JEE Main 2026 — Chemistry Inorganic Chemistry

2026integerhard

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

Official previous-year question

Held on 23 Jan 2026 · Verified 6 Jul 2026.

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Solution

$[Ni(CO)_4]$ :

Ni is zero-valent with configuration $[Ar]3d^{10}4s^2$ .

CO is a strong field ligand causing complete pairing in the filled $3d^{10}$ orbital.

Unpaired electrons = 0.

$[NiCl_4]^{2-}$ :

Ni is $+2$ with configuration $[Ar]3d^8$ .

Cl is a weak field ligand producing high-spin tetrahedral geometry with configuration $3d^8$ : five orbitals with two electrons paired, three electrons unpaired.

Unpaired electrons = 2.

$[PtCl_2(NH_3)_2]$ :

Pt is $+2$ with configuration $[Xe]4f^{14}5d^8$ .

Pt(II) adopts square planar geometry with strong-field splitting leading to complete pairing: $5d^8$ configuration places all electrons in bonding orbitals.

Unpaired electrons = 0.

$[Ni(CN)_4]^{2-}$ :

Ni is $+2$ with $3d^8$ configuration.

CN is a strong field ligand.

Square planar geometry is preferred, resulting in all electrons paired.

Unpaired electrons = 0.

$[Pt(CN)_4]^{2-}$ :

Pt is $+2$ with $5d^8$ configuration.

CN is a strong field ligand producing square planar geometry with complete pairing.

Unpaired electrons = 0.

Total unpaired electrons = 0 + 2 + 0 + 0 + 0 = 2.

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Solution

$[Ni(CO)_4]$ :

Ni is zero-valent with configuration $[Ar]3d^{10}4s^2$ .

CO is a strong field ligand causing complete pairing in the filled $3d^{10}$ orbital.

Unpaired electrons = 0.

$[NiCl_4]^{2-}$ :

Ni is $+2$ with configuration $[Ar]3d^8$ .

Cl is a weak field ligand producing high-spin tetrahedral geometry with configuration $3d^8$ : five orbitals with two electrons paired, three electrons unpaired.

Unpaired electrons = 2.

$[PtCl_2(NH_3)_2]$ :

Pt is $+2$ with configuration $[Xe]4f^{14}5d^8$ .

Pt(II) adopts square planar geometry with strong-field splitting leading to complete pairing: $5d^8$ configuration places all electrons in bonding orbitals.

Unpaired electrons = 0.

$[Ni(CN)_4]^{2-}$ :

Ni is $+2$ with $3d^8$ configuration.

CN is a strong field ligand.

Square planar geometry is preferred, resulting in all electrons paired.

Unpaired electrons = 0.

$[Pt(CN)_4]^{2-}$ :

Pt is $+2$ with $5d^8$ configuration.

CN is a strong field ligand producing square planar geometry with complete pairing.

Unpaired electrons = 0.

Total unpaired electrons = 0 + 2 + 0 + 0 + 0 = 2.

JEE Main 2026 — Chemistry Inorganic Chemistry

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Solution

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JEE Main 2026 — Chemistry Inorganic Chemistry

Did you get this right?

Solution

Your note

JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

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