Number of Geometrical Isomers in square planar [PdFClBrI]^2– are = 3

Hence, n = 3

[Fe(CN)₆]^3–

Fe³⁺ = 3d⁵, According to CFT configuration is ${\mathrm{t}}_{2\mathrm{g}}^{221}{\mathrm{e}}_{\mathrm{g}}^{00}$

$\text{μ=}\sqrt{\mathrm{n}(\mathrm{n}+2)}$

=1.73 B.M.

$\text{CFSE }=–0.\text{4}{\mathrm{\Delta }}_0\mathrm{ }\times {\text{ nt}}_{\text{2g}}+0.\text{6}{\mathrm{\Delta }}_0\times {\text{n}}_{\text{eg}}\mathrm{ }=–0.\text{4}{\mathrm{\Delta }}_0\mathrm{ }\times \text{ 5}=–\text{2}.0{\mathrm{\Delta }}_0$