JEE Main 2026 — Jan 23, Shift 2 — Chemistry solution

JEE Main2026Jan 23, Shift 2Chemistry

Total number of unpaired electrons present in the central metal atoms/ions of $[N i (C O)_{4}]$ , $[N i C l_{4}]^{2 -}$ , $[P tC l_{2} (N H_{3})_{2}]$ , $[N i (C N)_{4}]^{2 -}$ and $[P t (C N)_{4}]^{2 -}$ is ______.

Your answer

Solution

Analyse each complex:

1) $[N i (C O)_{4}]$ : Ni is in the $0$ state, $3 d^{8} 4 s^{2}$ . With strong-field $C O$ , the configuration rearranges to $s p^{3}$ tetrahedral with all electrons paired. Unpaired = 0.

2) $[N i C l_{4}]^{2 -}$ : $N i^{2 +}$ is $3 d^{8}$ . Weak-field $C l^{-}$ gives $s p^{3}$ tetrahedral; two unpaired electrons remain in the $d$ -shell. Unpaired = 2.

3) $[P tC l_{2} (N H_{3})_{2}]$ : $P t^{2 +}$ is $5 d^{8}$ . Strong field of $N H_{3}$ + larger CFSE for $5 d$ metals leads to square planar $d s p^{2}$ with full pairing. Unpaired = 0.

4) $[N i (C N)_{4}]^{2 -}$ : $N i^{2 +}$ is $3 d^{8}$ . Strong-field $C N^{-}$ gives square planar $d s p^{2}$ with full pairing. Unpaired = 0.

5) $[P t (C N)_{4}]^{2 -}$ : $P t^{2 +}$ is $5 d^{8}$ . Square planar $d s p^{2}$ , all paired. Unpaired = 0.

Total = $0 + 2 + 0 + 0 + 0 = 2$ .

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