Consider the following complex ions P = [ FeF _ 6 ] ^ 3 – , Q = [ V...

$[ {FeF}_{6}{]}^{3 –}: {Fe}^{+ 3}: [ Ar ] 3{d}^{5}$

$F$: Weak field Ligand 

No. of unpaired electrons $= 5$

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

$\mu = \sqrt{35}BM$

${\left[V{\left({H}_{2}O\right)}_{6}\right]}^{+ 2}: {V}^{+ 2}: 3{d}^{3}$ 

No. of unpaired electrons $= 3$

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

$\mu = \sqrt{15}BM$

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

${H}_{2}O$ : Weak field Ligand 

No. of unpaired electrons $= 4$

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

$\mu = \sqrt{24}BM$

Hence, the answer is option C.