Total number of unpaired electrons present in the central metal atoms/ions of $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right],\left[\mathrm{NiCl}_{4}\right]^{2-},\left[\mathrm{PtCl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\right],\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$ and $\left[\mathrm{Pt}\left(\mathrm{CN}_{4}\right)\right]^{2-}$ is \_\_\_\_.

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Coordination Compounds

Total number of unpaired electrons present in the central metal atoms/ions of

[Ni (CO)_{4}], [NiCl_{4}]^{2 -}, [PtCl_{2} (NH_{3})_{2}], [Ni (CN_{4})]^{2 -}

and

[Pt (CN_{4})]^{2 -}

is ____.

Answer: 2

Solution

[Ni (CO)_{4}]

has Ni(0) with strong-field CO forcing a

3 d^{10}

configuration (

s p^{3}

), so it has no unpaired electrons;

[PtC l_{2} (N H_{3})_{2}]

[Ni (CN)_{4}]^{2 -}

and

[Pt (CN)_{4}]^{2 -}

each have a

d^{8}

metal in a square-planar

d s p^{2}

field with all electrons paired, contributing none. Only

[NiC l_{4}]^{2 -}

, with Ni

^{2 +}

(

3 d^{8}

) in a weak-field tetrahedral (

s p^{3}

) field, retains two unpaired electrons. The total number of unpaired electrons is

2

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Integer answer

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