NCERT Class 11 Chemistry Chapter 3 Classification Of Elements And Periodicity In Properties Short Question And Answers

NCERT Solutions For Class 11 Chemistry Chapter 3 Elements And Periodicity In Properties Short Question And Answers

Question 1. Locate the position of an element in the long form of the periodic table.
Answer:

Electronic configuration: ls²2s²2p⁶3s²3p⁶3d¹⁰4s²4p³.

Period = 4th, Group No. = 10+ total no. of electrons present in 4s and 4p orbitals =10 + 2 + 3 = 15

Question 2. The atomic radius of elements in a period decreases with an increase in atomic number but for inert gases, it increases. Why?
Answer:

Atoms in inert gases are held together by weak van der Waals forces, not by covalent bonds. Thus, atomic radii and van der Waals radii of inert gases are equal. However, as the van der Waals radius is greater than the covalent radius, the atomic radius of elements across a period decreases with an increase in atomic number but for inert gases increases.

Question 3. The number of electrons in Sr²⁺ and Br^– are the same. Justify whether the atomic radii of these two ions will be the same or not.
Answer:

Although Sr²⁺ and Br^– have the same number of electrons, their proton number are different Number of protons in Sr²⁺ is greater than that of Br-‘; Thus, the nuclear pull on the electrons is greater in Sr²⁺ than in Br^–. So atomic radius or Sr^2-1, is less them that of Br^–.

Question 4. Which of the following oxides is the most acidic? Al₂O₃, SO₂, SO₃, P₄O₁₀ and CO.
Answer:

Out of the elements Al, S, P, and C, S lies at the extreme right of the periodic table and so it is the most electronegative. Again in SO₃ die oxidation state of S is the highest (+6). Hence, SO₃ is the most acidic compound. It reacts with water to give a strong acid.

Question 5. Arrange the elements according to the instructions given:

• Na, Cu, Zn (increasing order of electropositivity).
• Na, Cs, K, Rb, Li (increasing order of atomic volume)

Answer:

1. Cu < Zn < Na
2. Li < Na < K < Rb < Cs

Question 6. The first ionization enthalpy of Na is less than that of Mg, but the second ionization enthalpy of Na is higher than that of Mg. Explain.
Answer:

By losing an electron from its outermost shell. Na attains the stable electronic configuration of the nearest noble gas Ne. So, the first ionization enthalpy of Na is less than Mg.

On the other hand, the removal of one electron from the outermost shell of Mg⁺¹ leads to the attainment of the stable electronic configuration of Ne. So the second ionization enthalpy of Mg is less than that of Na.

Question 7. Ionization enthalpy values of Se, Br, Te andI are 869, 941, 1191, and 1142 kj. Mol ^-1. The values are not arranged in the correct order. Predict which element the ionization enthalpy value is 869kj. mol^-1 and 1142kJ . mol^-1 respectively.
Answer:

• Se and Te are found in the fourth and fifth periods of group 16 in the periodic table. On the other hand, Br and are found in the fourth and fifth periods of group 17.
• On moving across a period from left to right, ionization enthalpy increases whereas on moving down a group, its value decreases.
• So, the elements in increasing order of ionization enthalpy are Te <I < Se < Br. Therefore, ionization enthalpy values of Te and Br are 869 and 1142 kj-mol^-1 respectively.

Question 8. The ionization potential of hydrogen is 1312.0 kj mol^-1. Express the value in eV atom-1. (leV = 1.6 × 10¹⁹)
Answer:

Ionisation potential of hydrogen = 1312.0 kj-mol^-1

= \(\frac{1312 \times 10^5}{6.023 \times 10^{23}} \mathrm{~J} \cdot \mathrm{atom}^{-1} \)

= \(\frac{1312 \mathrm{w} \times 10^3}{6.023 \times 10^{23}} \times \frac{1}{1.6 \times 10^{-19}} \mathrm{eV} \cdot \mathrm{atom}^{-1}\)

= \(13.61 \mathrm{eV} \cdot \mathrm{atom}^{-1}\)

Question 9. Why is the size of F- smaller than that of O^2- ion? 
Answer:

O^2- and F^– ions are isoelectronic, but their nuclei contain 8 and 9 protons respectively.

