JEE Main 20 July 2021 Shift 2 Solved Paper

In presence of SFL Δ0>P means pairing occurs therefore
For Fe+2→3d6 ∴ No of unpaired e−(s)=0
∴µ=√n(n+2)BM=0
[n= No of unpaired e−(s)]
In NiCl2Ni+2 is having configuration 3d8
∴ Number of unpaired electron =2
After formation of oxidised product
[Ni(CN)6]−2Ni+4 is obtained
Ni+4⇒3d6 and CN−is strong field ligand
∴ number of unpaired electrons =0
∴ The charge is 2−0=2