressure from the beginning. So, gas A is present above its Boyle temperature.
2. According to the given figure, the depth of minimum in the PV vs P curve is maximum for the gas C. Hence, the compressibility of gas C is greater than that of either A or B gas. This implies that the forces of attraction between the molecules are stronger in gas C than other two gases. Again, the stronger the intermolecular forces of attraction in a gas, the easier it is to liquefy the gas. So, gas C can be liquefied more easily.

Question 47. Rubber balloon filled with H₂ gas gets deflated after some time—explain why.
Answer:

As rubber is a porous substance, a rubber balloon contains many invisible pores on its surface. The balloon contains H₂ gas at high pressure. But the pressure of air outside the balloon is comparatively lower.

As a result, H₂ gas escapes from the balloon by effusion. This is why the pressure inside the balloon gradually falls and after some time the balloon gets deflated.

Question 48. For a real gas which obeys the van der Waals equation, a graph is obtained by plotting the values of P Vm along the y-axis and the values of P along the x-axis. What is the value of the intercept on the y-axis of the graph?
Answer:

The plot of JPVM against P may give a graph like A or B. In both cases, the graphs intersect the y-axis at P = 0. Now at very low pressure (i.e., P), the van der Waals equation reduces to the ideal gas equation.

∴ PV = nRT

or \(P\left(\frac{V}{n}\right)=R T\)

or, PVm= RT, [Vm= molar volume]

So, the value intercept on the y-axis is RT.

Question 49. The molecular speeds of gas molecules are analogous to the speeds of rifle bullets. Why is the odour of a gas not detected so fast?
Answer:

Gas molecules move almost at the same speed as rifle bullets, but the molecules do not follow a straight line path. Since the molecules collide with each other at a very fast rate, the path becomes zig-zag. Hence, the odour of a gas can not be detected as fast as its molecules move.

Question 50. Assuming the same pressure in each case, calculate the mass of hydrogen required to inflate a balloon to a certain volume at 100°C if 3.5g He is required to inflate the balloon to half the volume at 25°C.
Answer:

Volume of 3.5gHe at 25°C and pressure P is \(V=\frac{n R T}{P}=\frac{3.5}{4} \times \frac{R T}{P}=\frac{3.5 \times 298 R}{4 P}\)

To fill the balloon with H₂ at 373 K, the volume of Hg gas required, VH = 2V.

Hence \(2 V=\frac{n R T}{P}=\frac{w}{2} \times \frac{R \times 373}{P}\)

Dividing [2] by [1] gives \(2=\frac{w \times 373}{2} \times \frac{4}{3.5 \times 298} \quad \text { or, } w=2.796 \mathrm{~g}\)

Question 51. The given figure indicates the plot of vapour pressure vs. temperature for the three liquids, A, B & C. Arrange them in the increasing order of their intermolecular forces of attraction and normal boiling points.
Answer:

According to the given plot, at a particular temperature, the vapour pressure of A is higher than that of B, which in turn is higher than C. Now, a liquid with weak intermolecular forces of attraction has high vapour pressure. Therefore, the order of intermolecular forces of attraction of the given liquids will be A < B < C.

The lower the vapour pressure of a liquid, the higher its normal boiling point. Alternatively, the higher the vapour pressure of a liquid, the lower its normal boiling point. Therefore, the order of normal boiling points of the given liquids will be: A < B < C

Question 52. Water spreads on a glass surface but It forms beads on a glass surface polished by paraffin—why?
Answer:

Adhesive forces between the molecules of glass and water are stronger than the cohesive forces between the water molecules. For this reason, water can spread on the glass surface.

On the other hand, adhesive forces between the molecules of paraffin (non-polar) and water are weaker than cohesive forces between the water molecules. For this reason, water forms small beads on a glass surface of polished paraffin.

Question 53. 4 gas-mixture consists of two gases, A and B, each with equal mass. The molar mass of B is greater than that of 4. Which one of the two gases will contribute more to the total pressure of the gas mixture?
Answer:

In a gas mixture, the component gas with higher partial pressure will have a greater contribution to the total pressure of the mixture. Masses of A and B in the mixture are the same but the molar mass of B is greater than that of A. Hence, in the mixture, the number of moles or the mole fraction (xA) of A will be greater than that (xB) of B. Suppose, in the mixture, the partial pressures of A and B are pA and pB respectively and the total pressure is P. According to Dalton’s law of partial pressures, pA = xAP and pB = xgP.

As xA>xB, pA will be greater than pB. Hence, the contribution of gas A to the total pressure of the mixture will be more than that of gas B.

Question 54. Under the same conditions of temperature and pressure, the rate of diffusion of hydrogen gas is four times that of oxygen gas—explain
Answer:

At constant temperature and pressure, rates of diffusion (r) of different gases are inversely proportional to the square roots of their molecular masses (M) \(r \propto \frac{1}{\sqrt{M}}\)

⇒ \(\frac{r_{\mathrm{H}_2}}{r_{\mathrm{O}_2}}=\frac{\sqrt{M_{\mathrm{O}_2}}}{\sqrt{M_{\mathrm{H}_2}}}=\frac{\sqrt{32}}{\sqrt{2}}=4 \text { or, } r_{\mathrm{H}_2}=4 \times r_{\mathrm{O}_2}\)

So, under the same conditions of temperature and pressure, the rate of diffusion of H2 gas is four times that of O₂ gas.

Question 55. Four tyres of a motor car were filled with nitrogen, hydrogen, helium and air. In which order are these tyres to be filled with the respective gases again
Answer:

According to Graham’s law of diffusion, under the conditions of temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

The order of molar masses of the given gases is:

⇒ \(M_{\mathrm{H}_2}<M_{\mathrm{He}}<M_{\mathrm{N}_2}<M_{\text {air }}\)

At the same temperature and pressure, the order of rates of diffusion of these gases would be—rH > rHe > rN > air After a certain time, the order of decrease in pressure in these three tyres tyre (air) < tyre (N₂) < tyre (He) < tyre (H₂). Hence, tyres are to be filled again with the respective gases in the following order, tyre(H₂) tyre(He), tyre(N₂), and tyre(air).

Question 56. When a football is pumped, both the volume and pressure of the gas inside it increase. Does this observation contradict Boyle’s law?
Answer:

According to Boyle’s law, at a constant temperature, the volume of a given mass of a gas is inversely proportional to its pressure. When the football is pumped, the quantity of air inside the football goes on increasing. As a result, the mass of air inside the football does not remain constant. Moreover, 2 pumping causes a rise in the temperature of air inside the football. Thus, neither the temperature nor the mass of the air remains constant. So, Boyle’s law is not applicable in this case.

Question 57. The molar mass of UF₆ is 176 times as high as that of H₂, yet at a particular temperature, the average kinetic energy of both is found to be the same—why?
Answer:

According to the kinetic theory of gas, the average kinetic energy of the molecules in a gas is directly proportional to the absolute temperature of the gas, and it does not depend upon the molar mass of the gas.

The average kinetic energy of a molecule of a gas at temperature TK is given by

⇒  \(\frac{3}{2} k T\) is Boltzmann constant].

Since the value of is constant at a given temperature and is independent of the mass of the gas molecule,

⇒   \(\frac{3}{2} k T\)

The average kinetic energy of the molecule of a heavier gas will be the same as that of the molecule of a lighter gas. Thus, at a given temperature, the average kinetic energy of a UF₆ molecule will be the same as that of a H₂ molecule.

Question 58. Two gases, obeying the van der Waals equation, have identical values of ‘b’ but different values of ‘a’. Which one of the two gases will occupy less volume under identical conditions? If the values of ‘a’ for the two gases are the same but the values of‘V are different, then under identical conditions which gas will be more compressible?
Answer:

The larger the value of ’a’ of a gas, the stronger the intermolecular forces of attraction in the gas. So, under identical conditions, a gas with a larger value of ‘a’ will be more compressible than that with a smaller value. Hence, between the two gases, the one with a higher value of ‘a’ will occupy less volume under identical conditions.

If the value of ‘a’ for two gases is the same the values of ‘b’ differ, then the gas with a smaller value of ‘b’ will be more compressible because a small value of ‘b’ for a gas signifies that the volume occupied by the molecules of the gas is small. So, this gas can be compressed to a greater extent.

Question 59. Under what conditions can a gas be liquefied?

• T = Tc and P < Pc
• T<TC and P – Pc

Answer:

It is possible to liquefy a gas by the application of pressure, provided that the temperature of the gas is equal to or less than its critical temperature. If the temperature (T) of a gas is equal to its critical temperature ( Tc), then it is possible to liquefy the gas if the pressure (P) of the gas is equal to or above its critical pressure (Pc).

So, under the condition, the gas cannot be condensed into a liquid. On the other hand, if the temperature (T) of a gas is below its critical temperature (Tc), then the gas can be transformed into liquid if the applied pressure on the gas is greater than or less than or equal to its critical pressure. So, under certain conditions, the gas can be liquefied.

Question 60. Derive the van der Waals equation for ‘n’ mol of a real gas from the equation for 1 mol of the gas.
Answer:

Van der Waals equation for 1 mol of real gas is given by:

⇒ \(\left(P+\frac{a}{v^2}\right)(\nu-b)=R T\) m……………………….(1)

If at a pressure P and a temperature T, the volume of ‘n’ moles of this gas is V, then

v = \(\frac{V}{n}\) Putting  v = \(\frac{V}{n}\) in equation n n [1], we have

⇒ \(\left(P+\frac{a}{\left(\frac{V}{n}\right)^2}\right)\left(\frac{V}{n}-b\right)=R T \text { or }\left(P+\frac{n^2 a}{V^2}\right)(V-n b)=n R T\)

Question 61. Between methanol (CH₃OH) and water (H₂O) whose surface tension is greater, and why?
Answer:

Molecules in methanol and water both are capable of forming hydrogen bonds. However, the number of hydrogen bonds per molecule in water is greater than that in methanol. So, the intermolecular forces of attraction are stronger in water than those in methanol. Again, the stronger the intermolecular forces of attraction in a liquid, the greater the surface tension of the liquid. So, the surface tension of water is greater than that of methanol.

Question 62. A liquid has a high normal boiling point. Will its viscosity and surface tension values be high or low?
Answer:

The high normal boiling point of a liquid indicates that the liquid possesses strong intermolecular forces of attraction. The values of viscosity and surface tension of a liquid depend upon the strength of intermolecular attractive forces in the liquid. The stronger the intermolecular attractive forces, the higher the values of viscosity and surface tension of the liquid. As the intermolecular forces of attraction are strong for the concerned liquid, the values of both surface tension and viscosity of the liquid will be high.

Question 63. How is the vapour pressure of a liquid affected if the surface area of the liquid is increased at a given temperature?
Answer:

• At a given temperature, the vapour pressure of a liquid does not depend upon the surface area of the liquid. If the surface area of a liquid is increased, keeping the temperature constant, then its rate of evaporation increases because at a particular time, more molecules are now able to leave the liquid surface and go to the vapour phase.
• The increased surface area of the liquid also causes the increase in the die rate of condensation because, at a particular time, more molecules are now able to reenter from vapour to the liquid phase.
• Therefore, the rates of both evaporation and condensation are increased to the same extent So, the vapour pressure remains the same at a constant temperature and is not affected by the surface area of the liquid

Question 64. Compare the viscosity coefficients of the following liquids at a particular temperature: Propanol (CH₃CH₂CH₂OH), ethylene glycol (HOCH₂—CH₂OH) and glycerol (HOCH₂—CHOH—CH₂OH).
Answer:

The number of OH groups in the molecules of CH₃CH₂CH₂OH, HOCH,—CH₂OH HOCH₂—CHOH—CH₂OH are 1, 2 and 3, respectively. So, the number of hydrogen bonds formed per molecule of CH₃CH₂CH₂OH HOCH₂—CH₂OH and HOCH₂ —CHOH —CH₂OH are 1, 2 and 3, respectively. Thus, the order of the intermolecular forces of attraction of the given liquids will be propanol < ethylene glycol < glycerol.

Again, the stronger the intermolecular forces of attraction of a liquid, the higher the value of its viscosity coefficient. Thus, the order of viscosity is propanol < ethylene glycol < glycerol.

Question 65. Find out the minimum pressure required to compress 2× M 500 dm³ of air at bar to 200 dm1 at 30nC
Answer:

In this process, the mass and temperature of the gas remain constant but the volume and pressure of the gas change.

Given P₁=1 bar, V₁= 500 dm³, v₂= 200 dm³.

Appyling Boyle’s law , pe \(P_2=\frac{P_1 V_1}{V_2}=\frac{1 \times 500}{200}=2.5 \text { bar }\)

So, the minimum pressure needed to V₂ compress 200 the gas is 2.5 bar.

Question 66. A vessel of 120 mi. capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 1 HO ml. at 35°C. What would be Its pressure?
Answer:

In this process, the pressure and volume ofthe gas change but its mass and temperature remain constant.

Applying Boyle’s law to the process, we have \(P_2=\frac{P_1 V_1}{V_2}\)

Given: p₁= 1.2 bar, V₁ = 120ml and V₂ = 180 ml

∴ \(p_2=\frac{1.2 \times 120}{180}=0.8 \text { bar }\)

Question 67. A real gas follows van der Waals equation. Find the compressibility for 1 mol of the gas at its critical temperature
Answer:

At the critical point

⇒ \(V=V_c=3 b ; P=P_c=\frac{a}{27 b^2} \text { and } T=T_c=\frac{8 a}{27 R b}\)

∴ \(Z=\frac{P V}{R T}=\frac{\frac{a}{27 b^2} \times 3 b}{R \times \frac{8 a}{27 R b}}=\frac{3}{8}\)

NCERT Solutions For Class 11 Chemistry Chapter 5 States Of Matter Gases And Liquids Warm-Up Exercise Question And Answers

Question 1. Under which of the following conditions will the density of a, fixed mass of SO₂ gas be higher?— STP 27°C and 3atm pressure.
Answer:

d = \(\frac{P M}{R T}\) and M is constant for a particular gas. Hence ∝PIT. The value of PIT at 3atm and 27°C is greater than that at STP.

Therefore, the density of SO₂ at 3atm and 27°C will be greater than that at STP.

Question 2. Determine the SI unit of
Answer:

⇒ \(\frac{P V^2 T^2}{n}=\frac{\mathrm{N} \times\left(\mathrm{m}^3\right)^2 \times \mathrm{K}^2}{\mathrm{~m}^2 \times \mathrm{mol}}=\mathrm{N} \cdot \mathrm{m}^4 \cdot \mathrm{K}^2 \cdot \mathrm{mol}^{-1}\)

Question 3. A gas mixture consists of three gases A, B, and C with the number of moles 1, 2, and 4, respectively. Which of these gases will have a maximum partial pressure if the total pressure ofthe mixture is Pata given temperature t what temperature will the average velocity of O₂ molecules be equal to that of U₂ molecules at 20K?
Answer:

In the mixture, the total number of moles of the constituent gases =(1 + 2 + 4) = 7mol.

The mole fractions of A, B, and C

⇒  \(\frac{1}{7}, \frac{2}{7} \text { and } \frac{4}{7} \text {, }\) respectively.

Since the mole fraction of C is the highest, its partial pressure will be the highest in the mixture.

Question 4. Rank Cl₂, SO₂, CO₂, and CH₄ gases in Increasing order of their rates of diffusion under identical set of conditions.
Answer:

The molar mass of the gases follows the order: \(M_{\mathrm{CH}_4}<M_{\mathrm{CO}_2}<M_{\mathrm{SO}_2}<M_{\mathrm{Cl}_2}\)

Hence, at constant temperature and pressure, the rates of diffusion will be the order: \(r_{\mathrm{Cl}_2}<r_{\mathrm{SO}_2}<r_{\mathrm{CO}_2}<r_{\mathrm{CH}_4}\)

Question 5. A closed vessel holds a gas mixture consisting of C₂H₆, C₂H₄, and CH₄, each with an amount of 2.5 mol. However, due to a pinhole in the vessel, the gas mixture undergoes effusion. What will be the order of partial pressures of the gases in the vessel after some time?
Answer:

Molar masses of C₂H₆, C₂H₄ and CH₄ follow the order:

⇒ \(M_{\mathrm{CH}_4}<M_{\mathrm{C}_2 \mathrm{H}_4}<M_{\mathrm{C}_2 \mathrm{H}_6}\)

Thus, at a certain temperature and pressure, their rates of effusion will be in the order \(r_{\mathrm{C}_2 \mathrm{H}_6}<r_{\mathrm{C}_2 \mathrm{H}_4}<r_{\mathrm{CH}_4}\)

Therefore, the order of partial pressure after some time will be:

⇒ \(p_{\mathrm{CH}_4}<p_{\mathrm{C}_2 \mathrm{H}_4}<p_{\mathrm{C}_2 \mathrm{H}_6}\)

Question 6. At constant temperature and pressure volume of an ideal gas (molecular mass 28 g. mol^-1) is 23.36 times greater than its mole number. Find out its density at the same temperature and pressure.
Answer:

Let us suppose, the tire volume and density of n mol of an ideal gas are V L and d g.L^-1at P atm and T K.

Now, PV = nRT and d \(=\frac{P M}{R T}\)i.e., \(d=M \times \frac{n}{V}\)

Question 7. Why cannot CO₂ gas be liquefied above 31.1°C?
Answer:

The critical temperature of CO₂ is 31.1°C. Above 31.1°C, due to the very high average kinetic energy of CO₂ molecules, the attraction between them becomes negligible. As a result, CO₂ gas cannot be liquefied above 31.1°C.

Question 8. Under what conditions will the value of- always be the same irrespective ofthe value of T?
Answer:

For a fixed mass of gas at constant volume, P ∝ T or, P = f Cx T (K = constant). Therefore, P/T = K. This relation indicates that the value of P/T is always constant for a fixed mass of a gas at constant volume irrespective of the value of T is

Question 9. Under what conditions will the value of PV always be the same irrespective of the value of P (or V)?
Answer:

According to Boyle’s law, for a fixed mass of gas at a constant temperature, PV = constant. Therefore, the value of PV is always constant for a fixed mass of a gas at a constant temperature irrespective ofthe value of P

Question 10. 1 mol of N₂ & 3 mol of O₂ are kept In two different containers with a volume of V at a fixed temperature. Compare—(1) the average kinetic energy and (11) the total kinetic energy ofthe molecules.
Answer:

The average kinetic energy of gas molecules depends only on the absolute temperature ofthe gas. Since both gases are at the same temperature, they will have equal average kinetic energy.

At T K, the total kinetic energy of nmol gas molecules

⇒ \(n \times \frac{3}{2} R T\) Hence, at T K, the total kinetic energy of 3mol O₂ molecules is 3 times that of 1 mol N₂ molecules.

Question 11. For a real gas, the van der Waals constant ‘a’ is zero. Can the gas be liquefied? Explain.
Answer:

The van der Waals constant ‘a’ measures the magnitude of intermolecular forces of attraction in a gas. Hence, a real gas with ‘a’= 0 signifies that there are no intermolecular attractive forces in the gas. Consequently, such a gas cannot be liquefied.

Question 12. State Gay Lussac’s law related to the pressure and temperature of a gas. 3.2 g of sulphur when vaporised, the sulphur vapour occupies a volume of 280.2 mL at STP. Determine the molecular formula of sulphur vapour under this condition. (S = 32)
Answer:

280.2 mL of sulphur weighs 3.2 g at STP. 22400 mL of sulphur weighs 255.8 g at STP. Let, the molecular formula of sulphur Sn. So, n × 32 = 255.8 or, n = 7.9 ~ 8 Molecular formula ofsulphur is Sg.

Question 13. At a constant temperature, a container of fixed volume holds NH3 and HCl gases. Can Dalton’s law of partial pressures be applied to this gas mixture?
Answer:

Dalton’s law of partial pressure applies only to a mixture composed of two or more non-reacting gases. NH₃ and HCl gases react together to produce NH₄Cl. So Dalton’s law of partial pressure will not be applicable in this case.