So, the nuclear attractive force acting on the electrons of the F^– ion is greater than that on the electrons of the O^2- ion. Consequently, the F^– ion is smaller than the O^2- ion.

Question 10. Compare the radii of K⁺ and Cl^– ions (each contain the same number of electrons)
Answer:

The nuclei of the isoelectronic ions K⁺ and Cl^– contain 19 and 17 protons respectively. So the magnitude of nuclear attractive force acting on the electrons of K⁺ ions is greater than that on the electrons of Cl^– ion. Consequently, the radius of the K+ ion is smaller than that of the Cl^– ion.

Question 11. The electronic configuration of the atom of an element is ls²2s²2p⁶3s²3p¹. Locate its position in the periodic table. Is it a metal or non-metal? What is its valency?
Answer:

The receives its last electron in 3p -orbital. So it belongs to p -p-block elements.

∴ Accordingly, its group number = 10 + no. of electrons in the valence shell = 10 + (2 + 1)= 13.

Again, the period of the element = several principal quantum numbers ofthe valence shell = 3.

It is metal because it belongs to the 13th group of the periodic table.

∴  Valency of the element = number of electrons in the valence shell = 2+1 = 3.

Question 12. Justify by mentioning two reasons, the inclusion of Ca and Mg in the same group of the periodic table.
Answer:

Both Ca and Mg have similar electronic configurations. Both of them belong to s -s-block.

Electronic configuration of Mg: ls²2s²2p⁶3s²

Electronic configuration of Ca: ls²2s²2p⁶3s²3p⁶4s²

Both are typical divalent metals and form stable ionic compounds,

For example: MgCl₂, CaCI₂; MgO, and CaO.

Question 13. Cu, Ag & Au are regarded as transition elements. Why?
Answer:

Despite having filled d -orbitals (d10), copper, silver, and gold are considered transition elements because at least in one stable oxidation state of the atoms of these elements, d -orbital is partially filled with electrons,

For example → The electronic configuration of Cu²⁺ is ls²2s²2p⁶3s²3p²3d94s².

Question 14. Can Cu (At. no. 29) and Zn (At. no. 30) be called transition elements? Explain.
Answer:

Although the Cu -atom in its ground state does not contain an incompletely filled d-orbital, Cu²⁺ has a partially filled d-orbital and so it is a transition element.

• Electronic configuration of Cu: ls²2s²2p⁶3s²3p⁶3d¹⁰4s¹
• Electronic confiuration of Cu²⁺: ls²2s²2p⁶3s²3p⁶3d⁹4S²

In the ground state or any stable oxidation state of Zn, the d orbital is filled. So, it is not a transition element.

Electronic config. of Zn: ls²2s²2p⁶3s²3p⁶3d¹⁰4s²

Electronle’dbnfig. of Zn²’: ls²2s²2p⁶3s²3p⁶3d¹⁰4s⁰

Question 15. Write down the electronic configuration of the element having atomic number 21. Name two other elements of the same series to which this element belongs. Why do they belong to the same series?
Answer:

Electronic configuration ofthe element (Atomicnumber= 21): ls²2s²2p⁶3s²3p⁶3d¹⁴s²

From the electronic configuration, it can be said that the element belongs to the first transition series as its 3d –orbital is partially filled.

Two other elements belonging to this series are Cr(24) and Mn(25). The 3d -orbitals of these elements are also partially filled.

Their electronic configurations are:

• Cr: ls²2s²2p⁶3s²3p⁶3d⁵4s¹
• Mn: ls²2s²2p⁶3s²3p²3d5⁴s²

Question 16. 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: ls²2s²2p⁶3s²3¹

₁₇17C: ls²2s²2p⁶3s²3p⁵

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

⇒ \(\begin{array}{|c|c|c|}
\hline \text { Element } & \text { Period number } & \text { Group number } \\
\hline \mathrm{A} & 2 & 10+2+5=17 \\
\hline \mathrm{B} & 3 & 10+2+1=13 \\
\hline \mathrm{C} & 3 & 10+2+5=17 \\
\hline
\end{array}\)

3.

1. 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.
2. Elements A and C are electronegative because they can accept one electron to attain a stable inert gas electronic configuration.
3. 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.

