JEE Main 2025 Jan 28 Shift 2, Chemistry Q11: Bonding Of Co-ordination Compounds

JEE Main2025Jan 28, Shift 2Chemistry

Match List-I with List-II.

List-I (Complex)	List-II (Hybridisation of central metal ion)
(A) $[C o F_{6}]^{3 -}$	(I) $d^{2} s p^{3}$
(B) $[N i C l_{4}]^{2 -}$	(II) $s p^{3}$
(C) $[C o (N H_{3})_{6}]^{3 +}$	(III) $s p^{3} d^{2}$
(D) $[N i (C N)_{4}]^{2 -}$	(IV) $d s p^{2}$

Choose the correct answer from the options given below :

A
(A)-(I), (B)-(IV), (C)-(III), (D)-(II)
B
(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
C
(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
D
(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Solution

Hybridisation is determined by the metal's d-electron count and the field strength of the ligand:

(A) $[C o F_{6}]^{3 -}$ : $C o^{3 +}$ is $d^{6}$ , $F^{-}$ is a weak-field ligand so no pairing. Outer-orbital complex uses $4 d$ orbitals giving $s p^{3} d^{2}$ hybridisation (octahedral, paramagnetic). (A) $\to$ (III).

(B) $[N i C l_{4}]^{2 -}$ : $N i^{2 +}$ is $d^{8}$ , $C l^{-}$ is weak field, giving $s p^{3}$ hybridisation (tetrahedral, paramagnetic). (B) $\to$ (II).

(C) $[C o (N H_{3})_{6}]^{3 +}$ : $C o^{3 +}$ ( $d^{6}$ ) with strong-field $N H_{3}$ ligands pairs electrons and uses inner $3 d$ orbitals, $d^{2} s p^{3}$ hybridisation (octahedral, diamagnetic). (C) $\to$ (I).

(D) $[N i (C N)_{4}]^{2 -}$ : $N i^{2 +}$ ( $d^{8}$ ) with strong-field $C N^{-}$ pairs the electrons in $3 d$ leaving one orbital free, $d s p^{2}$ hybridisation (square planar, diamagnetic). (D) $\to$ (IV).

$∴$ (A)-(III), (B)-(II), (C)-(I), (D)-(IV).

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JEE Main 2025 Jan 28 Shift 2, Chemistry Q11: Bonding Of Co-ordination Compounds

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