Match the species in List-I with their geometry in List-II. List-I: A. PCl₅, B. BrF₅, C. BF₄-, D. [Ni(CN)₄]²⁻. List-II: I. Tetrahedral, II. Square planar, III. Trigonal bipyramidal, IV. Square pyramidal.
- A.A-IV, B-III, C-I, D-II
- B.A-III, B-IV, C-I, D-II✓
- C.A-III, B-I, C-II, D-IV
- D.A-III, B-II, C-I, D-IV
Correct Answer
(B) A-III, B-IV, C-I, D-II
Solution & Explanation
Concept: use VSEPR / hybridisation to assign shapes; count bond pairs and lone pairs on the central atom. A. PCl₅ — P has 5 bond pairs, 0 lone pairs → sp³d → trigonal bipyramidal = III. B. BrF₅ — Br has 5 bond pairs + 1 lone pair (6 electron domains) → sp³d² with one lone pair → square pyramidal = IV. C. BF₄⁻ — B has 4 bond pairs, 0 lone pairs → sp³ → tetrahedral = I. D. [Ni(CN)₄]²⁻ — Ni²⁺ (d⁸) with strong-field CN⁻ → dsp² hybridisation → square planar = II. Trap: BrF₅ is square pyramidal (lone pair occupies one octahedral site), not octahedral or trigonal bipyramidal; and [Ni(CN)₄]²⁻ is square planar, not tetrahedral, because CN⁻ is a strong field ligand that pairs the d electrons. So A-III, B-IV, C-I, D-II. Hence the answer is B.
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