Question 14. At tC and t2°C, the values of viscosity coefficients of a liquid are x poise, and y poise respectively. If x>y, then which one is higher, t₁ or t₂?
Answer:

With the increase in temperature, the viscosity of a liquid decreases. Now, viscosity directly varies with the value of viscosity coefficients As the viscosity coefficient of the liquid at t2°C is smaller than that at t₁°C, t₂ > t₁.

Question 15. The equation of state of a real gas is P(V-b) = RT. Can the gas be liquified? Explain. Sketchlog P vlog V graph for a given mass of an ideal gas at constant temperature and indicate the slope.
Answer:

Value of van der Waals constant V= 0 for the given gas. Hence, there exists no force of attraction among the gas molecules. So, the gas cannot be liquefied.

Question 16. Determine the types of intermodular forces of attraction in the following instances. w-hexane, SO₂,CO₂, CHCI₃, (CH₃)₂CO, (CH₃)₂O
Answer:

London forces or instantaneous induced dipole instantaneous induced dipole attraction: n-hexane, CO₂, Dipole-dipole attraction: SO₂ CHCI₃, (CH₃)₂O. However, the London force also acts in this case.

Question 17. Which type of intermodular forces of attraction act between O₂ and water modules when O₂ is dissolved in water?
Answer:

O₂ is a non-polar molecule, whereas the H₂O molecule is polar. Hence, the force of attraction acting between O₂ and H₂O molecules is dipole-induced dipole attraction

Question 18. The critical temperatures of H₂, NH₃, and CO₂ gases are 5K, 405K, and 304K, respectively. Arrange them in the increasing order of their intermolecular forces of attraction.
Answer:

A gas with high critical temperature possesses strong intermolecular forces of attraction. The order of critical temperatures of the gases H2, NH₃, and CO₂ is NH₃ > CO₂ > H₂. So, the increasing order of their strength of intermolecular forces will be H₂<CO₂<NH₃.

Question 19. The critical temperatures of NH₃ and SO₂ gases are 405.0K and 430.3K, respectively. For which gas is the value of a der Waals constant greater, and why?
Answer:

The higher the critical temperature of a gas, the stronger its intermolecular forces of attraction, and hence the larger the value of the van der Waals constant the gas has. Thus, between NH₃ and SO₂, the value of a will be larger for SO₂ because its critical temperature is higher than that of NH₃.

Question 20. The critical temperatures of NH₃, CO₂, and O₂ gases are 405.6K, 304.1K, and 154.2K, respectively. If the gases are cooled from 500K to their respective critical temperatures, then which gas will be liquefied first?
Answer:

If the given gases are cooled from 500K, NH₃ gets liquefied first (critical temperature 405.6K). The reason is that the critical temperatures of CO₂ and O₂ are lower than that of NH₃. As a result, the liquefaction of either CO₂ or O₂ is not possible at the critical temperature of NH₃.

Question 21. At 20°C the increasing order of viscosity of acetic acid, acetone, and methanol is: acetone < methanol < acetic acid. Arrange the liquids according to their increasing intermolecular attractive forces.
Answer:

With the increase or decrease in the intermolecular force of attraction of a liquid, the value of viscosity confidence of the liquid increases or decreases. At 20°C, the increasing order of the values of the viscosity coefficient of the given liquids is acetone < methanol < acetic add. Thus, the order of increasing the intermolecular attractive force of these liquids is acetone < methanol < acetic add.

Question 22. Why does the surface energy increase on the dispersion of a liquid? large water drop into smaller droplets?
Answer:

When a large water drop disperses into smaller water droplets, the total surface area ofthe small water droplets becomes greater than the surface area of the large water drop. As the surface energy increases with an increase in surface area, the dispersion of a large water drop into smaller droplets leads to an increase in surface.

Question 23. \(c_{r m s}=\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 P V}{M}} .\) According to this equation, with the increase in pressure or volume, the value of arms increases. Justify this statement.
Answer:

The rms velocity of the molecules of a gas depends only on the temperature and molar mass of the gas. Its value does not depend on the pressure or volume of the gas because, at a constant temperature, PV for a given mass of gas always remains constant, irrespective of the values of pressure or volume.

Question 24. The volumes, energy numbers of the two ideal molecules gases A and and B average the same. What is the relationship between the pressures of these two gases?
Answer:

The average kinetic energy of gas molecules depends only on the temperature ofthe gas. As the molecules of A and B gases have the same values of average kinetic energy, both the gases have die same absolute temperature.

Since the volume, the number of molecules and the temperature of both gases are identical, the pressure will be the same.

Question 25. Any real gas behaves ideally at very low pressure and high temperature. Explain. The values of van der Waals constant ‘ a’ for N₂ and NH₃ are 1.37 and 4.30 L² -atm- mol¯² respectively. Explain the difference in values.
Answer:

Van der Waals constant ‘a’ denotes the magnitude of attractive forces between the molecules of real gases. The value of ‘a’ is higher in the case of NH₃ than N₂. That means attractive forces (London force, dipole-dipole attraction force) present among the molecules of NH₃ are stronger than those of N₂ (London force). Consequently, NH₃ can be liquefied more easily than N₂.

Question 26. Calculate the temperature of 4.0 mol of gas occupying 5 dm³ at 3.32 bar. (R = 0.083 bar. dm³– K^-1. mol^-1)
Answer:

To calculate the temperature ofthe gas, we apply the gas equation, PV = nRT. As given: P = 3.32 bar, V = 5 dm3 and n = 4 mol Putting these values into the equation, PV = nRT, we have

⇒ \(T=\frac{P V}{n R}=\frac{3.32 \times 5}{4 \times 0.083} \mathrm{~K}=50 \mathrm{~K}\)

Question 27. Calculate the total number of electrons present In 1.4 g of dinitrogen gas.
Answer:

⇒ \(1.4 \mathrm{~g} \text { of } \mathrm{N}_2=\frac{1.4}{28}=0.05 \mathrm{~mol} \text { of } \mathrm{N}_2\)

1 molecule of N₂ contains 14 electrons, Hence, 0.05 mol of N2, i.e., 0.05 × 0.022 × 10²³ molecules of = 4,2154 × 10²³ electrons. N2 contain 14 × 0,05 × 0,023 × 10²³

Question 28. When does the graph showing variation ofthe volume ofa given mass of gas with pressure at a constant temperature become linear?
Answer:

For a given mass of gas at a fixed temperature, PV = K (constant). That is, \(P=\frac{K}{V}\) This relation expresses an equation of a straight line passing through the origin. Therefore, if P is plotted against \(\frac{1}{V}\) for a given mass of a gas at a fixed temperature, a straight line will be obtained

Question 29. N₂ gas is present in a 1L desiccator at latm pressure. The pressure ofthe gas decreases to 78mmHg pressure when the desiccator is partially evacuated using a vacuum pump at a constant temperature. Find out the final volume of the gas.
Answer:

Since the volume of the desiccator is fixed, the final volume of the gas will be 1L even after the desiccator is partially evacuated. In this process, the number of moles of the gas decreases but its volume and temperature remain the same. As the pressure ofthe gas is reduced

Question 30. According to Boyle’s law, at a constant temperature, the volume ofa given mass of gas is inversely proportional to its pressure. But when a balloon is filled with air, both the volume and pressure ofthe gas inside it increase—Explain.
Answer:

When the balloon is pumped, the quantity of air inside the balloon goes on increasing. As a result, the mass of air inside it does not remain constant. Moreover, pumping causes a rise in the temperature of the air inside the balloon. Thus, neither the temperature nor the mass of the air remains constant. So, Boyle’s law is not applicable in this case.

NCERT Solutions For Class 11 Chemistry Chapter 5 States Of Matter Gases And Liquids Multiple Choice Questions

Question 1. Equal weight of CH₄ and H₂ are mixed in an empty container at 25°C. The fraction of the total pressure exerted by H₂ is

1. \(\frac{1}{9}\)
2. \(\frac{1}{2}\)
3. \(\frac{8}{9}\)
4. \(\frac{16}{17}\)

Answer: 3. \(\frac{8}{9}\)

The gas mixture contains equal masses of CH and H₂. Suppose, the mass of each of these gases = u>g so, in the mixture

⇒ \(x_{\mathrm{H}_2}=\frac{w / 2}{\frac{w}{2}+\frac{w}{16}}=\frac{8}{9} \text { and } x_{\mathrm{CH}_4}=1-x_{\mathrm{H}_2}=1-\frac{8}{9}=\frac{1}{9}\)

∴ The partial pressure of \(\mathrm{H}_2, p_{\mathrm{H}_2}=x_{\mathrm{H}_2} \times P=\frac{8}{9} P\) [p= total pressure of the mixture]

∴ \(\frac{p_{\mathrm{H}_2}}{P}=\frac{8}{9}\)

Question 2. Avan der Waals gas may behave ideally when

1. Volume is very low
2. Temperature is very high
3. The pressure is very
4. The temperature, pressure, and volume all are very high

Answer: 3. The pressure is very

A van der Waals gas behaves ideally when its temperature is very high or pressure is very low. At either of these two conditions, the volume of the gas becomes very large, which results in a large separation of gas molecules.

Thus, intermolecular forces of attraction become negligible, and gas behaves approximately like an ideal gas.

Question 3. Two gases X (mol. wt. Mx) and Y (mol. wt. My; My> Mx) are at the same temperature, in two different containers. Their root mean square velocities are Cx and CY respectively. If average kinetic energies per molecule of two gases X and Y are Ex and Ey respectively, then which of the following relation(s) is (are) true

1. Ex>EY
2. CX>CY
3. \(E_X=E_Y=\frac{3}{2} R T\)
4. \(E_X=E_Y=\frac{3}{2} k_B T\)

Answer: 3. \(E_X=E_Y=\frac{3}{2} R T\)

For 1 mol of a gas, van der Waals equation: \(\left(P+\frac{a}{V^2}\right)(v-b)=R T\) If the value of is negligible, then \(P+\frac{a}{V^2} \approx P\)

∴ P(V-b) = RT or, PV = RT + Pb \(\text { or, } \frac{P V}{R T}=1+\frac{P b}{R T} \quad \text { or, } Z=1+\frac{P b}{R T}\)

Question 4. The compressibility factor (Z) of one mole of a van der Waals gas of negligible ‘a’ value is-

1. 1
2. \(\frac{b P}{R T}\)
3. \(1+\frac{b P}{R T}\)
4. \(1-\frac{b P}{R T}\)

Answer: 3. \(1+\frac{b P}{R T}\)

For 1 mol of a gas, van der Waals equation: \(\left(p+\frac{a}{V^2}\right)(v-b)=R T\)

If the value of ‘a’ is negligible, the \(P+\frac{a}{V^2} \approx P\)

∴ P(V-b) = RT

⇒ \(\text { or, } P V=R T+P b \quad \text { or, } \frac{P V}{R T}=1+\frac{P b}{R T} \quad \text { or, } Z=1+\frac{P b}{R T}\)

Question 5. For one mole of an ideal gas, the slope of the V vs. T curve at a constant pressure of 2 atm is XL-moH-K¯¹. The value of the ideal universal gas constant ‘R’ in terms of X is

1. \(X \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)
2. \(\frac{X}{2} \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)
3. \(2 X \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)
4. \(2 X \mathrm{~atm} \cdot \mathrm{L}^{-1} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

Answer: 3. \(2 X \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

⇒ \(P V=n R T \text { or, } P\left(\frac{V}{n}\right)=R T\)

or,PVm = RT [Vm = molar volume \(\text { or, } V_m=\frac{R}{P} T\)

At constant pressure, \(\frac{R}{P}\) = constant = K. So, at constant pressure, for 1 mol of an ideal gas, Vm = KT. This relation represents a straight-line equation passing through the origin. So, for 1 mol of an ideal gas at constant pressure, the graph of Vm vs. Twill be a straight line with slope = K.

Given,K= XL.mol¯¹.K¯¹.

⇒ \(\text { or, } \frac{R}{P}=X \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \quad \text { or, } \frac{R}{2 \mathrm{~atm}}=X \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

∴ R = 2XL-atm-mol-1.K-1

Question 6. At a certain temperature, the time required for the complete diffusion of 200 mL of H₂ gas is 30 minutes. The time required for the complete diffusion of 50 mL of O₂ gas at the same temperature will be

1. 60 mins.
2. 30 mins.
3. 45 mins.
4. 15 mins.

Answer: 2. 30 mins.

According to Graham’s law \(\frac{V_{\mathrm{H}_2} / t_1}{V_{\mathrm{O}_2} / t_2}=\sqrt{M_{\mathrm{O}_2} / M_{\mathrm{H}_2}}\)

⇒ \(\text { or, } \frac{V_{\mathrm{H}_2}}{V_{\mathrm{O}_2}} \times \frac{t_2}{t_1}=\sqrt{\frac{M_{\mathrm{O}_2}}{M_{\mathrm{H}_2}}} \text { or, } \frac{200}{50} \times \frac{t_2}{30 \mathrm{~min}}=\sqrt{\frac{32}{2}}=4\)

∴ T₂ = 30 min

Question 7. Four gases P, Q, R, and S have almost the same values but their ‘a! values (a, b are van der Waals constants) are in the order Q<R<S<P. At particular temperatures, among the four gases, the most easily liquefiable one is

1. P
2. Q
3. R
4. S

Answer: 1. P

The van der Waals constant ‘a’ of a gas is a measure of the intermolecular forces of attraction in the gas. The larger the value of ‘a’, the stronger the intermolecular forces of attraction. Now, a gas with strong intermolecular forces of attraction can easily be liquefied. So, the most easily liquefiable gas is P.

Question 8. Units of surface tension and viscosity are

1. \(\mathrm{kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-1}, \mathrm{~N} \cdot \mathrm{m}^{-1}\)
2. \(\mathrm{kg} \cdot \mathrm{s}^{-2}, \mathrm{~kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-1}\)
3. \(\mathrm{N} \cdot \mathrm{m}^{-1}, \mathrm{~kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-2}\)
4. \(\mathrm{kg} \cdot \mathrm{s}^{-1}, \mathrm{~kg} \cdot \mathrm{m}^{-2} \cdot \mathrm{s}^{-1}\)

Answer: 2. \(\mathrm{kg} \cdot \mathrm{s}^{-2}, \mathrm{~kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-1}\)

⇒ \(\text { Surface tension }=\frac{\text { Force }}{\text { Length }}=\frac{\mathrm{N}}{\mathrm{m}}=\frac{(\mathrm{kg} \cdot \mathrm{m} \cdot \mathrm{s})^{-2}}{\mathrm{~m}}=\mathrm{kg} \cdot \mathrm{s}^{-2} \text {. }\)

Coefficient of viscosity =  \(\mathrm{N} \cdot \mathrm{m}^{-2} \cdot \mathrm{s}=\mathrm{kg} \cdot \mathrm{m} \mathrm{s}^{-2} \cdot \mathrm{m}^{-2} \cdot \mathrm{s}\)

Question 9. A gas can be liquefied at temperature T and pressure P if-

1. T= TC, P<PC
2. T<TC,P>PC
3. T>TC,P>PC
4. T>TC,P<PC

Answer: 2. Two important conditions for liquefying a gas are— temperature should be lower than critical temperature (T<TC) and pressure should be greater than critical pressure (P > Pc).

Question 10. The rms velocity of CO₂ gas molecules at 27°C is approximately 1000 m/s. For N₂ molecules at 600K the rms velocity approximately

1. 2000 m/s
2. 1414 m/s
3. 1000 m/s
4. 1500 m/s

Answer: 3. 1000 m/s

⇒ \(c_{r m s}=\sqrt{\frac{3 R T}{M}} \text {, so, } \frac{c_{r m s}(\mathrm{CO})}{c_{rms}\left(\mathrm{~N}_2\right)}=\sqrt{\frac{3 R T_{300}}{M_{\mathrm{CO}}} \times \frac{M_{N_2}}{3 R T_{600}}}\)

Question 11. Among tire following which should have the highest rms speed at the same temperature

1. SO₂
2. CO₂
3. O₂
4. H₂

Answer: 4. H₂

⇒ \(c_{r m s}=\sqrt{\frac{3 R T}{M}} \quad \text { or, } c_{r m s} \propto \frac{1}{\sqrt{M}}\)

Question 12. Which of the following has the dimension of ML⁰T^-2 

1. Coefficient of viscosity
2. Surface tension
3. Vapour pressure
4. Kinetic energy

Answer: 2. Surface tension \(\gamma=\frac{\text { force }}{\text { length }}\)

⇒ \(\text { or, } \gamma=\frac{\text { mass } \times \text { acceleration }}{\text { length }}\)

⇒ \(=\frac{\mathrm{M} \times \mathrm{LT}^{-2}}{\mathrm{~L}}=\mathrm{ML}^0 \mathrm{~T}^{-2}\)

Question 13. For the same mass of two different ideal gases molecular weights M₁ and M₂, plots of logV vs logP at a given constant temperature are shown.

Identify the correct option-

1. M₁>M₂
2. M₁=M₂
3. M₁<M₂
4. Can be predicted only if the temperature is known

Answer: 1. M₁>M₂ From the ideal gas equation

P V = n R T = \(frac{W}{M} R T\)

Or,PV= \(\frac{k}{M}\)

Where k = WRT

Or, \(\log P+\log V=\log \frac{k}{M}\)

Or,  \(\log V=-\log P+\log \frac{\kappa}{M}(y=m x+c)\)

According to the intercepts in the graph

⇒ \(\log \frac{k}{M_2}>\log \frac{k}{M_1}\) \(\text { or, } \frac{k}{M_2}>\frac{k}{M_1} \quad \text { or, } M_1>M_2\)

Question 14. Equal weights of ethane and hydrogen are mixed in an empty container at 25°C. The fraction of total pressure exerted by hydrogen is

1. 1: 2
2. 1:1
3. 1:16
4. 15:16

Answer: 4. 15:16

Let, wc₂H₆ = wH₂ = w

⇒ \(n_{\mathrm{C}_2 \mathrm{H}_6}=\frac{w}{30} \text { and } n_{\mathrm{H}_2}=\frac{w}{2}\)

∴ \(n_{\mathrm{C}_2 \mathrm{H}_6}=\frac{w}{30} \text {. and } n_{\mathrm{H}_2}=\frac{w}{2}\)

∴ \(x_{\mathrm{C}_2 \mathrm{H}_6}=\frac{n_{\mathrm{C}_2 \mathrm{H}_6}}{n_{\mathrm{C}_2 \mathrm{H}_6}+n_{\mathrm{H}_2}}=\frac{\frac{1}{30}}{\frac{1}{30}+\frac{1}{2}}=\frac{1}{16}\)

[n= number of moles, x = mole fraction]

Similarly \(x_{\mathrm{H}_2}=\frac{15}{16}\)

According to Dalton’s law of partial pressure

⇒ \(p_{\mathrm{H}_2}=x_{\mathrm{H}_2} \times P \quad[P=\text { total pressure }] \)

⇒  \(\frac{P_{\mathrm{H}_2}}{P}=x_{\mathrm{H}_2}=\frac{15}{16}\)

Question 15. Compressibility factor for a real gas at high pressure

1. 1
2. \(1+\frac{P b}{R T}\)
3. \(1-\frac{P b}{R T}\)
4. \(1+\frac{R T}{P b}\)

Answer: 2. \(1+\frac{P b}{R T}\)

Van der Waals equation \(\left(P+\frac{a}{V^2}\right)(V-b)=R T\)

At high pressure \(P \gg \frac{a}{V^2}\) hence, P(V-b) = RT

⇒ \(\text { or, } P V=R T+P b \text { or, } \frac{P V}{R T}=1+\frac{P b}{R T} \text { or, } Z=1+\frac{P b}{R T}\)

Question 16. ‘a’ and ‘b’ are van der Waals constant for gases. Chlorine is more easily liquefied them ethane because

1. A and B for Cl₂ < A and B for C₂H₆
2. A for Cl₂ < A for C₂H₆ but B for Cl₂ > B for C₂H₆
3. A for Cl₂ > A for C₂H₆ but B for Cl₂< B for C₂H₆
4. A and B for Cl₂ > A and B for C₂H₆

Answer: 3. A for Cl₂ > A for C₂H₆ but B for Cl₂ < B for C₂H₆

Van der Waals constants ‘a’ is a measure of intermolecular forces of attraction of a gas whereas ‘b’ is a measure of the size of gas molecules. Hence, more is the value of a more easily the gas will be liquefied.

