JEE Main 2026 — Jan 24, Shift 1 — Chemistry solution

JEE Main2026Jan 24, Shift 1Chemistry

Given below are two statements :

Statement-I Hybridisation, shape and spin only magnetic moment of $K_{3} [C o (C O_{3})_{3}]$ is $s p^{3} d^{2}$ , octahedral and 4.9 BM respectively.

Statement-II Geometry, hybridisation and spin only magnetic moment values (BM) of the ions $[N i (C N)_{4}]^{2 -}$ , $[M n B r_{4}]^{2 -}$ and $[C o F_{6}]^{3 -}$ respectively are square planar, tetrahedral, octahedral : $d s p^{2}$ , $s p^{3}$ , $s p^{3} d^{2}$ and 0, 5.9, 4.9.

In the light of the above statements, choose the correct answer from the options given below

A
Both statement-I and statement-II are false
B
Statement I is false but statement-II is true
C
Both statement-I and statement-II are true
D
Statement-I is true but statement-II is false

Solution

Statement-I: In $K_{3} [C o (C O_{3})_{3}]$ , Co is in $+ 3$ state. With the weak field $C O_{3}^{2 -}$ ligand, $C o^{3 +}$ ( $d^{6}$ ) remains high-spin with 4 unpaired electrons, giving $s p^{3} d^{2}$ hybridisation, octahedral geometry, and $μ = 4 (4 + 2) = 24 \approx 4.9$ BM. True.

Statement-II: Checking each ion:

$[N i (C N)_{4}]^{2 -}$ : $N i^{2 +}$ ( $d^{8}$ ) with strong-field $C N^{-}$ → $d s p^{2}$ , square planar, 0 unpaired electrons → $μ = 0$ BM. ✓

$[M n B r_{4}]^{2 -}$ : $M n^{2 +}$ ( $d^{5}$ ) with weak-field $B r^{-}$ → $s p^{3}$ , tetrahedral, 5 unpaired electrons → $μ = 35 \approx 5.9$ BM. ✓

$[C o F_{6}]^{3 -}$ : $C o^{3 +}$ ( $d^{6}$ ) with weak-field $F^{-}$ → $s p^{3} d^{2}$ , octahedral, 4 unpaired electrons → $μ \approx 4.9$ BM. ✓

All three match. True.

$∴$ Both statements are correct.

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