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JEE Main 202623 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Coordination Compounds

JEE Main 2026 · 23 January · Evening Shift · Chemistry · Coordination Compounds

Total number of unpaired electrons present in the central metal...

Total number of unpaired electrons present in the central metal atoms/ions of [Ni(CO)4],[NiCl4]2−,\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[Pt(CN)4]2−and \left[\mathrm{Pt}(\mathrm{CN})_{4}\right]^{2-} is ____\_\_\_\_.

Correct answer

2

Step-by-step solution

[Ni(CO)4][Ni(CO)_4]:
Ni is zero-valent with configuration [Ar]3d104s2[Ar]3d^{10}4s^2.
CO is a strong field ligand causing complete pairing in the filled 3d103d^{10} orbital.
Unpaired electrons = 0.

[NiCl4]2−[NiCl_4]^{2-}:
Ni is +2+2 with configuration [Ar]3d8[Ar]3d^8.
Cl is a weak field ligand producing high-spin tetrahedral geometry with configuration 3d83d^8: five orbitals with two electrons paired, three electrons unpaired.
Unpaired electrons = 2.

[PtCl2(NH3)2][PtCl_2(NH_3)_2]:
Pt is +2+2 with configuration [Xe]4f145d8[Xe]4f^{14}5d^8.
Pt(II) adopts square planar geometry with strong-field splitting leading to complete pairing: 5d85d^8 configuration places all electrons in bonding orbitals.
Unpaired electrons = 0.

[Ni(CN)4]2−[Ni(CN)_4]^{2-}:
Ni is +2+2 with 3d83d^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(CN)_4]^{2-}:
Pt is +2+2 with 5d85d^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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