Question 17. For the gaseous state, if the most probable speed is denoted by c, average speed by c, and mean square speed by c, then for a large number of molecules the ratios of these speeds are

1. c*:c:c = 1.225: 1.128: 1
2. c*:c:c = 1.128: 1.225: 1
3. c*:c:c = 1:1.128:1.225
4. c*:c:c = 1:1.225:1.128

Answer: 4. c*:c:c = 1:1.225:1.128

Question 18. If Z is a compressibility factor, the van der Waals equation at low pressure can be written as

1. \(Z=1+\frac{P b}{R T}\)
2. \(Z=1+\frac{R T}{P b}\)
3. \(Z=1-\frac{a}{R T V}\)
4. \(Z=1-\frac{P b}{R T}\)

Answer: 3. \(Z=1-\frac{a}{R T V}\)

Van der Waals constant ‘a’ is a measure of intermolecular forces of attraction of a gas whereas ‘b’ is a measure of the size of gas molecules. Hence, more is the value of a more easily the gas will be liquefied.

Question 19. The ratio of masses of oxygen and nitrogen in a gaseous mixture is 1:4 ratio of the number of their molecule is

1. 3:16
2. 1:4
3. 7:32
4. 1:8

Answer: 4. 1:8

Question 20. Intermolecular interaction that is dependent on the inverse cube of the distance between the molecules

1. London force
2. Hydrogen bond
3. ion ion interaction
4. ion-dipole interaction

Answer: 1. London force

1, 3, and 4, are not applicable as the interaction is intermolecular. 2 is not the correct choice as the hydrogen bond does not follow the relation mentioned above.

Question 21. Two closed bulbs of equal volume ( V) containing an Ideal gas Initially at pressure pt and temperature T₁ are connected through a narrow tube of negligible volume. The temperature of one of the bulbs is then raised to 7 2. The final pressure p₁ Is-

1. \(p_1\left(\frac{T_1 T_2}{T_1+T_2}\right)\)
2. \(2 p_i\left(\frac{T_1}{T_1+T_2}\right)\)
3. \(2 p_1\left(\frac{T_2}{T_1+T_2}\right)\)
4. \(2 p_i\left(\frac{T_1 T_2}{T_1+T_2}\right)\)

Answer: 3. \(2 p_1\left(\frac{T_2}{T_1+T_2}\right)\)

n₁+n₂=n₁‘+n’₂

∴ \(\frac{p_i V}{R T_1}+\frac{p_i V}{R T_1}=\frac{p_f V}{R T_1}+\frac{p_f V}{R T_2}\)

Question 22. A gas mixture was prepared by taking equal moles of CO and N₂. If the total pressure of the mixture was 1 atm, the partial pressure of N₂ in the mixture is

1. 0.5 atm
2. 0.8 atm
3. 0.9 atm
4. 1 atm

Answer: 1. 0.5 atm

In the mixture \(x_{\mathrm{N}_2}=\frac{1}{2} \text { and } x_{\mathrm{CO}}=\frac{1}{2}\)

∴ \(p_{\mathrm{N}_2}=x_{\mathrm{N}_2} \times P=\frac{1}{2} \times 1 \mathrm{~atm}=0.5 \mathrm{~atm}\)

Question 23. Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. Molar mass ofA is 49u. The molecular mass will be—

1. 50.00u
2. 12.25u
3. 6.50u
4. 25.00u

Answer: 2. 12.25u

⇒ \(\frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}} \text { or, } \frac{V / 20}{V / 10}=\sqrt{\frac{M_B}{49}} \text { or, } \frac{1}{2}=\sqrt{\frac{M_B}{49}}\)

∴ Mb = 12.25u

Question 24. By what factors does the average velocity of a gas molecule increase when the temperature (inK) is doubled

1. 2.0
2. 2.8
3. 4.0
4. 1.4

Answer: 4. 1.4

⇒ \(\bar{c}=\sqrt{\frac{8 R T}{\pi M}}\)

When Pis doubled, \(\bar{c}_1=\sqrt{\frac{8 R \times 2 T}{\pi M}}=\sqrt{2} \sqrt{\frac{8 R}{\pi M}}\)

∴ MB= 12.25u

Question 25. 50mL of each gas A and B takes 150s and 200s respectively for effusing through a pinhole under similar conditions. If the molar mass of gas B is 36, the molar mass of gas A—

1. 20.25
2. 64
3. 96
4. 128

Answer: 1. 20.25

⇒ \(\frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}} \text { or, } \frac{V_A}{t_A} \times \frac{t_B}{V_B}=\sqrt{\frac{M_B}{M_A}}\)

⇒ \(\text { or, } \frac{50}{150} \times \frac{200}{50}=\sqrt{\frac{36}{M_A}} \text { or, } \frac{4}{3}=\sqrt{\frac{36}{M_A}} \text { or, } \frac{16}{9}=\frac{36}{M_A}\)

∴ \(M_A=\frac{36 \times 9}{16}=20.25\)

Question 26.

1. Set-1: O₂, CO₂, H₂ and He,
2. Set-2: CH₄, O₂ and H₂.

The gases in set-I in increasing order of ’b’ and gases given in set-II in decreasing order of ‘a’ are arranged below here ‘a’ and ‘b’ are van der Waals constants. Select the Correct order from the following

1. O₂ < He < H₂ < CO₂ ; H₂ >O₂ > CH₂
2. H₂ < He < O₂ < CO₂; CH₄ >O₂ > H₂
3. H₂ < O₂< He < CO₂ ; O₂ > CH₄ > H₂
4. He < H₂ < CO₂ < O₂ ; CH₄ > H₂ > O₂

Answer: 2. H₂ < He < O₂ < CO₂ ; CH₄ > O₂ > H₂

A gas with strong intermolecular forces of attraction has a large value of a and a gas has a large value of ‘b’ if its molecules are big. The increasing order of sizes of H₂, He, O₂, and CO₂ molecules is H₂ < He <O₂ < CO₂. So, the increasing order of values for these gases is H₂ < He < O₂ < CO₂. CH₄, O₂, and H₂ are all non-polar molecules.

The only intermolecular forces of attraction that act in CH₄, O_2, and H₂ gases are London forces. The strength of London forces increases with molecular size.

So, the increasing order of intermolecular forces in CH_4, O₂, and H₂ gases will be CH₄ > O₂ > H₂. Again, the stronger the intermolecular forces of attraction in a gas, the larger the value of the gas. Therefore, the decreasing order of ‘b’ values for these gases will be CH₄ > O₂ > H₂.

Question 27. A certain gas takes three times as long to effuse out ns helium. Its molecular mass will be

1. 36u
2. 64u
3. 9u
4. 27u

Answer: 1. 36u

⇒ \(\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}} \text { or, } \frac{v / t_1}{v / t_2}=\sqrt{\frac{M_2}{M_1}} \text { or, } \frac{t_2}{t_1}=\sqrt{\frac{M_2}{M_1}}\)

Or, \(\frac{3}{1}=\sqrt{\frac{M_2}{4 \mathrm{u}}} \text { or, } M_2=36 \mathrm{u}\)

[t₁ and t₂ are the times for the diffusion of VmLHe and VmL unknown gas respectively. M₁ = molar mass of He, M₂ = molar mass of unknown gas]

Question 28. Maximum deviation from Ideal gas is expected in case of—

1. CH₄(g)
2. NH₃(g)
3. H₂(g)
4. N₂(g)

Answer: 2.

NH₃(g) CH₄, H₂ and N₂ are non-polar molecules. Only intermolecular forces that operate in CH₄, H₂, and N₂ gases are weak London forces.

As NH₃ is a polar molecule, besides weak London forces, relatively stronger dipole-dipole attractive forces also act among the molecules in NH₃ gas. So, among the given gases, intermolecular forces of attraction will be strongest in NH₃, and hence it will show maximum deviation from ideal behaviour.

Question 29. Equal masses of H₂, O₂, and methane have been taken in a container of volume V at a temperature of 27°C in identical conditions. The ratio of the volumes of gases H₂: O₂: methane would be –

1. 8:16:1
2. 16:8:1
3. 16:1:2
4. 8:1:2

Answer: 3. 16:1:2

Let the mass of each of the gases be wg. In the mixture

⇒ \(n_{\mathrm{H}_2}=\frac{w}{2} \mathrm{~mol}, n_{\mathrm{O}_2}=\frac{w}{32} \mathrm{~mol} \text { and } n_{\mathrm{CH}_4}=\frac{w}{16} \mathrm{~mol} \text {. }\)

Total number of moles in the mixture \(=\frac{w}{2}+\frac{w}{32}+\frac{w}{16}=\frac{19}{32} w\)

So,in the mixture, \(x_{\mathrm{H}_2}=\frac{w / 2}{19 w / 32}=\frac{16}{19}, x_{\mathrm{O}_2}=\frac{1}{19}\) and \(x_{\mathrm{CH}_4}=\frac{2}{19} .\)

The volume fraction of a component in the mixture = mole fraction ofthe component x total volume ofthe mixture

∴ \(V_{\mathrm{H}_2}=\frac{16}{19} \times V, V_{\mathrm{O}_2}=\frac{1}{19} \times V \text { and } V_{\mathrm{CH}_4}=\frac{2}{19} \times V\)

∴ \(V_{\mathrm{H}_2}: V_{\mathrm{O}_2}: V_{\mathrm{CH}_4}=16: 1: 2\)

Question 30. A gas such as carbon monoxide would be most likely to obey the ideal gas law at

1. High temperatures and low pressures
2. Low temperatures and high pressures
3. High temperatures and high pressures
4. Low temperatures and low pressures

Answer: 1. High temperatures and low pressures

At high temperatures and low pressures, real gases show ideal behaviour

Question 31. Equal moles of hydrogen and oxygen gases are placed in a container with a pinhole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape

1. \(\frac{1}{2}\)
2. \(\frac{1}{8}\)
3. \(\frac{1}{4}\)
4. \(\frac{3}{8}\)

Answer: 2. \(\frac{1}{8}\)

⇒ \(\text { } \frac{r_{\mathrm{O}_2}}{r_{\mathrm{H}_2}}=\sqrt{\frac{M_{\mathrm{H}_2}}{M_{\mathrm{O}_2}}} \text { or, } \frac{n_{\mathrm{O}_2} / t}{0.5 / t} \quad \text { or, } n_{\mathrm{O}_2}=\frac{1}{8}\)

Question 32. The correction factor ‘a’ to the ideal gas equation corresponds to—

1. Forces of attraction between the gas molecules
2. Density of the gas molecules
3. The electric field present between the gas molecules
4. The volume of the gas molecules

Answer: 1. Forces of attraction between the gas molecules

In the real gas equation \(\left(P+\frac{a n^2}{V^2}\right)(V-n b)=n R T \text {; }\) van der Waals constant ‘ a ‘ represents the intermolecular forces attraction between the molecules.

Question 33. Given van der Waals constant of NH₃, H₂, O₂, and CO₂ are respectively 4.17, 0.244, 1.36, and 3.59. Which one of the following gases is most easily liquefied—

1. NH₃
2. H₂
3. O₂
4. CO₂

Answer: 1. NH₃ The gases having strong intermolecular attraction have a value of van der Waals constant Such gases can be liquefied easily. Among the given gases NH₃ has the highest value of a.

Question 34. In the van der Waals equation, ‘ a ‘ signifies

1. Intermolecular attraction
2. Intramolecular attraction
3. The attraction between molecules and walls of the container
4. Volume of molecules

Answer: 1. Intermolecular attraction

In van der Waals equation, a signifies the intermolecular forces of attraction

Question 35. Arrange the following gases in order of their critical temperature: NH₃, H₂, CO₂, O₂

1. NH₃ > H₂O > CO₂ > O₂
2. O₂>CO₂>H₂O>NH₃
3. H₂O > NH₃ > CO₂ > O₂
4. CO₂ >O₂ > H₂O > NH₃

Answer: 3. H₂O > NH₃ > CO₂> O₂

The greater the intermolecular forces of attraction, the higher the critical temperature

Question 36. The density of gas A is thrice that of a gas B at the same temperature. The molecular weight of gas B is twice that of A. What will be the ratio of the pressures acting on B and A —

1. \(\frac{1}{4}\)
2. \(\frac{7}{8}\)
3. \(\frac{2}{5}\)
4. \(\frac{1}{6}\)

Answer: 4. \(\frac{1}{6}\)

⇒ \(\frac{d}{p}=\frac{M}{R T}\)

⇒ \(\frac{1}{6}\)

Let density ofgas B be d

∴ The density of gas A = 3d and molecular weight of A be M.

∴ Molecular weight of B = 2M Since, R is gas constant and T is the same for gases, so

⇒ \( p_A=\frac{d_A R T}{M_A} \text { and } p_B=\frac{d_B R T}{M_B}\)

⇒ \(\frac{p_B}{p_A}=\frac{d_B}{d_A} \times \frac{M_A}{M_B}=\frac{d}{3 d} \times \frac{M}{2 M}=\frac{1}{6}\)

Question 37. In van der Waals equation at constant temperature 300k, if a = 1.4 atm-L²-mol^-2, V = 100 mL, n = 1 mole, then what is the pressure of the gas

1. 42 atm
2. 210 atm
3. 500 atm
4. 106 atm

Answer: 4. 106 atm

At moderate pressure, the van der Waals equation is given

⇒ \(\left(P+\frac{a n^2}{V^2}\right)(V)=n R T\)

⇒ \(\left(P+\frac{1.4}{(0.1)^2}\right)(0.1)=1 \times 0.082 \times 300\)

or, (P+ 140) ×  0.1 = 24.6 or, 0.1, P+ 14 = 24.6

or, 0.1 P = 10.6 or, P = 106 atm

Question 38. When 1 g of gas A at 4 bar pressure is added to 2 g of gas B, the total pressure inside the container becomes 6 bar. Which of the following is true

1. MA= 2MB
2. MB=2MA
3. MA=4MB
4. MB=4MA

Answer: 4. MB=4MA

⇒ \(\frac{n_1}{p_1}=\frac{n_2}{p_2}\)

∴ \(\frac{\frac{1}{M_A}}{4}=\frac{\frac{1}{M_A}+\frac{2}{M_B}}{6}\)

⇒ \(\text { or, } \frac{6}{4 M_A}-\frac{1}{M_A}=\frac{2}{M_B}\)

⇒ \(\text { or, } \frac{6-4}{4 M_A}=\frac{2}{M_B} \quad \text { or, } \frac{1}{4 M_A}=\frac{1}{M_B} \quad \text { or, } M_B=4 M_A\)

Question 39. Gas in a cylinder is maintained at 10 atm pressure and 300 K temperature. The cylinder will explode if the pressure of the gas goes beyond 15 atm. What is the maximum temperature to which gas can be heated

1. 400k
2. 500k
3. 450k
4. 250k

Answer: 3. 450k

⇒ \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\)

∴ \(\frac{10}{300}=\frac{15}{T_2} \quad \text { or, } T_2=450 \mathrm{~K}\)

Question 40. Two separate bulbs contain gas A and gas B. The density of gas A is twice that of B. The molecular mass of A is half that of B. If temperature is constant, the ratio of the pressure of A and B is-

1. 1:1
2. 1:2
3. 4:1
4. 2:1

Answer: 3. 4:1

⇒ \(d=\frac{P M}{R T}\)

Given = \(\frac{d_A}{d_B}=2, \frac{M_A}{M_B}=\frac{1}{2} \)

= \(\frac{d_A}{d_B}=\frac{P_A M_A}{R T} \times \frac{R T}{P_B M_B}=2 \)

Or,  \(\frac{P_A}{P_B} \times \frac{M_A}{M_B}=2 \)

Or,  \( \frac{P_A}{P_B} \times \frac{1}{2}=2 \)

Or,  \(\frac{P_A}{P_B}\)

= 4: 1

Question 41. Which of the following does not change during compression of a gas at a constant temperature—

1. Density of a gas
2. Distance between molecules
3. The average speed of molecules
4. The number of collisions

Answer: 3. Average speed of molecules

Question 42. For which of the following gaseous mixtures, Dalton’s law of partial pressure is not applicable—

1. SO₂, HE, NE
2. NH₃, HBr, HC₁
3. O₂,N₂,CO₂
4. N₂,H₂,O₂

Answer: 2. NH₃, HBr, HCl

Question 43. The volume of a given mass of an ideal gas is VL at 27°C and 1 atm pressure. If the volume of the gas is reduced by 80% at constant pressure, the temperature of the gas will have to be—

1. -50°C
2. -127°C
3. -200°C
4. -213°C

Answer: 4. -213°C

Question 44. AT STP, the density of air is 1.3 × 10¯³g.cm¯³. The vapour density of air is—

1. 1.3
2. 14.6
3. 2.56
4. 10.8

Answer: 2. 14.6

Question 45. At a given temperature; the molar concentration of N₂ is greater than that of H₂ in a mixture of N₂ and H₂ gases present in a closed container. If the average kinetic energies of N₂ and H₂ molecules are xj and yj respectively, then

1. x>y
2. x<y
3. x = y
4. Impossible To Predict

Answer: 3. Impossible To Predict

Question 46. The density of gas A is dA at a temperature of TAK, and the density of gas B is dB at a temperature of TBK. The molar mass of A is 4 times that of B. If TA : TB = 2:1 and dA : dB = 1:2, the ratio of pressures of A to B is

1. 2:1
2. 1:8
3. 3:2
4. 1:4

Answer: 4. 1:4

Question 47. Two gases A and B have respective van der Waals constants a₂, bx and a₂, b₂. If ‘ A ’ is more compressible than ‘B,’ then which of the following conditions has to be satisfied

1. a₁ = a₂ and b₁> b₂
2. a₁ < a₂ and b₁> b₂
3. a₁ < a₂ and b₁ = b₂
4. a₁ > a₂and b₁ < b₂

Answer: 4. a₁ > a₂and b₁ < b₂

Question 48. The dimension of the coefficient of viscosity

1. MLT
2. ML^-1T^-1
3. MLT^-1
4. MLT^-2

Answer: 2. ML^-1T^-1

Question 49. The densities of water and water vapour are 1.0 g.cm¯³ and 0.0006 g.cm¯³ respectively at 100°C and 1 atm pressure. At this temperature, the total volume occupied by water molecules in 1L of water vapour is

1. 2.24 cc
2. 0.6 cc
3. 0.12 cc
4. 1.72 cc

Answer: 2. 0.6 cc

Question 50. The most probable velocities of the molecules of gas A (molar mass 16 g.mol¯¹) and that of the molecules of gas B (molar mass 28 g.mol¯¹) are the same. If the absolute temperatures of the gases A and B are T(A) and T{B) respectively, then

1. T{A) = 2T{B)
2. T(B) = 3T(A)
3. T(B) = 1.75 T (A)
4. T(B) = 2.5 T (A)

Answer: 3. T(B) = 1.75 T (A)

Question 51. At a given temperature and pressure, the volume of 1 mol of an ideal gas is 10L. At the same temperature and pressure, the volume of 1 mol of a real gas is VL. At this temperature and pressure, if the compressibility factor of the real gas is greater than 1, then

1. V- 10L
2. V< 10L
3. V> 10L
4. V< 10L

Answer: 3. V> 10L

Question 52. The pressure of a gas increases when its temperature is increased at constant volume. This is because with an increase in temperature—

1. The collision frequency of the gas molecules increases.
2. Motions of the gas molecules become more random
3. Gas molecules make more collisions with the walls of the container
4. The compressibility factor of the gas increases

Answer: 3. Gas molecules make more collisions with the walls of the container

Question 53. Under given conditions, the rate of diffusion of CH₄ gas is times that of f₂ gas. Gas 2 reacts with element A to form gaseous compounds AB₂ and AB₃. Under a given condition, the rate of diffusion of AB₂ is 1.12 times that of AB₃ The atomic mass of A (in g-mol^-1) is—

