JEE Main 2021 — Coordination Compounds Question with Solution

In presence of SFL $$\Delta$$₀ > P means pairing occurs therefore

For Fe⁺² $$\to$$ 3d⁶



$$\therefore$$ No of unpaired e^-(s) = 0

$$\therefore$$ $$\mu$$ = $$\sqrt{n(n+2)}$$BM = 0

[n = No of unpaired e^$$-$$(s)]

In NiCl₂, Ni⁺² is having configuration 3d⁸

$$\therefore$$ Number of unpaired electron = 2

After formation of oxidised product

[Ni(CN)₆]^$$-$$2$$\Rightarrow$$ Ni⁺⁴ is obtained

Ni⁺⁴ $$\Rightarrow$$ 3d⁶ and CN^$$-$$ is strong field ligand

$$\therefore$$ Number of unpaired electrons = 0

$$\therefore$$ The charge is 2 $$-$$ 0 = 2