Consider the following complex ions
P = [FeF₆]^3–
Q = [V(H₂O)₆]²⁺
R = [Fe(H₂O)₆]²⁺
The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :

[FeF₆]^3– : Fe⁺³ : [Ar]3d⁵
F : Weak field Ligand 
No. of unpaired electron’s = 5
$$\begin{gathered} \mathrm{\mu }=\sqrt{5(5+2)} \\ \begin{array}{l}\mathrm{\mu }=\sqrt{35}\mathrm{BM}\\ {\left[\mathrm{V}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{+2}:{\mathrm{V}}^{+2}:3{\mathrm{d}}^3\end{array} \end{gathered}$$

No. of unpaired electron’s = 3
$\begin{array}{l}\mathrm{\mu }=\sqrt{3(3+2)}\\ \mathrm{\mu }=\sqrt{15}\mathrm{BM}\\ {\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{+2}:{\mathrm{Fe}}^{+2}:3{\mathrm{d}}^6\end{array}$
H₂O : Weak field Ligand 
No. of unpaired electron’s = 4
$\begin{array}{l}\mathrm{\mu }=\sqrt{4(4+2)}\\ \mathrm{\mu }=\sqrt{24}\mathrm{BM}\end{array}$