Consider the following complex ions, P, Q, and R. $\text{P=}{\left[{\mathrm{FeF}}_6\right]}^{3-},\mathrm{Q}={\left[\mathrm{V}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+} \mathrm{and} \mathrm{R}={\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$, the correct order of the complex ions, according to their spin only magnetic moment values (in B. M.) is:

$\underset{\left(\mathrm{P}\right)}{{\left[{\mathrm{FeF}}_6\right]}^{3-}}\to {\mathrm{Fe}}^{3+}\to {\mathrm{d}}^5$ configuration, 

${\mathrm{F}}^{-}$ is weak field ligand 

Unpaired electrons = 5; $\text{∴\hspace{0.33em}μ=}\sqrt{5\left(5+2\right)}=5.95\mathrm{\hspace{0.33em}}\mathrm{B}.\mathrm{M}.$

$\underset{\left(\mathrm{Q}\right)}{{\left[\mathrm{V}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}}\to {\mathrm{V}}^{+2}\to {\mathrm{d}}^3$ configuration, 

${\mathrm{H}}_2\mathrm{O}$ is weak field ligand 

Unpaired electron = 3; $\therefore \mathrm{\mu }=\sqrt{3\left(3+2\right)}=3.87\mathrm{B}.\mathrm{M}.$

$\underset{\left(\mathrm{R}\right)}{{\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}}\to {\mathrm{Fe}}^{+2}\to {\mathrm{d}}^6$ configuration, 

${\mathrm{H}}_2\mathrm{O}$ is weak field ligand 

Unpaired electron = 4; $\therefore \mathrm{\mu }=\sqrt{4\left(4+2\right)}=4.90\mathrm{B}.\mathrm{M}.$

Therefore only spin magnetic moment $\left(\mathrm{\mu }\right)=\sqrt{\mathrm{n}\left(\mathrm{n}+2\right)}$

B.M. (n = unpaired electron) 

So, $\mu$ of P > R > Q.