1. 32
2. 16
3. 8
4. 24

Answer: 1. 32

Question 54. Two flasks are connected by a valve: One of them with volume 5L contains 0.1 mol of H₂ at 27°C and the other with volume 2L contains 0.1 mol of N₂ at the same temperature. If the valve is opened keeping temperature constant, then at equilibrium the contribution of H₂ gas to the total pressure of the gas mixture

1. Is the same as that of n₂ gas
2. Is greater than that of n₂ gas
3. Is less than that of n₂ gas
4. Cannot be predicted

Answer: 1. Is the same as that of n₂ gas

Question 55. A balloon filled with acetylene is pricked by a pin and dropped readily in a tank of H₂ gas under identical conditions. After a while the balloon will—

1. Enlarge
2. Shrink completely
3. Collapse remain
4. Unchanged in size

Answer: 1. Enlarge

Question 56. At STP, the density of a gas is 1.25g-l^-1. The molar concentration (mol^-1) of 0.7g of this gas at 27°C and a pressure of 2 atm is

1. 0.27
2. 0.08
3. 0.19
4. 0.64

Answer: 2. 0.08

Question 57. 100 persons are sitting at equal distances in a row XY. Laughing gas (N₂O) is released from side X and tear gas (mol. mass = 176) from side Y at the same moment and the same pressure. The person who will tend to laugh and weep simultaneously is

1. 34th from side X
2. 67th from side X
3. 76th from side X
4. 67th from side Y

Answer: 2. 67th from side X

Question 58. van der Waals constant, b of a gas is 4.42 centilitre – mol 1. How near can the centres of 2 molecules approach each other

1. 127.2pm
2. 427.2pm
3. 327.2pm
4. 627.2pm

Answer: 3. 327.2pm

Question 59. Which of the following liquids has the least surface tension

1. Acetic acid
2. Diethyl ether
3. Chlorobenzene
4. Benzene

Answer: 2. Diethyl ether

Question 60. At P atm pressure and TK, a spherical air bubble is rising from the depth of a lake. When it comes to the surface of the lake the percentage increase in the radius will be (assume pressure and temperature at the surface to be PI₄ atm and 27TC respectively)—

1. 100%
2. 50%
3. 40%
4. 200%

Answer: 1. 100%

Question 61. A given mass of a perfect gas is first heated in a small and then in a large vessel, such that their volumes remain unaltered. The P- T curves are

1. Parabolic with the same curvature
2. Linear with the same slope
3. Linear with different slopes
4. Parabolic with different curvatures

Answer: 3. Linear with different slopes

Question 62. At a given temperature, most of the molecules in a sample of oxygen gas move with a velocity of 4.08× 10⁴ cm. s^-1. The average velocity of the molecules of the gas at the same temperature is—

1. 1.7 × 10⁴ cm.s¯¹
2. 4.6 × 10⁴ cm.s¯¹
3. 5.0 × 10⁴ cm.s¯¹
4. 8.9 × 10³ cm.s¯¹

Answer: 2. 4.6 × 10⁴ cm.s¯¹

Question 63. There is a depression in the surface of the liquid inside a capillary tube when—

1. The cohesive force is greater than
2. The adhesive force the adhesive force is greater than
3. The cohesive force both adhesive and cohesive forces are equal
4. None of the above is true

Answer: 1. The cohesive force is greater than

Question 64. One mol of a real gas following the equation, P(V-b) = RT, has a compressibility factor of 1.2 at 0°C and 200 atm pressure. The value of ‘b’ for this gas is—

1. 0.03521 L-mol¯¹
2. 0.0224 L-mol¯¹
3. 0.04610 L-mol¯¹
4. 0.01270 L-mol¯¹

Answer: 2. 0.0224 L-mol¯¹

Question 65. At a given temperature, the root mean square velocity of O₂ molecules is times that of the molecules of a gas. The molar mass of the gas (in g-mol¯¹) is

1. 8
2. 64
3. 96
4. 16

Answer: 2. 64

Question 66. At a given condition, 20L of SO₂ gas takes 60 for its effusion. At the same condition, the volume of 09 gas that will effuse out in 30 seconds is

1. 12.4L
2. 10.9L
3. 14.1L
4. 6.8L

Answer: 3. 14.1L

Question 67. The average velocity of the molecules of a gas at T₁K will be the same as the most probable velocity of the molecules of the gas at T₂K when

1. T₁ > r₂
2. t₂ >T₁
3. t₁=t₂
4. t₁> r₂

Answer: 2. t₂ >T₁

Question 68. Two ideal gases A and B have molar masses MA and MB g-mol-1 respectively. Volumes of the same mass of A and B are the same, and the rms velocity of A molecules is twice that of the molecules of B. If MB: MA = 2:1, then the ratio of the pressures of A to B is—

1. 4:1
2. 8:1
3. 2:1
4. 1:6

Answer: 1. 4:1

Question 69. Containing gas molecules, the percentage of molecules moving with velocities 2× 10⁴cm.s^-1 and 1 × 10^-1cm s^-1 are 30% and 45% respectively, and the rest one moving with velocity 5 × 104 cm-s^-1. The root mean square velocity of the molecules is

1. 3.7 × 10⁴cm.s^-1
2. 1.8 × 10⁴ cm.s^-1
3. 6.2 × 10³ cms^-1
4. 2.8 × 10⁴ cms^-1

Answer: 4. 2.8 × 10⁴ cms^-1

Question 70. An open vessel has a temperature of TK. When the vessel is heated at 477°C, three-fifths of the air in the vessel escapes out. What fraction of air in the vessel would have been expelled out if the vessel were heated at 900K (assume that the volume of the vessel remains unchanged on heating)

1. 4
2. 3
3. 2
4. 5

Answer: 2. 2

Question 71. Critical temperatures of the gases A, B, C and D are 126K, 155K, 304K and 356K respectively. Among these gases, the one with the strongest intermolecular forces of attraction is—

1. A
2. B
3. C
4. D

Answer: 4. D

Question 72. The volumes of two gases A and B at 0°C and 200 atm pressure are 0.112L and 0.09L respectively. Which of the following comments is true for these gases at this temperature and pressure

1. The compressibility of gases a and b are the same
2. The compressibility of a is less than that of b
3. The compressibility of a is more than that of b
4. Both gases show positive deviation from ideality

Answer: 3. Compressibility of a is more than that of b

Question 73. Which of the following correctly represents the relation between capillary rise (h) and radius of the capillary (r) —

Answer: 2.

Question 74. For CO₂ gas the P vs V isotherms at temperatures above 31.1°C are

1. Straight line
2. Rectangular hyperbolic
3. Elliptical
4. Hyperbolic

Answer: 2. Rectangular hyperbolic

Question 75. At a certain temperature, lmol of chlorine gas at 1.2 atm takes 40 sec to diffuse while 1 mol of its oxide at 2 atm takes 26.5 sec. The oxide is

1. Cl₂O
2. ClO₂
3. Cl₂O₆
4. Cl₂O₇

Answer: 1. Cl₂O

Question 76. At 10 bar pressure, a 4:1 mixture of He and CH₄ is contained in a vessel. The gas mixture leaks out through a hole present in the vessel. The mixture effusing out has an initial composition of

1. 1:1
2. 2:1
3. 4:1
4. 8:1

Answer: 4. 8:1

Question 77. A gas mixture consisting of N₂ and 3 mol of O₂ had a pressure of 2 atm at 0 °C. Keeping the volume and the temperature of the mixture constant, some amount of O₂ was removed from the mixture. As a result, the total pressure of the mixture and the partial pressure of N₂ in the mixture became 1.5 atm and 0.5 atm respectively. The amount of oxygen gas removed was

1. 8g
2. 16g
3. 32g
4. 64g

Answer: 3. 32g

Question 78. The quantity — represents

1. Mass of a gas
2. Translation energy of a gas
3. Number of moles of a gas
4. Number of molecules in a gas

Answer: 4. Number of molecules in a gas

Question 79. At STP, O₂ gas present in a flask was replaced by SO₂ under similar conditions. The mass of SO₂ present in the flask will be

1. Twice that of O₂
2. Half that of O₂
3. Equal to that of O₂
4. One-third of O₂

Answer: 1. Twice that of O₂

Question 80. The relative densities of oxygen and carbon dioxide are 16 and 22 respectively. If 37.5cm³ of oxygen effuses out in 96s, what volume of carbon dioxide will effuse out in 75s under similar conditions

1. 25cm³
2. 37.5cm³
3. 14cm³
4. 30.8cm³

Answer: 1. 25cm³

Question 81. At 27°C, the average translational kinetic energies of the molecules in 8g of CH₄,8g of O₂ and 8g of He are, e₂ and e₂ respectively and the total kinetic energies of the molecules in these gases are E₁, E₂ and E₃ respectively. Which of the following is true

1. \(\bar{\epsilon}_1=\bar{\epsilon}_2=\bar{\epsilon}_3\)
2. \(\bar{\epsilon}_3=\bar{\epsilon}_2=\bar{\epsilon}_1\)
3. E₁ = E₂ = E₃
4. E₂<E₁<E₃

Answer: 1. \(\bar{\epsilon}_1=\bar{\epsilon}_2=\bar{\epsilon}_3\)

Question 82. Several molecules of an ideal gas present in a flask of volume 2L are 10²³. The mass of each gas molecule is 6.64 × 10^-23 g and the root mean square velocity of the molecules is 4.33 × 10⁴ cm-s-1. Hence

1. The pressure of the gas is 3.27 atm
2. The average kinetic energy of each molecule is 6.23 × 10¹⁴J
3. The total kinetic energy of the molecules is 6.23 × 10⁹J
4. The total kinetic energy of the molecules is 1.492 × 10⁹ J

Answer: 2. The average kinetic energy of each molecule is 6.23 × 10¹⁴J

Question 83. In which conditions does the most probable velocity of O₂ molecules have maximum value and in which conditions does it have minimum value

1. O₂ : P = 1 atm, d (density) = 0.0081 g mL^-1
2. O₂ : P = 4 atm, V = 2L and w (mass) = 4g
3. O₂ : r=300K
4. O₂: STP

Answer: 1. O₂ : P = 1 atm, d (density) = 0.0081 g mL^-1

Question 84. The time required to effuse V mL of H₂ gas through a porous wall at a constant temperature and pressure is 20 min. Under the same conditions time required to effuse V mL of the following gases is

1. He:28.28min
2. CO₂:90.82min
3. CH₄:60.52 min
4. N₂:74.83 min

Answer: 1. He:28.28min

Question 85. At a particular temperature and pressure, if the number of moles of an ideal gas is increased by 50%, then

1. The final volume of the gas will be 1.5 times its initial volume
2. The most probable velocity of gas molecules becomes 1.5 times its initial value
3. The total kinetic energy of the gas molecules becomes 1.5 times its initial value
4. The density of gas becomes 1.5 times its initial value.

Answer: 1. Final volume of the gas will be 1.5 times its initial volume

Question 86. The pressure and temperature of a gas are P and T respectively. If the critical pressure and critical temperature of the gas are Pc and Tc respectively, then liquefaction will be possible when

1. P<PC,T<TC
2. P = PC,T=TC
3. P = PC,T>TC
4. P>PC,T=TC

Answer: 2. P = PC,T=TC

Question 87. If the orders of the values of van der Waals constants a and b for three gases X, Y and Z are X < Y < Z and Z < Y < X respectively, then

1. Liquefaction will be easier for gas than gases and z.
2. The size of the molecule, y will be in between the sizes of the molecules x and z.
3. The order of the critical temperatures of these three gases is: x< y<z.
4. The gas, z, at 0°c and 1 atm will behave most ideally.

Answer: 2. The size of the molecule, y will be in between the sizes of the molecules x and z.

Question 88. Identify the correct statements

1. At a particular temperature, the vapour pressure of dimethyl ether is greater than water because the molar mass of dimethyl ether is higher than that of water.
2. The vapour pressure of a liquid remains the same when the surface area of the liquid is increased at a given temperature.
3. The correct order of viscosity coefficient is ethylene glycol < glycerol.
4. The surface tension of water at 30°c is greater than that at 20°c.

Answer: 2. The vapour pressure of a liquid remains the same when the surface area of the liquid is increased at a given temperature.

Question 89. P(V-b) = RT equation of state is obeyed by a particular gas. Which of the given statements is correct

1. For this gas, the isochoric curves have slope = \(=\frac{R}{V-b}\)
2. The compressibility factor of the gas is less than unity.
3. For this gas, the isobaric curves have slope = r/p
4. In this gas, the attraction forces are overcome by repulsive forces.

Answer: 1. For this gas, the isochoric curves have slope = \(=\frac{R}{V-b}\)

Question 90. Four gas balloons P, Q, R and S of equal volumes containing H₂, N₂O, CO, and CO₂ respectively were pricked with a needle and immersed in a tank containing CO₂. Which of them will shrink after some time?

1. P
2. Q
3. R
4. S

Answer: 1. P

Question 91. A liquid is in equilibrium with its vapour at its boiling point. On average, the molecules in the two phases have

1. Equal total energy and potential energy.
2. Equal kinetic energy different total
3. Energy and potential
4. Energy is different from kinetic energy.

Answer: 2. Equal kinetic energy different total

Question 92. The root mean square velocity of an ideal gas in a closed vessel of fixed volume is increased from 5× 10⁴cm.s^-1 to 10 × 10⁴cm.s^-1. Which of die following statements clearly explains how the change is accomplished 

1. By heating the gas, the die temperature is quadrupled.
2. By heating die gas, the temperature is doubled by heating the gas,
3. The pressure is quadrupled by heating the gas,
4. The pressure is doubled

Answer: 1. By heating the gas, the die temperature is quadrupled.

Question 93. Which of the following pairs of gases have the same type of intermolecular force of attraction

1. Ch4, CI₂
2. SO₂,CO₂
3. HC₁, CHCI₃
4. N₂,NH₂

Answer: 1. Ch₄, CI₂

Question 94. Select the correct orders

1. Critical temperature < boyleg’s temperature < inversion temperature
2. Van der waals constant ‘a’: H₂O> nh₃ > N₂ > ne
3. Van der waals constant ‘b’: CH₄> O₂ >H₂
4. Mean free path: he > H₂ >O₂ >  N₂ > CO₂
5. All the above

Answer: 4. All the above

Question 95. Which are responsible for the liquefaction of H₂ 

1. Coulombic forces
2. London forces
3. Hydrogen bonding
4. Van der Waals forces

Answer: 2. London forces

Question 96. Which of the following gases will have the same rate of effusion under identical conditions

1. CO
2. N₂O
3. C₂H₄
4. CO₂

Answer: 2. N₂O

Question 97. Select the correct statements

1. The presence of impurities invariably increases the viscosity of a liquid.
2. In the presence of impurities, the viscosity of a liquid remains unaltered
3. The viscosity coefficient of associated liquids is larger than that of non-associated liquids.
4. Viscosity coefficients of non-associated liquids are larger than those of associated liquids.

Answer: 1. Presence of impurities invariably increases the viscosity of a liquid.

Question 98. Select the correct statements

1. Surface energy of a liquid = \(\frac{\text { force } \times \text { distance }}{\text { area }}\)
2. Surface energy can be represented as force/area
3. The addition of NaCl increases and the addition of acetone decreases the surface tension of water.
4. The addition of NaCl decreases and the addition of acetone increases the surface tension of water.

Answer: 1. Surface energy of a liquid = \(\frac{\text { force } \times \text { distance }}{\text { area }}\)

Question 99. Precisely lmol of He and 1 mol of Ne are placed in a container. Select correct statements about the system

1. Molecules of the strike the wall more frequently
2. Molecules of he have greater average
3. The molecular speed molecule of the two gases strikes the wall of the container with the same frequency
4. He has a larger pressure

Answer: 1. Molecules of he strikes the wall more frequently

Question 100. Which of the following is correct for different gases under the same condition of pressure and temperature

1. Hydrogen diffuses 6 times faster than oxygen
2. Hydrogen diffuses 2.83 times faster than methane
3. Helium escapes at a rate 2 times as fast as sulphur dioxide does
4. Helium escapes at a rate 2 times as fast as methane does

Answer: 2. Hydrogen diffuses 2.83 times faster than methane.

Question 101. For a definite mass of ideal gas at constant temperature, V versus \(\frac{1}{p}\) plot is a

1. Parabola
2. Straight line
3. Hyperbola
4. Rectangular Hyperbola

Answer: 2. Straight line

Question 102. The surface tension of water with the increase of temperature may

1. Increase
2. Decreases
3. Remain same
4. Shows irregular behaviour

Answer: 2. Decreases

Question 103. Which of the following is the unit of van der Waals gas constant

1. L₂.mol
2. L.mol¯²
3. L.mol
4. L.mol¯¹

Answer: 4. L.mol¯¹

Question 104. The cause of the spherical drop of water is

1. Surface tension
2. Viscosity
3. Hydrogen bond
4. High critical temperature of h₂O vapour

Answer: 2. Viscosity

Question 105. Which gas among the following exhibits maximum critical temperature—

1. N₂
2. O₂
3. CO₂
4. H₂

Answer: 3. CO₂

Question 106. Indicate the correct answer: The rate of diffusion of helium gas at constant temperature and pressure will be four times the rate of diffusion of the following gases

1. CO₂
2. SO₂
3. NO₂
4. O₂

Answer: 2. SO₂

NCERT Solutions For Class 11 Chemistry Chapter 5 States Of Matter Gases And Liquids Long Question And Answers

Question 1. At a given pressure, the volume of a given amount of gas at 0°C is V0. Will the V vs t (celsius temperature) plot for this gas be linear? Will it be a straight line passing through the origin? If this straight line does not pass through the origin, then what will be its slope and intercept?
Answer:

According to Charles’ law, Vt = V₀ \(\left(1+\frac{t}{273}\right)\); where Vt and V₀ are the volumes of a given mass of gas at a temperature t°C respectively, when the pressure of the gas is kept constant. Hence, the above equation can be rewritten as

⇒  \(V_t=V_0+\frac{t}{273} V_0\)……………[1]

V₀ is a fixed quantity for a given mass of gas at a constant pressure. Thus, the equation [1] represents a straight-line equation. Hence, the Vt vs t plot will give a straight line.

Equation [1] does not represent an equation of a straight line passing through the origin. Hence, the Vf vs t plot will not be a straight line passing through the origin

According to the equation [1], the straight line resulting from the plot of V( vs t has a slope of V₀/273 and an intercept of V₀.

Question 2. A certain amount of an ideal gas is enclosed in a cylinder fitted with a movable piston. What would be the changes in the volume ofthe gain in the following processes?

1. The pressure of the gas is reduced by 25% at constant temperature.
2. The temperature of the gas is increased by 50% at constant pressure.