Question 17. Why is the atomic size of Ca²⁺ smaller than that of K⁺?
Answer:

• K+ and Ca²⁺ are isoelectronic (each contains 18 electrons). K⁺ contains 18 protons in its nucleus and Ca²⁺ contains 20 protons.
• Several protons being in Ca²⁺  ion, the electrons in the outermost shell of Ca²⁺  ion will experience greater attractive force by the nucleus compared to K+.
• Consequently, Ca²⁺ has a smaller ionic size than K⁺ ions.

Question 18. The atomic radius of the chlorine atom is 0.99 Å but the ionic radius of chloride (Cl^–) is 1.81 Å—explain.
Answer:

In an anion, the total number of electrons being greater than that of the number of protons, attraction of the nucleus for the outermost electrons decreases.

• Furthermore, due to mutual repulsion among electrons, the outermost orbit gets appreciably expanded.
• As a result, the anionic radius becomes more than the radius of the parent atom. On account of this, the radius of C^– (1.81Å) (18 electrons and 17 protons) is greater than the radius of the chlorine (0.99 Å) atom (17 electrons and 17 protons).

Question 19. First ionization energy of elements increases with the atomic numbers of the elements in a period—why? Cite an exception.
Answer:

In any period, with an increase in atomic number, the magnitude of the positive charge of the nucleus increases, but there is no addition of any new shell.

• In consequence, the attractive force of the nucleus for the outermost electrons increases.
• So, the amount of energy required for the removal of an electron from the outermost shell of the atom ( first ionization energy) in any period increases with the increase in atomic number.
• An exception to this generalization is the nitrogen-oxygen pair. In the second period, oxygen belonging to group VIA has a lower value of first ionization energy than nitrogen of group.

The reason can be ascribed to the stable electronic configuration of the nitrogen atom (ls²2s²2p³)

Question 20. What are d-block elements? Give their general electronic configuration.
Answer:

• The reasons behind placing Na and K in the same group of the periodic table are
• The valence shell electronic configurations are the same

Na: ls²2s²2p⁶3s¹ K: ls²2s²2p⁶3s²3p⁶4s¹.

Question 21. Calculate the energy (in kj unit) required to convert all sodium atoms into sodium ions, present in 3.45 mg of its vapor. [I.P of sodium 490kJ mol^-1 ]
Answer:

Sodium atoms present \(=\frac{3.45}{23} \times 10^{-3}=1.5 \times 10^{-4} \mathrm{~mol}\)

Na(g) + I₁ = Na+(g) + = 490 kj. mol^-1

The energy required for the ionization of 1 mol sodium is 490 kj.

The energy required for ionization of 1.5 × 10^-4 mol sodium is 490 × 1.5 × 10^-4 kj

= 73.5J

Question 22.

1. Write the name of the element which is diagonally related to the element beryllium.
2. Three elements A, B, and C have atomic numbers 11, 14, and 17 respectively.
   State the periodic table in which elements A and C belong. Write the formulas ofthe compounds formed between B and C and A and C. State the nature ofthe bonds.

Answer:

1. Aluminium (Al)

2.

• 11^A: ls²2s²2p⁶3s2³p²
• 17^C: ls²2s²2p⁶3s²3p⁵

From the electronic configuration is evident that A is an s – block element whereas C is a p – block element.

The compound formed between B and C has the formula BC₄ and the nature of the bond is covalent. The compound formed by a combination of A and C is AC (electrovalent).

Question 23.

1. Write the symbols of one transition and inner transition element.
   
2.  Indicate as directed:
   • Which has the highest ionic radius? Al³⁺, Mg²⁺, O^2-, F^–.
   • Which has the lowest electronegativity? H, Na, Si, Cl.
   • Which has the highest ionization energy? N, O, Ar, P
     

Answer: 

1.

• Symbol of one transition element — Fe.
• Symbol of one inner-transition element — Ce

2.

• Ionic radius is highest for O^2- for the given case.
• Electronegativity is lowest for Na for the given case.
• Ionization energy is highest for Ar for the given case.

Question 24. State the group number in the modem periodic table where solid, liquid, and gaseous elements are present at room temperature. Identify solid, and liquid elements. Indicate the given elements as alkali metal, alkaline-earth metal, coinage metal, chalcogen: Li, Ca, S, Cu.
Answer:

Group 17 of the modern periodic table contains solid (iodine, I₂), liquid (bromine, Br₂), and gaseous (chlorine, Cl₂) elements at the same time.

