The calculated spin-only magnetic moments B M of the anionic and cationic species of Fe H 2 O 6 + 2 and…

In both cases iron exhibits $+2$ oxidation state.

Electronic configuration of $F{e}^{+2}$ ion $=\left[\mathrm{A}\mathrm{r}\right]3{\mathrm{d}}^6$

In ${\left[\mathrm{F}\mathrm{e}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+},{\mathrm{H}}_2\mathrm{O}$ is a weak field ligand and hence, no pairing up.

$\therefore$ number of unpaired electron $\mathrm{n}=4$

$\Rightarrow$Spin magnetic moment $=\sqrt{4(4+2)}=\sqrt{24}=4.9\mathrm{B}\mathrm{M}$

In ${\left[\mathrm{F}\mathrm{e}{\left(\mathrm{C}\mathrm{N}\right)}_6\right]}^{4-},$ ${\mathrm{C}\mathrm{N}}^{-}$ is a strong field ligand and hence, pairing will occur.

$\therefore$ number of unpaired electron $\mathrm{n}=0$

$\Rightarrow$ spin magnetic moment = zero