Answer:

1. According to Boyle’s law, P₁V₁ = P₂V₂ when the mass and temperature of a gas are constant. After the reduction of initial pressure (P₁) by 25%, the final pressure (P₂) becomes

⇒ \(\left(P_1-P_1 \times \frac{25}{100}\right)=0.75 P_1\)

i.e., P₂ = 0.75P₁

∴ \(V_2=\frac{P_1 V_1}{P_2}=\frac{P_1 \times V_1}{0.75 P_1}=\frac{4}{3} V_1\)

Therefore, the change in volume \(=\frac{4}{3} V_1-V_1=\frac{V_1}{3}\)

2. According to Charles’ law, V₁T₂ = V₂T₁ when the mass and pressure of a gas remain constant. After increasing the initial temperature (T₁) by 50%, the final temperature (T₂) becomes

⇒  \(\left(T_1+T_1 \times \frac{50}{100}\right)=1.5 T_1 \text {, i.e., } T_2=1.5 T_1\)

∴ \(V_1 \times 1.5 T_2=V_2 T_2 \quad \text { or, } V_2=1.5 \times V_1\)

∴ The change in volume = 1.5 V₁– V₁ = 0.5V₁

Question 3. Determine the values of molar gas constant in the following units— mL torr K^-1mol^-1; kPa-L-K^-1-mol^-1.
Answer:

R =0.0821 L-atm -K^-1. mol^-1

= 0.0821 × 103 × 760 torr -mL .K^-1.mol^-1

[Since 1L = 10³mL and 1 atm = 760

= 6.23 ×  10⁴ torr -mL -K^-1.mol^-1

= 0.0821 L .atm -K^-1.mol^-1

= 0.0821 × 1.013 x 1 02 kPa . L K^-1. mol^-1

= 8.31 kPa-L-K^-1-mol^-1

[Since 1 atm = 1.013 ×10⁵ Pa = 1.013 × 10²kPa

Question 4. For the same mass of two ideal gases X and Y at the same temperature and pressure, the volume of Y is found to be three times as large as that of X. Compare the values of their molar masses.
Answer:

Px = Py, Tx = Ty, 3VX = Vy

According to the ideal gas equation, Pxvx = nXRTx and = P_x V_x = n_xRT_x

i.e., \(\frac{P_X}{P_Y} \times \frac{V_X}{V_Y}=\frac{n_X}{n_Y} \times \frac{T_X}{T_Y} \quad \text { or, } \frac{1}{3}=\frac{n_X}{n_Y}\)

Let, molar masses of X and Y be My and My respectively

∴ \(\frac{w / M_X}{w / M_Y}=\frac{1}{3} \quad \text { or, } \frac{M_X}{M_Y}=3\)

Question 5. When a flask of fixed volume is filled with – mol of an ideal gas A at a constant temperature, the pressure ofthe gas becomes 2 atoms. Adding 2y mol of another ideal gas B to the flask at the same temperature causes the pressure of the system to increase to 4.0 atm.
Answer:

For tyre gas. A: P = 2 atm and \(n=\frac{x}{2} \mathrm{~mol}\)

Hence, PV \(=n R T \text { or, } 2 \times V=\frac{x R T}{2}\)

For gas mixture: P=4atm \(\& n=\frac{x}{2}+2 y=\frac{1}{2}(x+4 y) \mathrm{mol}\)

∴ In case of gas mixture \(4 \times V=\frac{1}{2}(x+4 y) R T\)

Deriding [2] -r [1] we get \(\frac{4}{2}=\frac{x+4 y}{x} \text { or, } x=4 y\)

Question 6. At 27°C and 1 atm pressure, the volume of a 5.0g mixture of He and Ar gases is 10dm3. Find the mass per cent of the two gases in the gas mixture.
Answer:

Let, the amount of He and Ar in the mixture be ag and bg respectively. Hence, a + b = 5

⇒ \(\text { Now, } P V=\left(n_1+n_2\right) R T \text { or, } 1 \times 10=\left[\frac{a}{4}+\frac{b}{40}\right] \times 0.0821 \times 300\)

∴ 10 a +b = 16.24

By solving [1] and [2] we get, a = 1.25g and b = 3.75g

∴ \(\% \mathrm{He}=\frac{1.25}{5} \times 100=25 \text { and } \% \mathrm{Ar}=\frac{3.75}{5} \times 100=75\)

Question 7. A gas mixture consisting of O₂ and N₂ gases has a volume of 5 L at 25°C. In the mixture, if the mass of O₂ gas is twice that of N₂ gas, then which one of them will have a greater contribution to the total pressure ofthe mixture?
Answer:

In the mixture \(n_{\mathrm{O}_2}=\frac{2 w}{32}\)

=\(\frac{w}{16} \mathrm{~mol} \text { and } n_{\mathrm{N}_2}\)

=\(\frac{w}{28} \mathrm{~mol} \text {. }\)

∴ Total number of moles (n) \(=\frac{w}{16}+\frac{w}{28}=\frac{11 w}{112} \mathrm{~mol}\)

⇒ \(\text { Hence, } x_{\mathrm{N}_2}\)

= \(\frac{w}{28} \times \frac{112}{11 w}\)

= \(\frac{4}{11} \text { and } x_{\mathrm{O}_2}\)

= \(\frac{w}{16} \times \frac{112}{11 w}=\frac{7}{11}\)

Thus, xO₂ > xN₂. So, pO₂> PN₂

(Since pi = zip)

Question 8. In a gas mixture of H₂ and He, the partial pressure of H₂ is half that of He. Find the mole fractions of H₂ and He in the mixture.
Answer:

As given, p₂ = pH₂ or, pHe = 2 × PH₂

or, xHe × P = 2 × XH₂ × P or, xHe = 2 × xe = 2 × xH₂

Hence, 2 × xH₂+ xH₂ = 1

Or, \(x_{\mathrm{H}_2}=\frac{1}{3} \approx 0.34\)

∴ xHe = 2 × XH₂ = 0.68

Question 9. A closed vessel contains an equal mass of O₂ and CH₄ gases at 25°C. What fraction of the total pressure is contributed by CH₄ gas?
Answer:

⇒ \(n_{\mathrm{CH}_4}=\frac{w}{16}, n_{\mathrm{O}_2}=\frac{w}{32}\)

∴ Total number of moles = \(\frac{w}{16}+\frac{w}{32}=\frac{3 w}{32}\)

∴ \(x_{\mathrm{CH}_4}=\frac{w}{16} \times \frac{32}{3 w}=\frac{2}{3}\)

⇒ \(p_{\mathrm{CH}_4}=\frac{2}{3}\); Thus Memthane Contributes \(\frac{2}{3}\) rd of the total pressure.

Question 10. A mixture of O₂ and H₂ gases contains 20% of H₂ gas. At a certain temperature, the total pressure of the mixture is found to be 1 bar. What is the partial pressure of O₂ (in bar) in the mixture?
Answer:

In the mixture, 20% H₂ is present. Hence extent of oxygen = 80%

∴ \(n_{\mathrm{H}_2}=\frac{20}{2}=10 \text { and } n_{\mathrm{O}_2}=\frac{80}{32}=\frac{5}{2}\)

∴ Total number of moles \(=\left(10+\frac{5}{2}\right)=\frac{25}{2}\)

∴ \(p_{\mathrm{O}_2}=\frac{5}{2} \times \frac{2}{25} \times 1 \mathrm{bar}=0.2 \mathrm{bar}\)

Question 11. Under the same conditions, a gas diffuses times as fast as an SO₂ gas. Find the molecular mass ofthe gas.
Answer:

⇒ \(\frac{r_{\text {gas }}}{r_{\mathrm{SO}_2}}=\sqrt{\frac{M_{\mathrm{SO}_2}}{M_{\text {gas }}}}=\sqrt{2}\) \(\text { or, } \frac{M_{\mathrm{SO}_2}}{M_{\text {gas }}}=2 \text { or, } M_{\mathrm{gas}}=32 \mathrm{~g} \cdot \mathrm{mol}^{-1}\)

or, 8n-1 = 15 or, n = 2

Question 12. At what temperatures rms velocity, average velocity & most probable velocity of O₂ molecules will be 1500 m.s^-1?
Answer:

Molar mass (M) of O₂ – 32g.mol^-1 =0.032kg. mol^-1

In case of rms velocity:

⇒  \(\frac{3 R T_1}{M}=(1500)^2\)

⇒ \(\text { or, } \frac{3 \times 8.314 \times T_1}{0.032}=(1500)^2 \text { or, } T_1=2886.697 \mathrm{~K} \text {. }\)

In case of average velocity:

⇒ \(\frac{8 R T_2}{\pi M}=(1500)^2\)

⇒ \(\text { or, } \frac{8 \times 8.314 \times T_2}{0.032 \times 3.14}=(1500)^2 \text { or, } T_2=3399.085 \mathrm{~K} \text {. }\)

In case of most probable velocity: \(\frac{2 R T_3}{M}=(1500)^2\)

⇒ \(\text { or, } \frac{2 \times 8.314 \times T_3}{0.032}=(1500)^2 \text { or, } T_3=4330.045 \mathrm{~K} \text {. }\)

Question 13. At a given temperature and pressure, 1 mol of an ideal gas occupies a volume of 20.8 L. For mol of a real gas at the same temperature and pressure—

• Z will be equal to 1 if the volume ofthe real gas is…
  
• Z will be greater than 1 if the volume ofthe real gas is…
  
• Z will be less than 1 if the volume ofthe real gas is…

Answer: \(Z=\frac{V}{V_i}\); the volume of a certain amount of ideal gas at a given temperature and pressure and V = the volume of the same amount of a real gas at the same temperature and pressure.

As given, V- = 20.8L. Therefore, if

• V = 20.8L, then Z will be equal to1
• V> 20.8L, then Z will be greater than1
• V< 20.8L, then Z will be less than1

Question 14. Two ideal gases A and B are mixed at temperature T and pressure P. Show that \(d=\left(X_A M_A+X_B M_B\right) \frac{P}{R T}\) Id = density of the mixture, X₂ = mole fraction of A, XD = mole fraction of B, M A = Molar mass of A, Mlt = Molar mass of B]
Answer:

Suppose, the total volume of the gas mixture is V, and nA and nB are the respective number of moles of A and B in the mixture. If W A and be the masses of A and B respectively, in the mixture, then

⇒ \(n_A=\frac{w_A}{M_A} \text { and } n_B=\frac{W_B}{M_B}\)

Total mass ofA & Bin mixture = \(W_A+W_B=n_A M_A+n_B M_B\)

∴ Density Of The Mixture \(=\frac{W_A+W_B}{V}=\frac{n_A M_A+n_B M_B}{V}\) for the gas mixture

⇒ \(P V=\left(n_A+n_B\right) R T ;\)

∴ \(V=\left(n_A+n_B\right) \frac{R T}{P}\)

⇒ \(\text { So, } d=\frac{n_A M_A+n_B M_B}{n_A+n_B} \times \frac{P}{R T}=\left(\frac{n_A M_A}{n_A+n_B}+\frac{n_B M_B}{n_A+n_B}\right) \frac{P}{R T}\)

∴ \(\left.d=X_A M_A+X_B M_B\right) \frac{P}{R T}\left[X_A=\frac{n_A}{n_A+n_B} ; X_B=\frac{n_B}{n_A+n_B}\right]\)

Question 15. “The total kinetic energy of the molecules in an ideal gas with a volume V at pressure P and temperature T is equal to the total kinetic energy of the molecules present in the same volume of another ideal gas at the same pressure and temperature 2T”—Justify the statement.
Answer:

Suppose, the number of molecules present in volume V of the first gas =. If the root mean square velocity of the molecules in the first gas is cl, and the mass of each molecule is mx, then from the kinetic gas equation we have,

⇒ \(P V=\frac{1}{3} m_1 n_1 c_1^2 \quad \text { or, } P V=\frac{2}{3} n_1 \times \frac{1}{2} m_1 c_1^2\)

Or \(P V=\frac{2}{3} E_1 \text { [where } E_1=n_1 \times \frac{1}{2} m_1 c_1^2\)

Energy ofthe molecules.] Similarly, if the number of molecules in volume V of the second gas = n₂, the mass of each molecule = m2 and the root mean square velocity of molecules= c₂, then

⇒ \(P V=\frac{1}{3} m_2 n_2 c_2^2=\frac{2}{3} n_2 \times \frac{1}{2} m_2 c_2^2 \quad \text { or, } P V=\frac{2}{3} E_2\)

[where E₂ = total kinetic energy ofthe molecules of second gas] As P and V of both the gases are equal, so E₁ = E₂.

Question 16. Prove that at a certain pressure, the rate of diffusion of a gas is proportional to the square root of the I absolute temperature of the gas
Answer:

The rate of diffusion of a gas at constant pressure (r) oc root mean square velocity ofthe gas (Crms)

∴ \(r \propto c_{r m s} \text { or, } r \propto \sqrt{\frac{3 R T}{M}}\)

Since \(c_{r m s}=\sqrt{\frac{3 R T}{M}}\)

At constant pressure, if and r2 are the rates of diffusion of a particular gas at temperatures and T2, respectively, then

⇒ \(r_1 \propto \sqrt{\frac{3 R T_1}{M}} \text { and } r_2 \propto \sqrt{\frac{3 R T_2}{M}}\)

∴ \(\frac{r_1}{r_2}=\sqrt{\frac{T_1}{T_2}} \text { or, } r \propto \sqrt{T}\)

So, at constant pressure, the rate of diffusion of a gas is proportional to the square root of the absolute temperature of the gas.

Question 17. At constant temperature and pressure, the compressibility factor (Z) for one mole of a van der Waals gas is 0.5. If the volumes of the gas molecules are considered to be negligible, then show that \(a=\frac{1}{2} V_m\) where Vm and Tare the molar volume and temperature of the gas respectively.
Answer:

We known \(Z=\frac{P V}{n R T}\); Given Z= 0.5 and n=1

∴ \(P V=0.5 R T=\frac{1}{2} R T\)

The equation of state for mol of a van der Waals gas is, \(\left(P+\frac{a}{V_m^2}\right)\left(V_m-b\right)=R T\) [Vm= molar volume]

If volumes of the molecules are considered to be negligible, as per the given condition, then \(V_m-b \approx V_m\)

∴ \(\left(P+\frac{a}{V_m^2}\right) V_m=R T \quad \text { or, } P V_m+\frac{a}{V_m}=R T\)

Again, from equation [1] we get, \(P V=\frac{1}{2} R T\)

For 1 mol (n = 1) ofthe gas, V = Vm (molar volume)

Therefore \(P V_m=\frac{1}{2} R T\)

From equations [2] and [3] we have,

⇒ \(\frac{1}{2} R T+\frac{a}{V_m}=R T \quad \text { or, } a=\frac{1}{2} V_m R T\)

Question 18. At a given temperature and pressure, the volume fraction of an ideal gas is equal to its mole fraction in a mixture of ideal gases—is it true or false?
Answer:

Suppose, at a given temperature (T) and pressure (P), the volumes of two ideal gases are VA and Vg, respectively, and n_A and n_B are their respective number of moles. Let at the same temperature and pressure, the total volume ofthe mixture of these two gases is V.

∴ According to Amagat’s law of partial volume, \(V=V_A+V_B\)

So, the partial volume of A in the mixture \(=\frac{V_A}{V}\) and its mole fraction \(=\frac{n_A}{n_A+n_B}\)

Now, applying the ideal gas equation to each component gas as well as the gas mixture, we get

PV_A = n_ART ……………………(1)

PV_B=n_BRT ……………………(2)

PV = (n_A + n_B)RT ……………………(3)

Dividing equation no. [1] by equation no. [3], we have

⇒ \(\frac{P V_A}{P V}=\frac{n_A}{n_A+n_B}\)

∴ \(\frac{V_A}{V}=\frac{n_A}{n_A+n_B}\)

That, the volume fraction of A = the mole fraction of A

Similar \(\frac{V_B}{V}=\frac{n_B}{n_A+n_B}\)

[Dividing equation no. [2] by equation no. [3]

That is the volume fraction of B = the mole fraction of B. Therefore, at a given temperature and pressure, the volume fraction of an ideal gas in a mixture of ideal gases is equal to its mole fraction. Hence, the given statement is true.

Question 19. A 15.0 L vessel containing 5.6 g of N₂ is connected to a 5.0 L vessel containing 8.0 g of O₂ using a valve. After the valve is opened and the gases are allowed to mix, what will be the partial pressure of each gas in the mixture at 27°C?
Answer:

5.6g N₂ = \(\frac{5.6}{28}=0.2 \mathrm{~mol}\) N₂

8.0 g O₂ = \(\frac{8}{32}=0.25 \mathrm{~mol}\) O₂

= \(\frac{5.6}{28}=0.2 \mathrm{~mol} \mathrm{~N}_2, 8.0 \mathrm{~g} \mathrm{O}_2\)

=\(\frac{8}{32}=0.25 \mathrm{~mol} \mathrm{O}_2\) and

T = (273 + 27)K = 300 K.

After the opening of the valve, the total volume ofthe gas mixture, V = (15 + 5)L = 20 L

If the partial pressure of N₂ gas in the mixture is pN, then pN x V = nN RT or,

PN₂ × 20=0.2 × 0.0821× 300

∴ PN₂ = 0.2463 atm

If the partial pressure of O₂ gas in the mixture is pO₂

then XV= NO₂RT

Or, PO₂ × 20=0.25 × 0.00821 × 300

∴ PO₂= 0.30+748 atm

Question 20. Why is the quantity of air required to inflate the tyre of a car in summer less than that required in winter?
Answer:

• According to the kinetic theory of gases, the pressure of a gas originates due to the bombardment of the gas molecules on the walls ofthe container. At higher temperatures, the molecules in a gas have higher velocities or average kinetic energy, and they collide with the walls of their container more frequently and with greater force.
• As a result, the pressure of a gas increases with an increase in temperature if the volume of the gas remains fixed.
• The summer temperature is higher than the winter temperature, and the average kinetic energy of the air molecules in summer is comparatively more than that in the winter.
• As a result, the air molecules in the summer will exert a greater amount of force on the walls than that exerted by the same number of molecules in the winter.
• Therefore, if the volume remains fixed, the pressure of a certain amount of air in summer will be more than that in winter.

Question 21. Two flasks of equal volume are connected by a narrow tube of negligible volume and are filled with N₂ gas. When both the flasks are immersed in boiling water the gas pressure inside the system is 0.5 atm. Calculate the pressure of the system when one of the flasks is immersed in ice water while the other flask is in boiling water
Answer:

Temperature of the gas when flasks are in boiling water = 100 + 273 = 373 K and pressure = 0.5 atm

The average temperature of the gas when one flask is in ice and the other in boiling water

⇒ \(\frac{0+100}{2}=50^{\circ} \mathrm{C}=50+273=323 \mathrm{~K}\)

∴ The temperature of the gas when flasks are in boiling water = 100 + 273 = 373 K and pressure = 0.5 atm The average temperature of the gas when one flask is in ice and the other in boiling water

⇒ \(\frac{0+100}{2}=50^{\circ} \mathrm{C}=50+273=323 \mathrm{~K}\)

Question 22. Equal mass of two gases A and B are kept in two separate containers under the same conditions of temperature and pressure. If the ratio of molar masses of A and JB is 2:3, then what will be the ratio of volumes of the two containers?
Answer:

Let the volume of the container holding gas A = VA and that of the container holding gas B = VB. Suppose, the molar masses of A and B are MA and MB, respectively. Thus, for equal mass (say ‘m’) of each gas, the number of moles of

⇒ \(A\left(n_A\right)=\frac{m}{M_A} \text { and that of } B\left(n_B\right)=\frac{m}{M_B}\)

∴ \(\text { For } A: P V_A=\frac{m}{M_A} R T \text { and for } B: P V_B=\frac{m}{M_B} R T\)

∴ \(\frac{V_A}{V_B}=\frac{M_B}{M_A}=\frac{3}{2}\)

Hence, the ratio of volumes ofthe containers = 3:2

Question 23. At constant pressure for a given amount of gas, will the graphs obtained by plotting V vs t°C and V vs TK be different?
Answer:

According to Charles’ law, V = KT – [1] [at constant pressure for a given mass of a gas]

But, T = 273 + t, hence, V = K(273 + t) -[2] Both the equations [1] & [2] express equations for straight lines. Equation no. [1] represents a straight line passing through the origin. Equation no. [2] does not represent a straight line passing through the origin. The plot of V vs t results in a straight line that cuts the V-axis at 0°C. If this straight line is extrapolated backwards, it meets the y-axis at -273°C.

From the, above two graphs it is evident that there are no actual differences between the two graphs, because -273°C and OK express the same temperature.

