JEE Main 2026 — Coordination Compounds Question with 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.

$[NiC{l}_4{]}^{2-}$:
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.

$[PtC{l}_2(N{H}_3{)}_2]$:
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.