JEE Main 2021 — Coordination Compounds Question with Solution

According to VBT i.e. valence bond theory,

Electronic configuration of Ni = [Ar]3d⁸4s².

(a) NiCl₂ . 6H₂O

NiCl₂ . 6H₂O $$\to$$ NiCl₂ + 6H₂O

Oxidation number of Ni(x) = x + 2($$-$$1) = 0; where x, $$-$$1 and 2 are the oxidation number of Ni, oxidation number of Cl and number of Cl atoms respectively.

NiCl₂ $$\Rightarrow$$ x + ($$-$$2) = 0 $$\Rightarrow$$ x = 2

Electronic configuration of Ni²⁺ = [Ar]3d⁸4s⁰

Cl^$$-$$ is a weak field ligand. So, no pairing of electrons occurs.

For C.N. = 6


(b) K₂[Ni(CN)₄]

K₂[Ni(CN)₄] $$\to$$ 2K⁺ + [Ni(CN)₄]^2$$-$$

x + 4($$-$$1) $$-$$ ($$-$$2) = 0; where x, 4, $$-$$1 and $$-$$2 are the oxidation number of Ni, number of CN ligands, charge on one CN and charge on complex.

[Ni(CN)₄]^2$$-$$ $$\Rightarrow$$ x $$-$$ 4 + 2 = 0 $$\Rightarrow$$ x = +2

Electronic configuration of Ni²⁺ $$\Rightarrow$$ [Ar]3d⁸4s⁰

CN^$$-$$ is a strong field ligand. So, pairing of electrons occur. For C.N. = 4


(c) Ni(CO)₄

CO is neutral and strong field ligand. So, pairing of electrons occur.

Oxidation number of Ni is zero

Electronic configuration of Ni = [Ar]3d¹⁰4s⁰

For C.N. = 4


(d) Na₂[NiCl₄]

Na₂[NiCl₄] $$\to$$ 2Na + [NiCl₄]^2$$-$$

x + 4($$-$$1) $$-$$ ($$-$$2) = 0; where x, 4, $$-$$1 and $$-$$2 are the oxidation number of Ni, number of CN ligands, charge on one CN and charge on complex.

[NiCl₄]^2$$-$$ $$\Rightarrow$$ x $$-$$ 4 + 2 = 0 $$\Rightarrow$$ x = +2

Electronic configuration of Ni²⁺ = [Ar]3d⁸4s⁰

For C.N. = 4


Hence, only K₂[Ni(CN)₄] has dsp² hybridization.