The calculated spin-only magnetic moments $\left(\mathrm{B}\mathrm{M}\right)$ of the anionic and cationic species of ${\left[\text{Fe}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{+2}$ and ${\left[\text{Fe }{\left(\text{CN}\right)}_6\right]}^{4-}$ respectively, are:

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