JEE Main Chemistry Question (2026) — Solution

Step-by-step 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.