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

use the equation μ = √[n(n+2)] , where n = no. of unpaired electrons units = BM, Bohr Magneton)
no of unpaired electron in[FeF6]−3 is 5 thereforeμ=5.92BM
for[V(H2O)6]2+ no of unpaired d electron=2 μ=1.73BM
for[Fe(H2O)6]3+ no of unpaired d electron=4μ=4.89BM
therefore P>R>Q
ANS (B)