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Chemical Bonding and Molecular Structure question

2025 · 7 Apr · Shift 1 · Q6
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Chemical Bonding and Molecular Structure question

2025 · 7 Apr · Shift 1 · Q6

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1

Match the List I with List II.

List - I
Molecule/Ion		 List - II
Bond pair : lone pair
(on the central atom)
(A) ICl2−\mathrm{ICl}_2^{-}ICl2−​ (I) 4:24:24:2
(B) H2O\mathrm{H}_2 \mathrm{O}H2​O (II) 4:14:14:1
(C) SO2\mathrm{SO_2}SO2​ (III) 2:32:32:3
(D) XeF4\mathrm{XeF_4}XeF4​ (IV) 2:22:22:2

Choose the correct answer from the options given below:

  1. A
    A-IV, B-III, C-II, D-I
  2. B
    A-III, B-IV, C-I, D-II
  3. C
    A-II, B-I, C-IV, D-III
  4. D
    A-III, B-IV, C-II, D-I
View written solutionFree

Correct answer: D

  1. We need to match each molecule/ion with the ratio bond pair : lone pair on the central atom.\text{bond pair : lone pair on the central atom}.bond pair : lone pair on the central atom.

Let us determine the central atom and count bond pairs and lone pairs.

  1. For (A)  ICl2−(A)\; \mathrm{ICl_2^-}(A)ICl2−​
  • Central atom = Iodine.
  • Total valence electrons: 7+2(7)+1=227 + 2(7) + 1 = 227+2(7)+1=22
  • Two I–Cl bonds are formed, so iodine has 2 bond pairs.
  • After completing octets, iodine is left with 3 lone pairs.

So, ICl2−→2:3\mathrm{ICl_2^-} \rightarrow 2:3ICl2−​→2:3 This matches (III).

Thus, A→IIIA \to IIIA→III

  1. For (B)  H2O(B)\; \mathrm{H_2O}(B)H2​O
  • Central atom = Oxygen.
  • Oxygen has 6 valence electrons.
  • It forms two O–H bonds, so there are 2 bond pairs.
  • Remaining 4 electrons on oxygen form 2 lone pairs.

So, H2O→2:2\mathrm{H_2O} \rightarrow 2:2H2​O→2:2 This matches (IV).

Thus, B→IVB \to IVB→IV

  1. For (C)  SO2(C)\; \mathrm{SO_2}(C)SO2​
  • Central atom = Sulfur.
  • In VSEPR counting, each S=O double bond counts as one bonding domain, but here the question asks for bond pair : lone pair on the central atom.
  • Sulfur in SO2\mathrm{SO_2}SO2​ has two S=O double bonds, i.e. 4 bonding electrons pairs shared around sulfur, and 1 lone pair on sulfur.

Hence, SO2→4:1\mathrm{SO_2} \rightarrow 4:1SO2​→4:1 This matches (II).

Thus, C→IIC \to IIC→II

  1. For (D)  XeF4(D)\; \mathrm{XeF_4}(D)XeF4​
  • Central atom = Xenon.
  • Xenon forms four Xe–F bonds, so there are 4 bond pairs.
  • Xenon also has 2 lone pairs.

So, XeF4→4:2\mathrm{XeF_4} \rightarrow 4:2XeF4​→4:2 This matches (I).

Thus, D→ID \to ID→I

  1. Final matching:

A→III,B→IV,C→II,D→IA \to III, \quad B \to IV, \quad C \to II, \quad D \to IA→III,B→IV,C→II,D→I

This corresponds to Option D.

  1. Comparison with stored answer:

Stored correct answer = D

Our derived answer = D

So, they agree.

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