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

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.