• Group 17 of the modern periodic table contains solid
• (iodine, I₂), liquid (bromine, Br₂) and gaseous
• (chlorine, Cl₂) elements at the same time.

Question 25. Which has greater ionization energy and why—S or P?
Answer:

Phas greater ionization energy than S.

• According to Hund’s rule, the half-filled electronic configuration is highly stable, and the outermost 3p -orbital of the P atom is half-filled.
• To produce P+ ions by removing an electron from the 3p orbital of Prequireshuge energy resulting in a very high value of ionization energy.
• On the other hand, the S atom has a partially filled 3p -orbital which attains a stable half-filled electronic configuration in the S+ ion.
• Therefore, to produce S to S+ comparatively lower energy is required resulting in a lower value of ionization energy.

Question 26. Why is the electron-gain enthalpy of oxygen less than that of sulfur? Arrange the following metal oxides in terms of ascending order of basicity: ZnO, MgO, CaO, CuO.
Answer:

Electron-gain enthalpy of O is less than that of S as the 2p -orbital of the outermost shell of O-atom is much smaller in size than the 3p -orbital of the outermost shell of the S-atom.

• So, the additional electron-electron repulsive force produced due to the addition of one electron to the 2p -orbital of the oxygen atom from outside is more than the additional electron-electron repulsive force developed by the addition of one electron to the 3p -orbital of the S-atom.
• So the electron-gain enthalpy of the O-atom becomes less than that S-atom.

⇒ \(\mathrm{CuO}<\mathrm{ZnO}<\mathrm{MgO}<\mathrm{CaO}\)

Question 27. Why is the first ionization enthalpy of helium maximum among all the elements? Arrange the given compounds in terms of ascending order of oxidizing property: HCl, HBr, HI, HF
Answer:

Configuration of He is Is₂, i.e., all electrons of He are present in the Is -orbital. These electrons are attracted very strongly by the nucleus. Further, there are no inner electronic orbitals to shield these electrons from the pull of the nucleus.

So removal of an electron from the Is -orbital requires a large amount of energy. Therefore, the first ionization enthalpy of He is maximum. Hydrogen halides do not show oxidizing properties.

However, the reducing power of the hydrogen halides follows the sequence: HF <HCl < HBr < H

Question 28.

1. Which of the following two elements has a diagonal relationship? Li, Be, Al, and Si
2. Between ₂₉Cu and ₁₉K which one has higher ionization enthalpy and why

Answer:

1. Be and Al

2.

• Electronic configuration of 19K: ls²2s²2p⁶3s²3p⁶4s¹
• Electronic configuration of 2gCu: ls²2s²2p⁶3s²3p⁶3d¹⁰4s¹

The nucleus of the Cu-atom contains 10 more protons than that of the K-atom.

Additional nuclear pull on the outermost electron (4s¹) in Cu-atom is not counter-balanced by the shielding effect of ten 3d-electrons because orbitals have poor screening effects. Thus the effective nuclear charge acting on the 4s-electron of copper is greater than that acting on the 4s-electron of

Question 29. Which is more stable between BCl₃ and TlCl₃ and why? What is the oxidation state of Zn in Zn-Hg?
Answer:

Because of the poor shielding effect by the inner d – and f – electrons, the inert pair effect is maximum for Tl. Thus the most stable oxidation state of Tl is +1 and not +3. Therefore TIClg is unstable. B does not exhibit an inert pair effect and thus BCl₃ is stable.

Question 30. In terms of period and group, where would you locate the element with Z =
Answer:

It is known that the difference between atomic numbers of the successive members of any group is 8,8,18,18 and 32 (from top to bottom). So, the element with atomic number 114 will lie just below the element with atomic number (114- 32) = 82.

The element with atomic number 82 is lead (Pb), which is a member of the 6th period belonging to group number 14 (p -p-block element). Thus the element with atomic number 114 takes its position in the 7th period and group- 14 (p -block element) of the periodic table.

Question 31. Write the atomic number of the element present in the third period & seventeenth group of the periodic table.
Answer:

The general electronic configuration of the valence shell of the elements of group-17 (halogens) is ns²np⁵. For the third period, n = 3.