Question 24. Under similar conditions of temperature and pressure, if the time taken for effusion of the same volume of H₂, N₂ and CO₂ gas through the same porous wall are t₁, t₂ and t₃  respectively, then arrange t₁, t₂ and t₃ in their increasing order.
Answer:

At constant temperature and pressure, if V volume of H₂ gas effuses in time t j, then according to Graham’s law, \(\frac{V}{t_1} \propto \frac{1}{\sqrt{M_{\mathrm{H}_2}}}\)

For N₂ gas and CO₂ gas, the equations are respectively,

⇒ \(\frac{V}{t_2} \propto \frac{1}{\sqrt{M_{\mathrm{N}_2}}}\)

⇒ \(\frac{V}{t_3} \propto \frac{1}{\sqrt{M_{\mathrm{CO}_2}}}\)

Volume of gas effused in each case is the same From equations [1] & [2]; from equations [2] & [3] we get \(\frac{t_2}{t_1}=\sqrt{\frac{M_{\mathrm{N}_2}}{M_{\mathrm{H}_2}}}; \frac{t_3}{t_2}=\sqrt{\frac{M_{\mathrm{CO}_2}}{M_{\mathrm{N}_2}}}\)

Since, MH₂ < MN₂ < Mc₂, therefore, t₂ > and t₂ > t₂ Hence, t₁,t₂ and t₂ will be in order: t₁<t₂<t₂

Question 25. If, at a given temperature, the total kinetic energy of the molecules in a unit volume of an ideal gas is E, show that the pressure of the gas, P = 2/3E.
Answer:

According to the kinetic gas equation, \(\mathrm{PV}=\frac{1}{3} \mathrm{mnc}^2\)

Suppose, m = mass of each molecule of the gas and n = number of molecules present in V volume of the gas, c = root mean square velocity of gas molecule

∴ \(P V=\frac{2}{3} \times n \times \frac{1}{2} m c^2 \quad \text { or, } P=\frac{2}{3} \times\left(\frac{n}{V}\right) \times \frac{1}{2} m c^2\)

where \(\frac{n}{V}\) the number of molecules per unit volume and \(\frac{1}{2} m c^2\) = the average kinetic energy of each molecule

∴ \(\left(\frac{n}{V}\right) \times \frac{1}{2} m c^2\) hre total kinetic energy of the molecules present per unit volume = E

∴ P = 2/3E (Proved)

Question 26. For the molecules of a given gas at a constant temperature, arrange the most probable velocity (cm), root mean square velocity (CRM) and average velocity (CFL) in the order of their increasing values. With the increase in temperature, will the ratio of these velocities increase, decrease or remain constant? What will the effect of increasing temperature be on the value of (crnls- cm) for a given gas?
Answer:

1. At temperature T, the most probable velocity (cm) ofthe gas molecules = \(\sqrt{\frac{2 R T}{M}}\) [M – molar mass ofthe gas] rms velocity \(\left(c_{r m s}\right)=\sqrt{\frac{3 R T}{M}}\) average velocity \(\left(c_a\right)=\sqrt{\frac{8 R T}{\pi M}}\)

∴ \(c_{r m s}>c_a>c_m\)

2.  Since the value of each of the velocities, crms’ ca and cm’ proportional to Jf, their ratio remains unaltered with temperature rise.

3. Suppose, at temperature, riK, the root mean square velocity (arms) and most probable velocity (cm) of the molecules of a gas are (arms)j and (cm) respectively. If the difference between the values of these velocities is Axx, then

⇒ \(\Delta x_1=\sqrt{\frac{3 R T_1}{M}}-\sqrt{\frac{2 R T_1}{M}}=\sqrt{\frac{R T_1}{M}}(\sqrt{3}-\sqrt{2})\)

Let the temperature of the gas be raised to T2K and at this temperature, the difference between the values of these two velocities is Ax2, then,

⇒ \(\Delta x_2=\left(c_{r m s}\right)_2-\left(c_m\right)_2=\sqrt{\frac{R T_2}{M}}(\sqrt{3}-\sqrt{2})\)

Since T₂ > , Ax₂ > Ax₁. Hence, the difference between the values of these velocities increases with the temperature rise.

Question 27. A and B are closed flasks having the same volume. In flask A, O₂ gas is present at TK and 1 atm pressure. In flask B, H₂ gas is present at 1 atm pressure. If these gases behave ideally, then compare their O total kinetic energies, total number of molecules, and root mean square velocities.
Answer:

Suppose, M₁ and n₂ are the number ofmoles ofthe gases present in flask A and flask B, respectively. Applying the ideal gas equation to the gases O₂ and H₂, we obtain

⇒ \(P V=n_1 R T \text { and } P V=n_2 R \frac{T}{2}\)

∴ \(\frac{n_1}{n_2}=\frac{1}{2}\)

The total kinetic energy of the molecules of mol O₂ in flask A, \(E_1=n_1 \times \frac{3}{2} R T\) and the total kinetic energy of the molecules of n₂ mol H₂ in flask B, \(E_2=n_2 \times \frac{3}{2} R \times \frac{T}{2}\)

[ since Total kinetic energy of the molecules of1mol gas =[Rx absolute temperature]

∴ E₁=E₂

So, the total kinetic energy of the molecules of O₂ gas = the total kinetic energy of the molecules of H₂ gas

If n₁ and n₂ be the number of molecules of O₂ gas and H₂ gas respectively then n1 = nxN and n₂ = n₂xN [AT= Avogadro’s number]

∴ \(\frac{n_1^{\prime}}{n_2^{\prime}}=\frac{n_1 \times N}{n_2 \times N}=\frac{1}{2}\)

∴ Number of molecules in H₂ gas = 2 x number of molecules in oxygen gas

ms velocity of O₂, Crms \(=\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 R T}{32}}\)

rms velocity of H₂, Cms \(=\sqrt{\frac{3 R \frac{T}{2}}{2}}=\sqrt{\frac{3 R T}{4}}\)

∴ \(\frac{c_{r m s}\left(\mathrm{O}_2\right)}{c_{r m s}\left(\mathrm{H}_2\right)}=\sqrt{\frac{4}{32}}=\sqrt{\frac{1}{8}}=\frac{1}{2 \sqrt{2}}\)

Question 28. Which one of the gases, under the given conditions, exhibits real gas behaviour?

1. 0.25 mol CO₂, T = 1200K, 1. = 24.63 atm, V=1 L
2. 1.0 mol SO₂, T = 300 K, P = 50 atm, V = 0.35 L

Answer:

At a certain temperature and pressure, if the compressibility factor (Z) of a gas is less than or more than 1, then at that temperature and pressure the gas behaves as a real gas. If Z = 1, then the gas exhibits ideal behaviour

1. For CO₂ gas \(Z=\frac{P V}{n R T}=\frac{24.63 \times 1}{0.25 \times 0.0821 \times 1200}=1\)

2. For SO₂ gas ,Z \(=\frac{P V}{n R T}=\frac{50 \times 0.35}{1 \times 0.0821 \times 300}=0.71\)

Under the given conditions, for CO₂ gas, Z = 1 and hence it shows ideal behaviour. In the case of SO₂, Z < 1. So, under the given conditions, SO₂ behaves as a real gas.

Question 29. Write the van der Waals equation for a real gas containing n molecules.
Answer:

If v is the volume of mol of a real gas at a pressure P and temperature T then the van der Waals equation for the gas is

⇒ \(\left(P+\frac{a}{v^2}\right)(v-b)=R T\)

Suppose, the volume for a real gas containing n molecules = V. So, volume for n/N moles of real gas = V [N= Avogadro’s number]

Therefore, the volume for 1 mol of real gas \(=\frac{V \times N}{n}=v\)

Substituting the value of v in equation [1], we obtain \(\left(P+\frac{n^2 a}{V^2 N^2}\right)(V N-n b)=n R T\)

This is the van der Waals equation for a real gas containing n molecules

Question 30. What will the value of compressibility factor (Z) be for a gas if the pressure correction term in the van der Waals equation for the gas is neglected?
Answer:

For 1 mol of a real gas \(\left(P+\frac{a}{V^2}\right)(V-b)=R T\)

If the pressure correction term is neglected, then \(P+\frac{a}{V^2} \approx P\)

∴ P(V-b) = RT or, PV -Pb = RT

Or, \(P V=R T+P b \quad \text { or, } \frac{P V}{R T}=1+\frac{P b}{R T}\)

∴ \(Z=1+\frac{P b}{R T}\)

Question 31. The value of van der Waals constant ‘a’ for nitrogen gas is 1.37 L²-atm mol^-2, but that for ammonia gas is 4.30 L_²-atm-mol^-2 What is the reason for this large difference? Which one of these two gases would you expect to have a higher critical temperature?
Answer:

Since N₂ is a non-polar molecule, the only attractive forces that operate between the molecules in N₂ gas are weak London forces. On the otherhand, NH₃ is apolarmolecule. So, the molecules in NH₃ gas experience dipole-dipole attractive forces in addition to London forces.

Hence, the intermolecular forces of attraction in NH₃ gas are much stronger than those in N₂ gas, and this makes the value of ‘a’ for NH₃ gas higher. The critical temperature of a gas depends upon the strength of intermolecular forces of attraction in the gas.

The stronger the intermolecular forces of attraction in a gas, the higher the critical temperature of the gas. Since the intermolecular forces of attraction are stronger in NH₃ gas than those in H₂ gas, the critical temperature of NH₃ will be higher than that of H₂.

Question 32. The values of ‘a’ and ‘b’ for three A real gases B A, Band C Care—

1. Which one of these gases has the largest molecular size?
2. Which one of these will behave most like an ideal gas at STP?

Answer:

1. The volume of a molecule depends on its radius. Now we know b ∝ r₃. Hence, the molecular size will be the largest for the gas which has the highest value of ‘b’. From die given values, it is found that gas B has the highest value of ‘b’ So, so the molecule ofthe gas will be the largest.

2. A gas with small values of ‘a’ and ‘b‘ behaves close to an ideal gas. For gas A, these two quantities have the smallest values. Hence at STP gas A will show the most ideal behaviour.

Question 33. At 0°C, the density of a certain oxide of a gas at 2 bar is the same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Answer:

The equation expressing the relation between density (d), pressure (P), absolute temperature (T) and molar mass (M) of a gas is d \(=\frac{P M}{R T}\)

Density ofthe oxide of the gas, \(d_1=\frac{2 \times M}{R \times 273}\)

Density of N₂ gas, d₂ \(=\frac{5 \times 28}{R \times 273}\)

It is mentioned that under the given conditions, d₁ = d₂

∴ \(\frac{2 \times M}{n \times 273}=\frac{5 \times 28}{R \times 273} \quad \text { or, } M=70\)

Therefore, the molecular mass ofthe oxide is 70 g-mol^-1

Question 34. The pressure of Ig of an Ideal gas A at 27°C Is found to be 2 bar. When 2 g of another ideal gas B Is introduced in the same flask at the same temperature, the pressure becomes 3 bar. Find a relationship between their molecular masses
Answer:

Suppose, the molar masses of gases A and B are M_A and M_B g-mol-1 respectively.

⇒ \(\lg \text { of } A=\frac{1}{M_A} \text { mol of } A \text { and } 2 \mathrm{~g} \text { of } B=\frac{2}{M_B} \text { mol of } B\)

Using the ideal gas equation (PV = NRT) for the gas A and the gas mixture of A and B, we have—

⇒ \(2 \times V=\frac{1}{M_A} R T\)……………………………..(1)

⇒ \(B=\frac{2}{M_B}\)……………………………..(2)

Dividing Equation (2) by (1), we have

1.5 = \(\frac{M_B+2 M_A}{M_B}\)

Or,\(2 M_A=0.5 M_B\)

Or,\(\frac{M_A}{M_B}=\frac{1}{4}\)

Question 35. The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20°C and lbar will be released when 0.15g aluminium reacts?
Answer:

The following reaction involving aluminium (Al) and caustic soda (NaOH) produces H₂ gas

⇒ \(\begin{array}{ll}
2 \mathrm{Al}(s)+2 \mathrm{NaOH}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow 2 \mathrm{NaAlO}_2(a q)+ & 3 \mathrm{H}_2(g) \\
2 \times 27 \mathrm{~g} & 3 \times 1 \mathrm{~mol}
\end{array}\)

According to this reaction 54g of Al = 3 mol of H₂

∴ \(0.15 \mathrm{~g} \text { of } \mathrm{Al} \equiv \frac{3}{54} \times 0.15 \equiv 8.33 \times 10^{-3} \mathrm{~mol} \mathrm{H}_2\)

To calculate the volume of the liberated H₂ gas, we apply the ideal gas equation, PV = nRT.

Given:

P = bar and T = (273 + 20)K =293K

The number of moles of liberated H₂ gas (n) =8.33 × 10^-3 mol

∴ \(V=\frac{n R T}{P}=\frac{8.33 \times 10^{-3} \times 0.0821 \times 293}{0.987} \mathrm{~L}=0.203 \mathrm{~L}=203 \mathrm{~mL}\)

Question 36. What will be the pressure exerted by a mixture of 3.2g of methane and 4.4g of carbon dioxide contained in a 9dm3 flask at 27°C?
Answer:

Total number of moles (n) in the mixture

⇒ \(\left(\frac{3.2}{16}+\frac{4.4}{44}\right)\) 0.3 mol

[Molar mass: \(\mathrm{CH}_4 \Rightarrow 16, \mathrm{CO}_2 \Rightarrow 44\)

As given, V = 9 dm3, T = (273 + 27)K = 300K, P = ?

So, \(p=\frac{n R T}{V}=\frac{0.3 \times 0.0821 \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 300 \mathrm{~K}}{9 \mathrm{dm}^3}\)

= 0.821 aztm [ 1=1dm³]

= 8.316 × 10⁴

pa[1 atm=1.013 × 10⁵pa]

Question 37. What will be the pressure of the gaseous mixture when 0.5L of H₂ at 0.8 bar and 2.0L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C?
Answer:

To calculate the number of moles of unmixed H₂ and O₂ m gases, we apply the ideal gas equation, PV = nRT

In the case of H₂, P = 0.8bar, V = 0.5L and T = (273 + 27)K = 300K

And in case of O₂, P = 0.7bar, V = 2.0L and T = (273 + 27)K = 300K

∴ \(n_{\mathrm{H}_2}=\frac{P V}{R T}=\frac{0.8 \times 0.5}{R \times 300}\)

= \(\frac{0.4}{300 R} \mathrm{~L} \cdot \text { bar }\) and

⇒ \(n_{\mathrm{O}_2}=\frac{P V}{R T}=\frac{2 \times 0.7}{R \times 300}\)

=\(\frac{1.4}{300 R} \mathrm{~L} \cdot \text { bar }\)

In the mixture of H₂ and O₂, the total number of mol

= \(n_{\mathrm{H}_2}+n_{\mathrm{O}_2}\)

=\(\frac{1}{300 R}(0.4+1.4) \mathrm{L} \text { bar }\)

=\( \frac{1.8}{300 R} \mathrm{~L} \cdot \text { bar }\)

For this mixture V = 1L and T = (273 + 27)K = 300K

If the pressure of the mixture is P, then

P = \(\frac{n R T}{V}\)

= \(\frac{1.8}{300 R} \times \frac{R \times 300}{1}\) bar

=1.8bar

Question 38. The density of a gas is found to be 5.46 g/dm₃ at 27°C and 2 bar pressure. What will be its density at STP?
Answer:

Density (d), pressure (P), absolute temperature (T) and molar mass of a gas (M) are related by,

d = PM/RT

Under the conditions of T = (273 + 27)K = 300K and

P = 2 bar, the density of the gas

(d) = 5.46 g.dm^-3

∴ = 5.46 g.dm^-3 = \(\frac{(2 \times 1.013) \mathrm{atm} \times M}{0.0821 \mathrm{~atm} \cdot \mathrm{dm}^3 \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 300 \mathrm{~K}}\)

M = 66.37 g.mol¯¹

At STP, if the density of the gas is d₁, then

⇒ \(d_1=\frac{P M}{R T}=\frac{1 \times 66.37}{0.0821 \times 273}\)

At STP, T = 273K P =  1 atm

= \(2.96 \mathrm{~g} \cdot \mathrm{L}^{-1}\)

= \(2.96 \mathrm{~g} \cdot \mathrm{dm}^{-3}\)

Question 39. 4.05mL of phosphorus vapour weighs 0.0625g at 546°C and 0.1 bar pressure. What is the molar mass of phosphorus?
Answer:

Suppose, the molar mass of phosphorus =M g-mol^-1

So, 0.0625g of phosphorus

= \(\frac{0.0625}{\mathrm{M}} \mathrm{mol}\) of phosphorus As given, P = 0.1 bar, T = (273 + 546)K

= 819K, V = 34.05 mL = 34.05 × 10^-3L

Therefore, 0.1 bar × 34.05 × 10^-3L.

⇒\(\frac{0.0625}{M} \mathrm{~mol} \times 0.082 \mathrm{~L} \cdot \mathrm{bar} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 819 \mathrm{~K}\)

or, M = 1232.71, Hence, Molar mass = 1232.71g.mol^-1

Question 40. A student forgot to add the reaction mixture to the round-bottomed flask at 27°C but instead, he/she placed the flask on the flame. After a lapse of time, he realised his mistake, and using a pyrometer he found the temperature of the flask was 477°C. What fraction of air would have been expelled out?
Answer:

Suppose the number of mol of air inside the flask at 27°C is n1 and that at 477°C is n₂. Since the flask is opened to air, the pressure of air inside the flask at 27°C and 477°C is the same as that of the atmospheric pressure. Let this pressure be P. Again on heating the volume of a round bottom flask remains the same. Hence, the volume ofthe flasks the some at both 27°C and 477°C. Let this volume be V.

Applying the ideal gas equation we have, PV= n₁ R(273 + 27)

= HJ × 300R n₂PV = n₂ (273 + 477) = n₂ × 750R

Hence, n₂ × 300R = n₂ × 750R

Or, n₂ = \(\frac{2}{5} n_1\)

∴ Fraction of air that would have been expelled out = \(\frac{n_1-\frac{2}{5} n_1}{n_1}=\frac{3}{5}\)

Question 41. Payload Is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the payload when a balloon of radius 10 m, mass 100 kgs filled with helium at 1.66 bar at 27°C. (R = 0.083 bar. dm³. K mol^-1 and density of air = 1.2kg.m^-3
Answer:

Volume (V) of the balloon =\(\frac{4}{3} \pi r^3=\frac{4}{3} \pi \times(10)^3 \mathrm{~m}^3\)

= 4186.66m³

= 4186.66 × 10³ dm³

To calculate the number of moles of He gas enclosed in the balloon, we apply the ideal gas equation, PV = nRT

Given: P = 1.66 bar, T = (273 + 27)K = 300K So, according to the equation, PV = nRT,

n = \(\frac{P V}{R T}=\frac{1.66 \times 4186.66 \times 10^3 \mathrm{bar} \cdot \mathrm{dm}^3}{0.083 \mathrm{bar} \cdot \mathrm{dm}^3 \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 300 \mathrm{~K}}\)

= \(279.11 \times 10^3 \mathrm{~mol}\)

Hence, the mass of 1<< gas enclosed in the balloon

= 4  × 279. 11 × 10³g = 1,110 × 10⁶g 1116kg

Therefore, the mass of the balloon filled in with He gas

= (100+ 1116)1kg = 1216kg

The volume of air displaced by the balloon = Volume of the balloon

= 4186.66 × 10³ dm³

Density of air – 1,2 kg. m^-3 = 1.2 × 10^-3 kg.dm^-3

The mass of air displaced by the balloon

= 4180,60 × 10³  × 1,2 × 10^-3 kg =5024kg

Hence, payload = the mass of displaced air- the mass of the balloon = (5024- 1216)kg = 3808 kg

Question 42. Calculate the volume occupied by 8.8 g of C02 at 31.1 °C &1 bar pressure. t=0.083 bar -L-K^-1mol^-3
Answer:

⇒ \(8.8 \mathrm{~g} \mathrm{CO}_2=\frac{8.8}{44}=0.2 \mathrm{~mol} \mathrm{CO}_2\)

Applying the ideal gas equation (PV = NRT) to calculate volume, we have

⇒  \(V=\frac{n R T}{P}=\frac{0.2 \times 0.083 \times(273+31.1)}{1}=5.048 \mathrm{~L}\)

∴ The volume of 8.8 g CO₂ at 31.1°C and 1 bar pressure = 5.048L.

