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Chemical Bonding and Molecular Structure · Question 2

JEE Advanced 2026 Paper 1MatchingMedium📈 Topic in 17% papers+4−1
1

Consider the following species: SOCl2, XeOF4, ClF3, ClF5, XeF5+, SO32-, XeF3+, SF4 List-I contains different molecular shapes and List-II contains total number of species with the same molecular shapes from the given species. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

  1. P → 1; Q → 2; R → 5; S → 3
  2. P → 5; Q → 4; R → 2; S → 3
  3. P → 3; Q → 2; R → 1; S → 4
  4. P → 1; Q → 3; R → 5; S → 4

Answer

(1) P → 1; Q → 2; R → 5; S → 3

Solution

Shapes from VSEPR (bond pairs + lone pairs on the central atom):

SOCl2 (3 bp + 1 lp): trigonal pyramidal. SO32- (3 bp + 1 lp): trigonal pyramidal.

XeOF4, ClF5, XeF5+ (5 bp + 1 lp): square pyramidal, so 3 species.

ClF3, XeF3+ (3 bp + 2 lp): T-shaped, so 2 species. SF4 (4 bp + 1 lp): see-saw, so 1 species.

None of the species is trigonal planar, so 0.

P → 1, Q → 2, R → 5, S → 3

Answer: (A)

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  2. NO2-<NO3-<NO2<NO2+
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  2. The highest occupied molecular orbital (HOMO) of F2 is σ-type.
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  2. NO2, O3, HCl and H2SO4
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