• Therefore, the electronic configuration of the valence shell of the element of the third period and group-17 Is 3ia³p1² and the complete electronic configuration of this element Is ls²2s²2p⁶3s²3p⁵.
• There are a total of 17 electrons In this element.
• Thus, the element In the third period and seventeenth group of the periodic table has atomic number = 17.

Question 32. Consider the given species: N^2-, O^2-, I^1-, Nⁿ⁺, Mg²⁺ and Al³⁺. What Is Common In them? Arrange them in the order of increasing ionic radii.
Answer:

Each of the given ions has 10 electrons. Hence, they are all isoelectronic species.

• Ionic radii of isoelectronic ions decrease with an increase in the magnitude of the nuclear charge.
• The order of increasing nuclear charge of the given isoelectronic ions is N^3- < O^2- <F^– < Na⁺ < Mg²⁺ < Al³⁺.
• Therefore, the order of increasing ionic radii is: Al³⁺ < Mg²⁺ < Na+ < F- < O^2- < N^3-.

Question 33. What is the basic theme of organization in the periodic table?
Answer:

The basic theme of organization in the periodic table is to study different physical and chemical properties of all the elements and their compounds simply and systematically.

• Elements belonging to the same group have similar physical and chemical properties.
• So, if the physical and chemical property of any one element of a group is known, then it is possible to predict the physical and chemical properties of the remaining elements of that group.
• Therefore, it is not important to keep in mind the physical and chemical properties of all elements in the periodic table.

Question 34. How would you react to the statement that the electronegativity of N on the Pauling scale is 3.0 in all the nitrogen compounds?
Answer:

The electronegativity of any element depends on the hybridization state and oxidation state of that element in a particular compound, i.e., the electronegativity of an element varies from compound to compound. For example, the electronegativity of Natom varies as sp³ —N < sp²—N < sp—N. So, the, given statement is not correct.

Question 35. Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Answer:

Isotopes of an element have the same number of electrons and similar electronic configurations. So their nuclear charge and atomic radii are identical. Consequently, two isotopes of the same element are expected to have the same ionization enthalpy.

Question 36. What are the major differences between metals and non-metals?
Answer:

Metals have a strong tendency to lose electrons to form cations they are strong reducing agents, have low ionization enthalpies, have less negative electron gain enthalpies, low electronegativity, and form basic oxides and ionic compounds.

Non-metals have a strong tendency to accept electrons to form anions. They are strong oxidizing agents, have high ionization enthalpies, have high negative electron-gain enthalpies, and high electronegativity, and form acidic oxides and covalent compounds.

Question 37. Formulas of oxide and chloride of an element A are A205 and AClg respectively. Which group of Mendeleev’s periodic table will the element belong to? State whether it is a metal or a non-metal.
Answer:

Hiking oxygen as standard, the valency ofthe element A is 5, and taking chlorine as standard, the valency of A is 3. Since oxygen-based valency (8-chlorine-based valency) is the same as that ofthe group number of the element, the element will be of group Mendeleev’s periodic table. Thus, itis a non-metal.

Question 38. A and B are two elements with atomic numbers 9 and 17 respectively. Explain why the element A is a more powerful oxidizing agent than the element B.
Answer:

Electronic configuration of ₉A: ls²2s²2p⁵.

Electronic configuration of  ₁₇B: ls²2s²2p⁶3s²3p⁵.

Both A and B can accept electrons to form the anions of A^– and B having inert gas electronic configuration. However, the anions A^– are more stable than B^– because of the smaller atomic size of A. So A is a stronger oxidizing agent than B.

Question 39. Which one is more basic and why—MgO & Al₂O₃?
Answer:

MgO is more basic than Al₂O₃. Mg & Al are elements of group-2A and 3A of period-3 respectively. Since in a periodic table, metallic property decreases along a period from left to right, the metallic property of Mg is greater than that of Al. Again the oxide of an element with more metallic character is more basic than that with less metallic character. Thus, the basic character of MgO will be more than that of Al₂O₃.

Question 40. LiCO₂ despite being an alkali metal carbonate, is sparingly soluble in water like MgCO₃ — explain.
Answer:

The electronegativities of Li and Mg are very close to each other. Furthermore, both Li+ and Mg2+ have similar ionic potential) values and both of them have high polarising power.

So Li and Mg show some similarities in their chemical properties. Due to their small ionic size and high polarising power, their carbonates are somewhat covalent. So, both LiCO₃ (alkali metal carbonate) and MgCO₃ are sparingly soluble in water.