Question 43. 2.9 g of gas at 95°C occupied the same volume as 0.184g of dihydrogen at 17°C, at the same pressure. What is the molar mass of the gas?
Answer:

Let the molar mass ofthe unknown gas =M g-mol^-1 Number ofmoles for 2.9g ofthe gas \(=\frac{2.9}{M} \mathrm{~mol}\)

Number of moles for 0.184g of H₂ gas \(=\frac{0.184}{2}=0.092 \mathrm{~mol}\)

As both the gases have the same volume at the same pressure and specified temperature, they will have the same value of PV. To calculate PV, we apply the equation

⇒  PV = nRT. For unknown gas \(P V=\frac{2.9}{M} R(273+95)\)

And for He Gas, Pv [= 0.0932R(273+17)

Therefore \(\frac{2.9 \mathrm{R}}{M} \times 368=0.092 R \times 290 \quad \)

Hence, the molar mass ofthe unknown gas = 40g-mol^-1

Question 44. Oxygen gas is present in a 1L flask at a pressure of 7.6 x 10¯1° mm Hg at 0°C. Calculate the number of O₂ molecules in the flask
Answer:

P= \(\frac{7.6 \times 10^{-10}}{760}\)

∴ \(n=\frac{P V}{R T}=\frac{7.6 \times 10^{-10} \times 1}{760 \times 273 \times 0.082}\)

[V = 1L, T = 273K, n = 4.46× 10^-14]

Question 45. Sketch PIT versus T plot for an ideal gas at constant volume. Indicate the value ofthe slope (mass fixed). Under the same conditions of temperature and pressure NH₃, Cl₂ and CO₂ gases are allowed to diffuse through a porous wall. Arrange these gases in the increasing order ofthe rate of diffusion.
Answer:

According to Gay-Lussac’s law, \(\frac{P}{T}=K\) (constant), when the mass and volume of a gas are constant. Hence, the plot of \(\frac{P}{T}=K\) versus T indicates a straight line parallel to the X -axis and the slope ofthe curve is tan 0 = 0.

The order of molar masses of NH₃ Cl₂ and CO₂ gases are—NH₃ < CO₂ < Cl₂.

Hence, the increasing order of their rates of diffusion is Cl₂ < CO₂ < NH₂.

The value of ‘R ‘ in J-K^-1-mol^-1 unit is 8.314

Question 46. A gas of molar mass 84.5g/mol is enclosed in a flask at 27°C has a pressure of 2atm. Calculate the density of the gas. [R = 0.082L.atm- K^-1.mol^-1 ]
Answer:

The density of a gas,

d =  \(\frac{P M}{R T}=\frac{2 \times 84.5}{0.082 \times 300} \mathrm{~g} \cdot \mathrm{L}^{-1}\)

=\(6.87 \mathrm{~g} \cdot \mathrm{L}^{-1}\)

Question 47. For which property of the liquid the shape of a liquid drop is spherical? A 10-litre volumetric flask contains 1 gHe and 6.4 g of O₂ at 27°C temperature. If the total pressure of the mixture is 1.107 atm, then what is the partial pressure ofHe and O₂?
Answer:

Surface tension.

⇒ \(n_{\mathrm{He}}=\frac{1}{4}=0.25 \mathrm{~mol}, n_{\mathrm{O}_2}=\frac{6.4}{32}=0.2 \mathrm{~mol}\)

∴ Total number of moles =(0.25 + 0.2) = 0.45

⇒\(x_{\mathrm{He}}=\frac{0.25}{0.45}=0.5555 \text { and } x_{\mathrm{O}_2}=\frac{0.2}{0.45}=0.4444\)

pHe = 0.5555 × 1.107 atm = 0.615 atm

Po2 = 0.4444 × 1.107 atm =0.491 atm

Question 48. What will be the ratio of \({ }^{235} \mathrm{UF}_6\) diffusion of And \({ }^{235} \mathrm{UF}_6\)
Answer:

Let, the rate of diffusion of \({ }^{235} \mathrm{UF}_6 \text { and }{ }^{238} \mathrm{UF}_6\) be r1 and r2.

According to Graham’s law, \(\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}\)

⇒ \(M_1=\text { molecular mass of }{ }^{235} \mathrm{UF}_6=349 \mathrm{~g} \cdot \mathrm{mol}^{-1}\)

⇒ \(M_2=\text { molecular mass of }{ }^{238} \mathrm{UF}_6=352 \mathrm{~g} \cdot \mathrm{mol}^{-1}\)

∴ \(\frac{r_1}{r_2}=\sqrt{\frac{352}{349}}=1.004\)

Question 49. Determine the volume of 2.2 g of carbon dioxide at 27°C and 570 mmHg pressure.
Answer:

⇒ \(2.2 \mathrm{~g} \text { of } \mathrm{CO}_2 \equiv \frac{2.2}{44}=0.05 \mathrm{~mol} \text {. }\)

According to the ideal gas equation (taking CO2 as an ideal gas), PV = nRT

⇒ \(\text { or, } \frac{570}{760} \times V=0.05 \times 0.082 \times 300\)

or, v= 1.64 The volume of CO₂ at the given condition is 1.64L.

CBSE Class 11 Chemistry Notes For Chapter 6 Chemical Thermodynamics Introduction

Energy is the ability of a physical system to do work. It always remains conserved in any event, i.e., it cannot be created or destroyed. However, it can take different forms such as mechanical energy, chemical energy, electrical energy, heat energy, light energy, sound energy, etc. Energy can be converted from one form to another.

Thermodynamics is the study to find out the conditions of interconversion of different forms of energy as well as to predict the extent of their conversions.

The branch of science which deals with the interconvertibility of different forms of energy (mainly heat and work) is called thermodynamics. The Greek words ‘thermo’ means heat and ‘dynamics’ means power.

Thermodynamics is based on four fundamental rules or laws namely zeroth law, first law, second law, and third law of thermodynamics. These laws are based on natural experience gathered over centuries. As the laws of thermodynamics are derived from the direct human experience, it is sometimes called axiomatic science.

Class 11 Chemistry Notes For Chapter 6 Chemical Thermodynamics

Energy which we generally use in our daily lives is mostly obtained from chemical reactions. Thus the study of energy changes in chemical reactions is considered to be an important topic in chemistry.

The branch of thermodynamics that deals with energy evolved or absorbed during physical or chemical transformation is called chemical thermodynamics or chemical energetics.

Importance Of Thermodynamics

Some important applications of thermodynamics are:

• In thermodynamics, we study the inter-relationships among various macroscopic variables like pressure, temperature, volume, etc., and the changes in these variables as a result of various processes.
• Energy changes associated with physical or chemical transformations can be explained using thermodynamics.
• Under a given set of conditions, the feasibility of a physical or chemical transformation can be predicted from thermodynamics.
• Thermodynamics can predict the extent to which a physical or chemical transformation occurs before it reaches an equilibrium state.
• The relative amount of the reactants and products at the equilibrium of a reaction as well as the value of the equilibrium constant of the reaction can be determined by thermodynamics.
• Under a given set of working conditions, the maximum efficiency of a heat engine can be determined by thermodynamics.
• The ideal condition(s) for the transformation of different types ofenergy can be determined by thermodynamics.

Limitations of thermodynamics

1. Classical thermodynamics applies only to the macroscopic system (For example in the case ofa few grams of ice) and not to the microscopic system (For example in the case of a few molecules or atoms). So it cannot give us any idea regarding the structure of matter.
2. Thermodynamics does not give us any information about the rate or velocity ofa process (such as chemical reaction, osmosis, etc.). Also, it cannot give us any idea about the mechanism or path ofa chemical reaction.

Class 11 Chemistry Notes For Chapter 6 Terms And Concepts Related To Thermodynamics

Some terms are frequently used in thermodynamics. One should have a good knowledge of these terms before going through the subject of thermodynamics. Here are these terms and their brief explanations.

System, Surroundings, and Boundary

System:

In thermodynamics, a system is defined as the part of the universe under study, which is separated from the rest of the universe by real or imaginary boundaries- System

System, Surroundings, and Boundary Explanation:

If we study something 2 about the human body, then the M human body is considered to be the system. Similarly, if we perform an experiment living cell or one mole of water, then the living cell or one mole of water will be the system.

Surroundings:

Everything outside the system In the universe Is called the surroundings. Once the system Is defined, the surroundings will be defined automatically.

System + Surroundings=Universe

Boundary:

Heal or Imaginary surface that separates a system from its surroundings is called the boundary of the system.

Explanation:

If a certain amount of water taken In a beaker is considered to be the material of our experiment, then water will be our system.

• The rest of the universe including the beaker will be the surroundings of the lire system.
• The interface between water and glass and that between water and air are the boundaries between the system and its surroundings.
• If a gas enclosed in a cylinder fitted with a piston is considered to be the object of our analysis, then the gas will be our system, and the rest of the universe including the cylinder and the piston will constitute the surroundings of the system.

1 lore, the inner surfaces of the walls of the die cylinder, and the piston are the boundaries between the system and its surroundings.

Role of boundary:

The boundary of a system plays an important role when there occurs an interaction between the system and its surroundings. During the interaction, a system exchanges energy or matter or both matter and energy with Its surroundings through its boundary.

Characteristics of the boundary of a system:

• The boundary of the n system may be real or imaginary.
• It may be rigid or non-rigid, in the case of a rigid boundary, the volume of the system does not change, while In the case of a non-rigid boundary, the volume of the system can change.
• Tim boundary of a system may be permeable or Impermeable. In the case of a permeable boundary, the exchange of matter takes place between the system and its surroundings through the boundary, whereas in the case of an impermeable Imundury, no exchange of matter can take place through the boundary.
• Exchange of beat may take place between a system and US surroundings through the boundary. A boundary that permits the cycling of heat between a system and Its surroundings Is called an antennal or diathermic boundary. On the other hand, a boundary that does not permit the exchange of heat between the system and its surroundings is called an adiabatic boundary.
• In reality, a perfect adiabatic boundary is not possible. The wall of a Dewar flask nearly behaves as an adiabatic boundary. According to the definition, the surroundings mean everything outside the system.
• But in thermodynamics, the surroundings are considered as the portions of the universe around the system, upto which a change occurring in the system has its influence, For Example, the burning of a candle produces light and heat.
• When we consider the effect of light, the tire space enclosing the candle, which is illuminated by Uie light, will be the surroundings. On the other hand, if we consider the effect of heat (produced), a narrow space affected by the produced heat, will be the surroundings

Types of system

Depending on the nature of its boundary, a system may or may not exchange matter energy, or both with its surroundings. Based on the exchange of matter and energy with the surroundings, systems may be classified as open systems, closed systems, and isolated systems.

1. Open system:

A system that can exchange both energy and matter with its surroundings is called an open system.

Examples: Every living being in nature:

Every living being (system) takes food (matter) and excretes waste materials (matter) to the surroundings. They (systems) also exchange heat (energy) with the surroundings.

Some water (or any other liquid) in an open container:

• In an open atmosphere, water (system) continuously evaporates and water vapor (matter) escapes from the container to the air (surroundings). Also, O₂ or CO₂ (matter) from the air may dissolve into water (system).
• So, the exchange of matter takes place between the system and its surroundings.
• If the temperature of water (system) is different from that of its surroundings, then there occurs an exchange of heat (energy) between water and its surroundings.

The ocean:

The Ocean Is a perfect example of an open system. Water (matter) evaporates from the ocean to the atmosphere and then again is added to it in time of rain. The ocean also absorbs heat (solar energy) and releases its energy in the form of latent heat.

2. Closed system:

A system that can exchange energy with its surroundings but not matter is called a closed system.

Examples: Boiling of water in a closed glass/metallic container:

• During boiling, water vapor (matter) cannot escape to the surroundings from the container, also any matter from the surroundings cannot enter into the system.
• So, the exchange of matter is not possible between the system and its surroundings. But if the temperature of the water (system) is different from the surroundings, then heat (energy) will be exchanged between the system and its surroundings through the boundary wall of the container.
• So, the exchange of energy is possible between the system and its surroundings.

A gas enclosed in an impermeable metallic cylinder fitted with a piston:

Since the cylinder and piston are impermeable, matter will not be exchanged between the system and its surroundings.

• But heat (energy) will be exchanged between the system and its surroundings through the wall and piston of the cylinder.
• If a pressure greater than the pressure of the system is applied on the piston, work is done on the system. As a result, energy in the form of work is transferred to the system from its surroundings.
• On the other hand, if the pressure on the piston is kept lower than that of the system, the system does the work on its surroundings, and energy in the form of work is transferred from the system to its surroundings.

Isolated system:

A system that can exchange neither energy nor matter with its surroundings is called an isolated system.

An isolated system does not interact with its surroundings as its boundary is impervious to all matter and does not permit energy to pass through it As a result, any change occurring in an isolated system does not influence the surroundings, and vice-versa

Examples: Hot tea kept in a sealed & thermally insulated flask (Dewar flask):

Since the flask is closed, matter cannot be transferred from the system (hot tea) to its surroundings, and vice-versa. As the walls ofthe flask are thermally insulated, the system cannot exchange energy with its surroundings.

Water taken in a closed container having rigid, impermeable, and thermally insulated walls:

Since the container is closed and its walls are impermeable, no matter can exchange between the system (water) and its surroundings.

Furthermore, the container is insulated. So, no heat can flow into or out of the system. As the walls are rigid, there will be no change in the volume of the system. So, no energy in the form of work can be exchanged between the system and its surroundings. A perfectly isolated system is a hypothetical concept because no wall is perfectly adiabatic.

Based on physical properties and chemical composition, a system may further be classified as—

• Homogeneous system: If the physical properties and chemical compositions are uniform throughout a system, then it is called a homogeneous system. For example, pure solid, liquid or gaseous substance, gas mixture, 2 completely miscible liquids like water, alcohol, etc.
• Heterogeneous system: If the physical properties and the chemical compositions are different in different parts of a system, then it is called a heterogeneous system. For example, the mixture of two immiscible liquids (water and benzene), a mixture of two solids (sugar and NaCl) etc.
• Extensive and intensive properties of a system: A system consisting of a large number of atoms, ions, or molecules is called a macroscopic system. For example, a certain amount of water, a certain volume of solution, a certain mass of sodium chloride, etc., are macroscopic systems. The properties associated with a macroscopic system are called macroscopic properties. For example, temperature, pressure, concentration, mass, density, composition, etc., are the properties ofa macroscopic system.

The macroscopic properties of a system can be classified into two categories:

1. Extensive properties
2. Intensive properties.

Extensive property:

Tile property which depends upon the mass (or size) of the system i.e., the quantity of matter present in the system is called an extensive property. Extensive properties are additive, i.e., the total value of an extensive property of a system is equal to the sum of the extensive properties of different parts ofthe system.

Examples:

Mass, volume, internal energy, enthalpy, entropy, heat capacity, Gibbs free energy, etc., are the extensive properties of a system.

The volume of a system:

If V is the volume of lg substance (system) at a particular temperature and pressure, then the volume of 5 g of the same substance will be 5 x V. So the volume of a system is an extensive property.

The number of moles ofa system:

If the number of moles of1 g of a substance (system) is ‘ri then the number of moles for 2 g of that substance will be ‘ 2n! So, several moles of a system is an extensive property.

The internal energy of a system:

When water transforms to ice, its internal energy in the form of heat is liberated. Under an identical set of conditions, the heat liberated in the transformation of 5 g of water to 5 g of ice is found to be five times as much as that liberated when 1 g of water transforms into 1 g of ice. This means that a 5 g sample of water contains five times as much internal energy as a 1 g sample of water does. Thus, the internal energy ofa system is an extensive property.

Intensive property:

The property that does not depend upon the mass (or size) of the system i.e., the amount of matter present in the system, is called an intensive property. Intensive property has the same magnitude at every point in a homogeneous system under equilibrium.

Examples:

Temperature, pressure, density, viscosity, molar heat capacity, refractive index, boiling point, freezing point, surface tension, Viscosity coefficient, molar volume, molar internal energy, molar enthalpy, molar entropy, molar free energy, mole fraction, color, concentration, thermal conductivity, specific rotation, standard reduction potential, etc.

The boiling point of a liquid:

If we take different amounts of pure water in two different containers and determine their boiling points at a given pressure, we get the same boiling point for both although the amount of water is different. Thus, the boiling point ofa liquid does not depend upon the amount of the liquid, implying that it is an intensive property.

The density of a substance:

At a given temperature, the density of 1 kg of a pure sample of copper is the same as the density of lg of the same sample. Thus, the density of a substance does not depend upon the amount of substance, indicating that it is an Intensive property.

The concentration of a homogeneous solution:

If the concentration of a homogeneous solution is 1 g.L-1 at a particular temperature, then the concentration ofa drop of that solution will also be 1 g.L-1 at that temperature. Thus the concentration of a homogeneous solution does not depend upon the amount of solution. So, it is an intensive property.

Important points regarding intensive & extensive properties:

1. The ratio of two extensive properties is always an intensive property.

Explanation:

• The mass (m) and volume (V) of a system are extensive properties. But the ratio of these two, i.e., density \(\left(\frac{m}{V}=d\right)\) is an intensive property.
• Extensive property becomes intensive when It Is expressed in terms of per unit mass or unit mole.

Explanation:

Ifx is an extensive property for moles of a system, then the value of X per mole, \(X_m=\frac{X}{n}\), will be an intensive property because it denotes the value of X for 1 mol of the system and is independent of the amount of substance.

• The internal energy of a system is an extensive property, but internal energy per mole (molar internal energy) or internal energy per gram (specific internal energy) is an intensive property.
• The volume of a system is an extensive property, but volume per mole (molar volume) or volume per gram (specific volume) is an intensive property.
• Similarly, the heat capacity of a system is an extensive property, but heat capacity per mole (molar heat capacity) or heat capacity per gram (specific heat capacity) is an intensive property.

Class 11 Chemistry Notes For Chapter 6 Chemical Thermodynamics – Internal Energy And Its Change

Heat is released during the solidification of a liquid (for example when 1 g of water is transformed into ice at 0°C and late pressure, it releases 80 cal heat.) When coal burns, it produces both light and heat.

Mechanical work is done during the expansion of water vapor from high pressure to low pressure. We have many other examples like these, where we see that the system does work or produces heat or other form of energy out of its own intrinsic or inherent energy, without taking aid from an external energy source.

These observations indicate that every system contains some amount of energy intrinsically associated with it This intrinsic energy associated with every system or substance, is called its internal energy. Internal energy is denoted by the symbol either U or E.

Internal Energy And Its Change Definition

Every system, for its existence, is associated with an amount of energy in exchange for which it can do work or produce heat or another form of energy without the help of external energy. This energy is termed internal energy.

Origin of internal energy:

The constituent particles in a system possess kinetic and potential energies arising from different internal modes of motion such as translational motion, vibrational motion, rotational motion, and electronic motion. In addition, they also possess nuclear energy, bond energy, and energy due to intermolecular attractions or repulsions.

• All these energies contribute to the internal energy ofthe system.
• Therefore, the sum of all forms of energies ofthe constituent particles in a system gives rise to the internal energy ofthe system.
• The absolute value of the internal energy of a system cannot be measured experimentally because it is not possible to determine all the types ofenergy associated with the internal energy of a system.
• However, in a process, the change in the internal energy (AU) of a system can be determined experimentally.

Some important points about internal energy

1. Internal energy is an extensive property:

Explanation:

Internal energy ofa system is an extensive property because it increases as the amount of substance present in the system increases. For example, the internal energy of 5g of water is five times that of lg of water.

2. Internal energy depends upon the nature of the system:

Explanation:

The constituent particles (atoms, ions, and molecules) are different for different systems. This makes magnitudes of translational energy, rotational energy, vibrational energy, electronic energy, binding energy, etc. different for different systems.

Therefore, the values of internal energy will be different for different systems, even under identical conditions. For example, under identical conditions of temperature and pressure, the internal energy of one mole of O₂ gas Is different from that of one mole of N₂ gas.

3. Internal energy Increases with an increase in temperature:

Explanation:

With increasing temperature, the magnitudes of the translational motion, rotational motion, vibrational motion, etc., of constituents of a system increase. This results in an increase in energies associated with these motions. Consequently, the internal energy of the system increases.

4. Internal energy ofa system is a state function:

Explanation:

In a process, the change in internal energy (All) ofa system only depends upon the initial and final states of the system. It does not depend upon the path followed for carrying out the process. A process with a given initial and final states can be carried out in different paths but the change in internal energy will be the same in all paths.

For example, the combustion of glucose,

⇒  \(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(s)+6 \mathrm{O}_2(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(l)\),

Can be carried out either by its direct combustion in a reaction vessel, or by oxidizing it in presence of enzyme. However, the change in internal energy in each case will be the same as the initial and final states for both processes are identical.

