JEE Main 2025 Apr 4 Shift 2, Chemistry Q11: Colour, Magnetism And Stability Of Co-ordination Compounds

JEE Main2025Apr 4, Shift 2Chemistry

The correct order of $[F e F_{6}]^{3 -}$ , $[C o F_{6}]^{3 -}$ , $[N i (C O)_{4}]$ and $[N i (C N)_{4}]^{2 -}$ complex species based on the number of unpaired electrons present is :

A
$[F e F_{6}]^{3 -} > [C o F_{6}]^{3 -} > [N i (C N)_{4}]^{2 -} > [N i (C O)_{4}]$
B
$[N i (C N)_{4}]^{2 -} > [F e F_{6}]^{3 -} > [C o F_{6}]^{3 -} > [N i (C O)_{4}]$
C
$[C o F_{6}]^{3 -} > [F e F_{6}]^{3 -} > [N i (C O)_{4}] > [N i (C N)_{4}]^{2 -}$
D
$[F e F_{6}]^{3 -} > [C o F_{6}]^{3 -} > [N i (C N)_{4}]^{2 -} = [N i (C O)_{4}]$

Solution

Count the unpaired $d$ -electrons in each complex.

$[F e F_{6}]^{3 -}$ : $F e^{3 +}$ is $d^{5}$ . $F^{-}$ is a weak-field ligand, so high spin: 5 unpaired electrons.

$[C o F_{6}]^{3 -}$ : $C o^{3 +}$ is $d^{6}$ . With weak-field $F^{-}$ , high spin gives $t_{2 g}^{4} e_{g}^{2}$ : 4 unpaired electrons.

$[N i (C N)_{4}]^{2 -}$ : $N i^{2 +}$ is $d^{8}$ . $C N^{-}$ is strong-field and the geometry is square planar; all electrons paired: 0 unpaired.

$[N i (C O)_{4}]$ : $N i$ is in zero oxidation state and the $4 s$ electrons promote into $3 d$ to give $3 d^{10}$ in a tetrahedral $s p^{3}$ complex: 0 unpaired.

$∴$ $[F e F_{6}]^{3 -} > [C o F_{6}]^{3 -} > [N i (C N)_{4}]^{2 -} = [N i (C O)_{4}]$ .

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JEE Main 2025 Apr 4 Shift 2, Chemistry Q11: Colour, Magnetism And Stability Of Co-ordination Compounds

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