P d F C l B r I 2 - has n number of geometrical isomers. Then, the spin-only magnetic moment and crystal…

Number of Geometrical Isomers in square planar ${\left[\mathrm{P}\mathrm{d}\mathrm{F}\mathrm{C}\mathrm{l}\mathrm{B}\mathrm{r}\right]}^{2-}$ are $=3$
Hence, $\mathrm{n}=3$

${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}$
${\mathrm{Fe}}^{3+}=3{\mathrm{d}}^5,$ According to CFT, configuration is ${\mathrm{t}}_{2\mathrm{g}}^{221}{\mathrm{e}}_{\mathrm{g}}^{00}$; it means number of unpaired electron $\left(\mathrm{n}\right)=1$

So spin-only magnetic moment
$\mathrm{\mu }=\sqrt{\mathrm{n}\left(\mathrm{n}+2\right)}=1.73 \mathrm{BM}$

$\mathrm{CFSE}=\left(-0.4{\mathrm{\Delta }}_0\times {\mathrm{nt}}_{2\mathrm{g}}\right)+\left(0.6{\mathrm{\Delta }}_0\times {\mathrm{n}}_{\mathrm{eg}}\right)$ $=-0.4{∆}_0\times 5=-2.0{∆}_0$