Change in the internal energy of a system in a process Definition

Process, the change in internal energy of a system is defined as the difference in internal energies between the final and initial states of the system in the process.

Let us consider a process in which the internal energy of the system at the initial and final states are U₁ and U₂, respectively. Therefore, the change in internal energy of the system, ΔU = U₂-U₁

Change in internal energy of a system in a chemical reaction:

In a chemical reaction, the system is considered to be made up of reactants and products involved in the reaction.

At the beginning of the reaction, the system contains only reactants, and after the completion of the reaction, the system contains only products. Therefore, the change in internal energy in a chemical reaction equals the difference between the internal energies of the products and the reactants.

Let us consider a reaction:

A →B The change in internal energies of the reaction is given by,

⇒ \(\Delta \boldsymbol{U}=\overline{\boldsymbol{U}}_{\boldsymbol{B}}-\overline{\boldsymbol{U}}_A ; \text { where } \bar{U}_A \text { and } \bar{U}_B\)

Are the molar internal energies (internal energies per mole) of the product (B) and reactant (A), respectively

1. If then ΔU = negative, indicating an exothermic reaction.
2. If then ΔU = positive, indicating an endothermic reaction.

Both calorimeters are used to determine the change in internal energy in a combustion reaction.

• The change In internal energy in an isothermal expansion or compression of an ideal gas (system) is zero.
• When an ideal gas undergoes isothermal expansion or compression, the average distance between the molecules in the gas changes, but this change does not affect the internal energy of the gas because the molecules of an ideal gas do not experience intermolecular forces of attraction.
• Therefore, an isothermal expansion or compression of an ideal gas (system) does not cause any change in the internal energy of the gas.
• The change in internal energy of an ideal gas depends only upon temperature. At a constant temperature, the internal energy of an ideal gas is independent of its volume or pressure

Class 11 Chemistry Notes For Chapter 6 Chemical Thermodynamics – Heat Change In A Chemical Reaction

Like other processes, heat exchange with the surroundings also occurs in case of chemical reactions. The amount of heat absorbed or given off during a reaction depends on the conditions under which the reaction is carried out.

Generally, reactions are carried out either under the condition of constant volume or under the condition of constant pressure. However, the fact is that carrying out a reaction at constant volume is not as convenient as that at constant pressure.

In the laboratory, reactions are carried out in containers open to the atmosphere, so they occur under the condition of constant atmospheric pressure. According to the first law of thermodynamics, if a process involving only pressure-volume work occurs at constant volume, on heat change (qv) in the process equals the change in internal energy (AU) of the system.

Therefore, ΔU = qv or, = -qy. On the other hand, if a process occurs at constant pressure according to the first law of thermodynamics, the heat change in the process (qp) becomes equal to the change in enthalpy (ΔH) of the system. Therefore, ΔH = qp or, -ΔH= -qP

In a process, the change in internal energy is ΔU = U₂-U₁, and the change in enthalpy is ΔH – H₂– H₁ where subscripts 1 and 2 denote the initial state and the final state of the system, respectively, in the process. In a reaction system, only reactants are present at the beginning of the reaction and only products are present at the end of the chemical reaction.

Therefore, the internal energy or enthalpy at the beginning of a reaction means the total internal energy or enthalpy ofthe reactants undergoing the reaction. Similarly, the internal energy or enthalpy at the end of a reaction means the total internal energy or enthalpy of the products formed in the reaction. Therefore,

1. For a chemical reaction occurring at constant volume, the heat change (qv) = AU = (total internal energy of the products – total Internal energy of the reactants) \(=U_{\text {products }}-U_{\text {reactants }}=U_P-U_{R^*}\)

2. For a reaction occurring at constant pressure, the heat Change \(\left(q_P\right)=\Delta H=H_{\text {products }}-H_{\text {reactants }}=H_P-H_R \text {. }\)

As most of the reactions are carried out at constant pressure and heat change associated with constant pressure equals the change in enthalpy, the heat change of the reaction in a reaction usually signifies the change in enthalpy (ΔH) is the reaction unless the constant-volume condition is stated.

Exothermic and endothermic reactions

Exothermic reaction:

Reactions associated with the evolution of heat are called exothermic reactions. In an exothermic reaction, heat is released from the reacting system to the surroundings.

Thus, when we touch the reaction container (which is the part of the surroundings), in which the exothermic reaction is taking place, we feel warm. Since heat is released in an exothermic reaction, the total energy of the reactants is greater than that of the products.

Let us consider an exothermic reaction:

A + B→C + D. If the heat released in the reaction is’ q ‘ then, the total energy of the reactants = total energy of the products + q t.e., the total energy of A and H = the total energy of C and D + q.

If we imagine ‘q’ as the part of the die product, then the above equation can be written as

A + B →C + D + q The amount of heat evolved is usually written with a positive sign on the right-hand side of the balanced equation of the reaction.

Endothermic reaction) Reactions Associated with the absorption of hoot are called endothermic reactions, In an endothermic reaction. In the absorbed by the reading system to the surroundings. Thus, when we touch the reaction container (which Is the pan of the surroundings), in which an endothermic reaction Is taking place, we feel cold.

Since the absorption of bent occurs In mi endothermic reaction, the total energy of products will be greater than the total energy of the reactants. Let us consider an endothermic reaction; A + B→C+D.

Heat absorbed In this reaction he then the total energy of the reactants s the total energy if the products total energy of A and If = total energy of C and D- q.

If imagine ‘q’ as the part ofthe reactant, then the above equation can be written as:

A+B+QC+D Or, A+B+C+D-q

The amount of heat absorbed is usually written with a negative sign on the right-hand side of the balanced equation ofthe reaction.

Enthalpy change in exothermic reactions:

In an exothermic reaction, heat is released by the reacting system. Therefore, for any exothermic reaction occurring at constant pressure qp<0.

Since qp=ΔH, ΔH<0 Or, \(\Sigma H_{\text {products }}-\Sigma H_{\text {reactants }}<0 \quad \text { or, } \quad \Sigma H_{\text {products }}\) Thus the total enthalpy ofthe products is less than that ofthe reactants.

In an exothermic reaction, the enthalpy of the reaction system decreases (ΔH < 0). Enthalpy change in endothermic reactions:

In an endothermic reaction, heat is absorbed by the reaction system. Therefore, for any endothermic reaction occurring at constant pressure, qo>0.

Since \(q_p=\Delta H, \Delta H>0 \text { or, } \Sigma H_{\text {products }}-\Sigma H_{\text {reactants }}>0\text { or, } \Sigma H_{\text {products }}>\Sigma H_{\text {reactants }}\)

In an endothermic reaction, the enthalpy of the reaction system Increases (ΔH > 0).

Enthalpy diagram of exothermic and endothermic reactions In an exothermic reaction

⇒ \(\Sigma H_{\text {products }}<\Sigma H_{\text {reactants }}\) So, in its enthalpy diagram ,\(\Sigma H_{\text {products }}\) lies below the \(\Sigma H_{\text {reactants }}\) hand,

In an endothermic reaction,

⇒ \(\Sigma n_{\text {products }}>\Sigma H_{\text {reactants }}\) So, in Its enthalpy diagram, \(\Sigma H_{\text {products }}\) lies above the \(\Sigma H_{\text {reactions }}\)

A few examples of exothermic and endothermic reactions

1. Exothermic reactions and the values of All (at 25°C and 1 atm)

CH₄(s) + 2O₂ (g)→ CO₂ (g) + 2H₂O(l) ; ΔH = -890 kJ

C(graphite, s) + O₂(g) → CO₂ (g) ; ΔH = – 393.5 kJ

2H₂(g) + O₂ (g)→ H₂O(l) ;ΔH = – 571.6 kJ

H₂ (g) + Cl₂(g) → 2HCl(g) ; ΔH = -185 kJ

2. Endothermic reactions and the values of All (at 25°C and 1 atm)

N₂ (g) + O₂ (g)→ 2NO(g) ; ΔH = +180.5 kJ

CaCO₃(s) → CaO(s) + CO₂ (g) ; ΔH = +178.3V 0

C(coal, s) + H₂O(g)→ CO(g) + H₂(g) ; ΔH = +130 kJ

Thermochemical equations

Thermochemical equations Definitions

The thermochemical equation is a balanced chemical equation, in which the physical states of the reactant(s) and product(s) as well as the amount of heat evolved or absorbed in the reaction are mentioned.

Conventions for writing a thermochemical equation:

To indicate the physical states ofthe reactant (s) and the product(s), symbols s, l, and g are used for solid, liquid, and gaseous states, respectively. These symbols are to be placed within parentheses just after the chemical formulae ofthe substances concerned.

For any reactant or product dissolved in aqueous solution, the term ‘ aq ’ (short form of the word aqueous) is to be placed within a parenthesis just after its formula.

The amount of heat evolved or absorbed (or the enthalpy change) in a reaction is to be written with a proper sign (+ or -) on the right-hand side immediately after the balanced equation.

In an exothermic reaction, if x kj of heat is evolved, then +x kj or ΔH = -x kj is to be written on the right-hand side immediately after the balanced chemical equation.

Example:

⇒  \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)+890.3 \mathrm{~kJ}\)

Or, \(\text { or, } \mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

Δ H =-890.3 kJ

In an endothermic reaction, if +x kj of heat is absorbed, then -x kj or AH = +xkj is to be written on; the right hand immediately after the balanced equation.

Example:

⇒\(\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g)-180.5 \mathrm{~kJ}\) or, \(\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g)\)

Δ H=+180.5 kJ

The coefficients of the reactants and the products in a thermochemical equation indicate their respective number of moles This allows us to use fractional coefficients for reactants and products. Because ΔH is an extensive property, when the balanced equation is multiplied by a factor, the value of ΔH will also be multiplied by that factor.

⇒ \(2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l)\)

ΔH=-571.6 kJ,

Or, \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l), \Delta H=-285.8 \mathrm{~kJ}\)

Unless otherwise stated the value of ΔH mentioned in a thermochemical equation is considered to be the value at standard states.

A thermochemical equation can be written in this manner. When this is done, the magnitude of AH remains the same but its sign becomes the opposite

Example:

⇒ \(\mathrm{N}_2(g)+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})\)

ΔH=+180.5 kJ

Or,  \(2 \mathrm{NO}(g) \rightarrow \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) ; \Delta H=-180.5 \mathrm{~kJ}\)

Class 11 Chemistry Notes For Chapter 6 Chemical Thermodynamics Numerical Examples

Question 1. At a fixed temperature & pressure, the heat released in the formation of 3 mol SO₃(g) from SO₂(g) and 2(S) is 291 kj. What will be the change in enthalpies in the formation of1 mol & 4 mol SO₃(g)?
Answer:

The thermochemical equation for the formation of 3 mol of SO₃(g) from the reaction between SO₂(g) and O₂(g) is given

⇒ \(3 \mathrm{SO}_2(\mathrm{~g})+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 3 \mathrm{SO}_3(\mathrm{~g}), \Delta H=-291 \mathrm{~kJ}\)

The thermochemical equation for the formation of 1 mol of

⇒  \(\mathrm{SO}_3(g) \text { is: } \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g}) \text {, }\)

⇒ \(\Delta H=\frac{1}{3}(-291) \mathrm{kJ}=-97 \mathrm{~kJ}\)

Therefore, the change in enthalpy for the formation of lmol and 4mol SO₃(g) will be -97 kJ and 4 ×(-97)

=- 388 kj, respectively.

Question 2. At a particular temperature and pressure, the heat produced in the formation of 2 mol of C₂H₆(g) from the reaction between C₂H₂(g) and H₂(g) is 626 kj. What amount of H₂(g) will react with the required amount of C₂H₂(g) to produce 939 kj of heat at the same temperature and pressure?
Answer:

The Thermochemical equation for the formation of 2 mol of 2H₆(g) from the reaction between C₂H₂(g) and H₂(g) is

⇒ \(2 \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})+4 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}) ; \Delta H=-626 \mathrm{~kJ}\)

∴ 626 kJ of heat = 4 mol H₂(g)

∴ \(939 \mathrm{~kJ} \text { of heat } \equiv \frac{4}{626} \times 939 \equiv 6 \mathrm{~mol} \mathrm{H}_2(\mathrm{~g})\)

= 6 × 2

= 12g H₂(g)

Therefore, if 12 g H₂(g) reacts with the required amount of C₂H₂(g) to produce C₂H₆(g), then 939 kJ of the heat is evolved.

Question 3. At a particular temperature and pressure, N₂(g) and O₂(g) react to form 4 mol of N₂O. The heat absorbed in this reaction is 328 kj. What would the change in enthalpy be due to the formation of 2 mol of N₂(g) and 1 mol of O₂(g) from N₂O(g) at the same temperature and pressure?
Answer:

The thermochemical equation for the formation of

⇒ \(\mathrm{N}_2 \mathrm{O}(g): 4 \mathrm{~N}_2(g)+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{~N}_2 \mathrm{O}(g) ; \Delta H=+328 \mathrm{~kJ}\)

Writing this equation in the reverse manner, we obtain,

⇒ \(4 \mathrm{~N}_2 \mathrm{O}(\mathrm{g}) \rightarrow 4 \mathrm{~N}_2(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g})\)

ΔH=-328 kJ

Therefore, the thermochemical equation for the formation of 2 mol of N₂(g) and 1 mol of O₂(g) from N₂O(g) will be:

⇒  \(2 \mathrm{~N}_2 \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{~N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g});\)

⇒\(\Delta H=-164 \mathrm{~kJ}\)

Thus, the change in enthalpy for the formation of 2 mol of N₂(g) and 1 mol of O₂(g) is -164 kJ.

Class 11 Chemistry Notes For Chapter 6 Chemical Thermodynamics – Various Types Of Heat Of Reaction Or Enthalpy Of Reaction

Heat or enthalpy of the formation of a compound

Heat or enthalpy of the formation of a compound Definition

At a given temperature and pressure, the enthalpy of formation of a compound is defined as the enthalpy change for the reaction in which one mole of the compound is formed from its constituent elements

The enthalpy of formation of a compound is denoted by the symbol ΔHf, where subscript ‘f’ stands for formation. To compare the enthalpy of formation for different compounds, the enthalpy of formation at a standard state is calculated for the compounds. The enthalpy of formation for any compound in the standard state is known as the standard enthalpy of formation of that compound.

Standard enthalpy of formation:

The standard enthalpy of formation of a compound is defined as the change in enthalpy for the reaction in which one mole of the compound in its standard state (i.e., at a particular temperature and 1 atm pressure) is produced from its constituent elements in their standard states

The standard enthalpy of formation for any compound is denoted by \(\Delta H_f^0\), where subscript ‘f’ stands for formation and superscript ‘0’ indicates the standard state. The value of \(\Delta H_f^0\) may be positive or negative

In the thermochemical equation representing the formation reaction of a compound, one mole of the compound is formed. For this reason, the unit of ΔH0f is expressed in kJ.mol-1 (or J.moI-1) or kcal.mol^-1 (or cal. mol^-1).

Example:

⇒ \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H_f^0=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Generally, the standard enthalpy of formation \(\Delta H_f^0\) of a compound is quoted at 25°C (or 298 K). If temperature is not mentioned, then 25°C temperature is to be considered. At a given temperature, the standard state of an element or a compound is the most stable and purest state of that element or compound at 1 atm pressure and at that temperature. The most stable forms of some elements at 25°C and 1 atm are given below.

1. Dihydrogen: H₂(g)
2. Dioxygen: O₂(g)
3. Dinitrogen: N₂(g)
4. Sodium: Na(s)
5. Chlorine molecule: Cl₂(g)
6. Bromine molecule: Br₂(Z)
7. Iodine molecule: I₂(s)
8. Carbon: C (graphite, s)
9. Sulfur: S (rhombic, s), etc.

The constituent elements of a compound must be present in their standard states in the equation representing the formation reaction of the compound.

At 25° nC, the standard enthalpy of formation of water =-205.8kI.mol^-1:

At 25 °C and 1 atm, the heat evolved due to the formation of 1 mol of H₂O(/) from the reaction between 1 mol of H₂(g) and 1/2 mol of O₂(g) is 285.8 kJ. Alternatively, it can be said that the change in enthalpy of the given reaction at 25 °C and 1 atm is -285.8 kj

⇒ \(\underbrace{\mathrm{H}_2(g)+5^{\circ} \mathrm{C} \& 1 \mathrm{~atm}}_{25^{\circ} \mathrm{C} \& 1 \mathrm{~atm}} \underbrace{\frac{1}{2} \mathrm{O}_2(\mathrm{~g})} \rightarrow \Delta H_f^0=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

At 25°C, The standard enthalpy of formation of nitric acids

⇒ \([\mathrm{NO}(g)]=+90.3 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

At 25°C and 1 atm, the heat evolved due to the formation of 1 mol of NO(g) from the reaction between 1/2 mol of N₂(g) and 1/2 mol of 02(g) is 90.3kJ. mol^-1 . Alternatively, it is said that the change in enthalpy of the following reaction at 25°C and I atm is +90.3 kj.

⇒ \(\underbrace{\frac{1}{2} \mathrm{~N}_2(g)+\frac{1}{2} \mathrm{O}_2(g)}_{25^{\circ} \mathrm{C} \& 1 \mathrm{~atm}} \rightarrow \underbrace{\mathrm{NO}(g)}_{25^{\circ} \mathrm{C} \& 1 \mathrm{~atm}}\)

⇒ \( \Delta H_f^0=+90.3 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Important points about standard enthalpy of formation:

The standard enthalpy of formation of a compound is not always equal to the value of AH° indicated in the thermochemical equation representing the reaction in which the compound in its standard state is formed from its stable constituent elements in their standard states.

Explanation:

At 25°C and 1 atm, two chemical equations for the formation of H₂O(Z) from their stable constituent elements H₂(g) and O₂(g) are given below.

⇒ \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-285.8 \mathrm{~kJ}\)

⇒ \(2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-571.6 \mathrm{~kJ}\)

The ΔH° in equation [1] indicates the standard enthalpy of the formation of H₂O(Z) because, in this reaction, one mole of H₂O(l) is formed from its stable constituent elements H₂(g) and O₂(g). On the other hand, in equation (2), two moles of H₂O(Z) are formed from the stable constituent elements H₂(g) and O₂(g).

Thus, according to the definition of the enthalpy of formation, the value of AH° from equation [2] does not indicate the standard enthalpy of formation of H₂O(Z).

The standard enthalpies of formation of all elements in their standard states are conventionally taken as zero. In the case of an element having different allotropes, the standard enthalpy of formation of the most stable allotropes form in the standard state is considered zero.

Explanation:

The most stable forms of hydrogen, oxygen, nitrogen, sodium, etc. at 25 °C and 1 atm are H₂(g), O₂(g), N₂(g), Na(s), etc. Thus, the standard enthalpies of formation of H₂(g), O₂(g), N₂(g), Na(s), etc. are zero.

According to the definition, at 25°C and 1at, the standard enthalpy of formation of hydrogen is the same as the standard enthalpy change ofthe following reaction \(\mathrm{H}_2\left(\mathrm{~g}, 25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right) \rightarrow \mathrm{H}_2\left(\mathrm{~g}, 25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)\)

No change has occurred in this process. Hence ΔH°= 0. Therefore, at 25°C, for hydrogen \(\Delta H_f^0=0\).

For the same reason \(\Delta H_f^0\) is zero for N₂(g), O₂(g), etc. Diamond and graphite are the two allotropic forms of solid carbon.

Between these two forms, graphite is the most stable form at 25°C and 1 atm. Thus the standard enthalpy of formation

⇒  \(\Delta H_f^0\) of solid graphite [C (graphite, s)] at 25°C is zero. But, for diamond [C (diamond, s)], at 25°C \(\Delta H_f^0 \neq 0\) A

At 25°C, the standard enthalpy of formation of diamond

⇒  \(\Delta H_f^0 \neq 0\) is equal to the standard enthalpy of reaction for the following change \(\left.\mathrm{C} \text { (graphite, } s, 25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right) \longrightarrow \mathrm{C} \text { (diamond, } s, 25^{\circ} \mathrm{C}, 1 \mathrm{~atm} \text { ) }\)