NIOS Class 12 Chemistry Chapter 18 General Characteristics of the p-block Elements Solutions English Medium As Per New Syllabus to each chapter is provided in the list so that you can easily browse throughout different chapters NIOS Class 12 Chemistry Chapter 18 General Characteristics of the p-block Elements Notes in English and select need one. NIOS Class 12 Chemistry Solutions English Medium Download PDF. NIOS Study Material of Class 12 Chemistry Notes Paper Code: 313.
NIOS Class 12 Chemistry Chapter 18 General Characteristics of the p-block Elements
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General Characteristics of the p-block Elements
Chapter: 18
| Module – VI: Chemistry of Elements |
INTEXT QUESTIONS 18.1
1. Which of the following atoms is expected to have smaller size?
(i) 9F and 17CI.
(ii) 6C and 14Si.
(iii) 5B and 6C.
(iv) 6C and 7N.
Ans: (i) 9F (ii) 6C (iii) 6C (iv) 7N.
2. Which atom in the following pairs of atoms is expected to have higher ionization enthalpy?
(i) 4Be and 5B.
(ii) 16S and 17CI.
(iii) 2He and 10Ne.
(iv) 😯 and 16S.
Ans: (i) 4Be (ii) 17Cl (iii) 2He (iv) 8O.
3. Arrange the following atoms in order of increasing ionization enthalpy: 2He,4Be, 7N, 11Na.
Ans: Na < Be < N < He.
4. How does the ionization enthalpy vary in general in a group and in a period of the p-block elements?
Ans: In a group, it decreases down the group and it usually increases along a period.
INTEXT QUESTIONS 18.2
1. Why does fluorine have electron gain enthalpy lower than chlorine?
Ans: The unexpectedly low value of electron gain enthalpy for F as compared to that of Cl atom may be attributed to the extremely small size of F atom visa-vis Cl atom. The addition of an electron produces a negative ion which has a high electron density and leads to increased interelectronic repulsions.
2. Which atom in the following pair of atoms has greater electron gain enthalpy?
(i) F, Cl
(ii) Br, I
(iii)I, Xe
(iv) O, F
(v) O,S
Ans: (i) Cl (ii) Br (iii) I (iv) F (v) S
3. Give two reasons for the fact that the first element in each group of p-block exhibits unique behaviour.
Ans: (i) Small size (ii) absence of d-orbitals.
4. Explain why oxygen exists as a gas whereas sulphur exists as a solid.
Ans: Because oxygen can form multiple bonds (O = O).
5. Mention two reasons which are responsible for the so called “inert pair effect”.
Ans: (i) Lower bond energy in the compounds of heavier atoms and.
(ii) the higher energy involved in promotion from the ground state (s²p¹) to the valence state (s¹p²).
6. What is the consequence of “inert pair effect” on the oxidation states of Tl and Pb?
Ans: Lower oxidation states become more stable. For Tl, +1 and for Pb, +2.
INTEXT QUESTION 18.3
1. Which of the following oxides is the most acidic?
(i) Al2₂O₃
(ii) CO₂
(iii) SO₂
Ans: (iii) SO₂
2. Which of the following hydrides of main group elements is the most acidic?
(i) H₂Se
(ii) H₂O
(iii) HCI
(iv) HI
Ans: (iv) HI.
3. Arrange the following in the increasing order of covalent character.
SiCl₄, CCl₄, SnCl₄, GeCl₄.
Ans: SnCl₄ < Ge Cl₄ < SiCl₄ < CCl₄
4. What happens when SiCl₄ reacts with water? Write complete chemical equation for the reaction.
Ans: SiCl₄+ 4H₂O → Si(OH)₄ + 4HCl
5. How do the bond angles vary among the following hydrides NH₃, PH₃, AsH₃, SbH₃.
Ans: The bond angle decreases from 107º to almost 90º.
6. Give equations for the formation of the following from the elements:
(i) Al₂O₃
(ii) SiCl₄
(iii) CCl₄
Ans: (i) 4Al(s) + 3O₂(g) → 2Al₂O₃ (s)
(ii) Si(s) + 2Cl₂(g) → SiCl₄(l)
(iii) C(s) + 2Cl₂(g) → CCl₄(l)
7. Which is more covalent in each of the following pairs?
(i) AlCl₃ and BCl₃
(ii) PbCl₂ and PbCl₄
Ans: (i) BCl₃
(ii) PbCl₄

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