Question 41. Compare the atomic radii of fluorine and neon.
Answer:

Fluorine and neon are the members of the second period having atomic numbers 9 and 10 respectively. The covalent radius of fluorine (halogen) is expressed in terms of its covalent radius, while that of neon (noble gas) is expressed in terms of its van der Waals radius. Since by definition, van der Waals radii are always greater than covalent radii, therefore, the atomic radius of neon is greater than that of fluorine.

Question 42. Give one example of each of the following and indicate their positions in the periodic table (long form)

1. Liquid non-metal
2. Liquid metal
3. Radioactive halogen
4. Radioactive inert gas
5. Radioactive alkali-metal.

Answer:

1. Liquid non-metal: Bromine (Br), 17th group in 4th period,
2. Liquid metal: Mercury (Hg), 12th group in 6th period,
3. Radioactive halogen: Astatine (At), 17th group in 6th period,
4. Radioactive inert gas: Radon (Rn), 18th group in 6th period,
5. Radioactive alkali metal: Francium (Fr), 1st group in 7th period.

Question 43. Why is the ionization enthalpy of oxygen less than those of nitrogen and fluorine?
Answer:

For the explanation ofthe ionization enthalpy of oxygen is less than nitrogen, the  Ionisation enthalpy of oxygen is less than fluorine because—

1. Nuclear charge increases from oxygen to fluorine,
2. The number of shells remains the same and the addition of differentiating electrons occurs in the same shell of fluorine,
3. The atomic size of oxygen is greater than fluorine.

Question 44. What are rare-earth elements? Why are they so called?
Answer:

• The 14 elements from cerium (₅₈Ce) to lutetium (₇₁Lu) ofthe periodic table are called rare-earth elements.
• These elements are so named because most of these elements occur in very small amounts in the earth’s crust.

Question 45. What is the basic difference between electron-gain enthalpy and electronegativity of an element?
Answer:

• Electron-gain enthalpy means the enthalpy change involved when an electron is added to an isolated gaseous atom in its lowest energy state.
• Whereas electronegativity means the tendency of an atom to attract the shared pairs of electrons toward its nucleus when the atom is covalently bonded in a molecule.

Question 46. How does the basicity of the oxides of representative elements vary on moving a group in the periodic table? On moving across a period from left to right, how does die acidity of the oxides of representative elements vary?
Answer:

• On moving down a group in the periodic table, the basicity of the oxides of representative elements increases.
• On moving across a period from left to right, the acidity of the oxides ofthe representative elements increases.

Question 47. Formulas of oxide and chloride of an element M are M₂O₅ & MCI₃ respectively. State the group to which the element belongs to. Determine whether it is metal or non-metal.
Answer:

• The valency of the element M is 5 when oxygen Is considered the standard and the valency of M Is 3 when chlorine Is considered as the standard.
• Since, the oxygon-butted valency of an element Is equal to Its group number, the group to which M belongs Is VA(15). Element  M Is a nonmetal as It Is located In group VA of the periodic table

NCERT Solutions For Class 11 Chemistry Chapter 3 Classification Of Elements And Periodicity In Properties Warm-Up Exercise Question And Answers

Question 1. Three elements X, Y, and Z follow Dobereiner’s law of triads. If the atomic masses of X and Z are 7 and 39 respectively, then determine the atomic mass of Y.
Answer:

The atomic mass of Y = arithmetic mean of atomic masses of X and Z. Therefore, the atomic mass of Y = 7 + 39/2 = 23.

Question 2. Which properties of the elements are dependent on their electronic configurations and which are not
Answer:

The chemical properties and some physical properties of elements are dependent on their electronic configurations whereas nuclear properties (like radioactivity) do not depend on their electronic configurations.

Question 3. The outermost electronic configuration of the atom of an element is 3s-3p³. Mention the position of the element in the long periodic table.

Answer:

It is a p -block element. Hence period no. = principle quantum no. ofthe outermost shell= 3 Its group no = (total no. of electrons in s-and p orbital) +10 = 2 + 3 + 10 = 15

Question 4. Elements of the 4th, 5th, and 6th periods of Mendeleev’s periodic table were divided into even and odd series—why?
Answer:

Due to the difference in properties of each pair of elements belonging to the same period and group, the elements of the 4th, 5th, and 6th periods of Mendeleev’s periodic table were divided into even and odd series.

