Consider the following complex ions P = [ FeF 6 ] 3 – , Q = [ V ( H 2 O ) 6 ] 2 + , R = [ Fe ( H 2 O ) 6 ] 2…

$[{\mathrm{FeF}}_6{]}^{3–} : {\mathrm{Fe}}^{+3} : [\mathrm{Ar}] 3{\mathrm{d}}^5$

$\mathrm{F}$: Weak field Ligand 

No. of unpaired electrons $=5$

$\mathrm{\mu }=\sqrt{\left(5\right(5+2\left)\right) }$

$\mathrm{\mu }=\sqrt{35}\mathrm{BM}$

${\left[\mathrm{V}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{+2}: {\mathrm{V}}^{+2}: 3{\mathrm{d}}^3$ 

No. of unpaired electrons $=3$

$\mathrm{\mu }=\sqrt{3(3+2)}$

$\mathrm{\mu }=\sqrt{15}\mathrm{BM}$

${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{+2}: {\mathrm{Fe}}^{+2}:3{\mathrm{d}}^6$

${\mathrm{H}}_2\mathrm{O}$ : Weak field Ligand 

No. of unpaired electrons $=4$

$\mathrm{\mu }=\sqrt{4(4+2)}$

$\mathrm{\mu }=\sqrt{24}\mathrm{BM}$

Hence, the answer is option C.