Question 5. Identify the elements eka-aluminum and eka-silicon. What was the reason for such naming?
Answer:

Ga and Ge. The elements were so named by Mendeleev because he predicted that the properties of ekaaluminium and eka-silicon would be similar to those of aluminum and silicon respectively.

Question 6. State the reason for the repetition of properties ofthe elements after certain regular intervals of their atomic numbers.
Answer:

After certain regular intervals of their atomic numbers, elements show similar electronic configurations in their valence shell. For this reason, repetition of properties of the elements is observed.

Question 7. In the modern periodic table, period indicates the value of atomic number atomic mass principal quantum number azimuthal quantum number
Answer:

Each period in the modern periodic table begins with the filling of a new shell. So, the period indicates the value of a principal quantum number.

Question 8. Cu in the +1 oxidation state and Hg in the +2 oxidation state resemble each other in their properties. Explain.
Answer:

Valence shell configurations of Cu+([Ar]3d¹⁰) and Hg²⁺([Xe]5d¹⁰) are similar. So they resemble each other in their properties.

Question 9. Write the formula of the compound formed by the most electronegative and least electronegative elements.
Answer:

The most electronegative element is F while the least electro-negative element is Cs. Formula ofthe compound is CsF. 85

Question 10. Calculate the atomic volume of sodium (atomic mass – 23) if its density is 0.972 g-cm^-3.
Answer:

Atomic volume of sodium \(=\frac{23 \mathrm{~g}}{0.972 \mathrm{~g} \cdot \mathrm{cm}^{-3}}=23.66 \mathrm{~cm}^3\)

Question 11. Determine the position of an element in the long form of the periodic table if its electronic configuration is [₁₈Ar]3d¹⁰4s²
Answer:

It is a d -block element. Hence period no. = principle quantum no. of the outermost shell = 4 and its group no. = total electronin 3d and 4s -orbital = 10 + 2 = 12.

Question 12. Mention the name and the position of two elements, one which is most electronegative and the other most positive in the periodic table.
Answer:

• Most electronegative element =F (group 17, period-2)
• Most electropositive element = Cs (group 1, period-6)

Question 13. The outermost electronic configurations of the two elements are 2s² and 2s²2p¹ respectively. Which has greater ionization enthalpy Which has the highest ionization enthalpy: N, O, Ar, P?
Answer: The element with the outermost electron configuration 2s² has greater ionization enthalpy.

Question 14. Electron-gain enthalpy of N is less than that of O. Explain.
Answer:

Nitrogen (2s²2p³) has a stable outer electronic configuration with a half-filled 2p -subshell. So, it is reluctant to accept an additional electron. Thus the electron-gain enthalpy of nitrogen is less than that of oxygen.

Question 15. Consider the set of ions (Na⁺, N^3-, Mg²⁺, O^2-, F^-, and Al³⁺) and answer the following questions: What is the common factor associated with the species? Arrange the ions in order of increasing radii.
Answer:

All are isoelectronic species.

Al³⁺+ < Mg²⁺ < Na⁺ < F^– < O^2- < N^–

Question 16. Which products are liberated at the cathode and anode when molten ionic hydrides are electrolyzed?
Answer:

When molten Ionic hydrides are electrolyzed, the metal Ions are discharged at the cathode while hydrogen gas is liberated at the anode.

Question 17. Arrange according to the instruction given: Al₂O₃, P₂O5,  Cl₂O₇, SO₃   (increasing order of acidity)  MgO, ZnO, CaO, Na₂O, CuO (increasing order of basicity).
Answer:

Al₂O₃< P₂O5 < SO₃ < Cl₂O₇

CuO < ZnO < MgO < CaO < Na₂O

Question 18. What do you understand by the negative value of electron-gain enthalpy of an element?
Answer:

It signifies that energy is released when an isolated gaseous atom of the element under consideration accepts an electron to form a monovalent gaseous anion.

Question 19. Why is the value of electron-gain enthalpy negative?
Answer:

When an electron is added to a neutral gaseous atom to form a negative gaseous ion, energy’ is usually liberated, i.e., the enthalpy change in the process is usually negative. So electron-gain enthalpies of most elements have negative values.