Question

Consider the following complex ions P, Q and R. $P=[Fe{F}_6{]}^{3-},Q=[V(H_{2}O{)}_6{]}^{2+}\text{and}R=[Fe(H_{2}O{)}_6{]}^{2+}$ 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

Answer 1

The correct option is B Q < R < P

$\text{Spin only magnetic moment}=\sqrt{n(n+2)}B.M.$
$n:$ number of unpaired electrons
$P:[Fe{F}_6{]}^{3-}$
$F^{-}:$ weak field ligand
${\Delta }_0<P$

number of unpaired electron = 5
$\mu =\sqrt{5(5+2)}=\sqrt{35}B.M.$

$Q:\left[V\right(H_{2}O{)}_6{]}^{2+}$
$H_{2}O:$ (weak field ligand)
${\Delta }_0<P$

number of unpaired electron = 3
$\mu =\sqrt{3(3+2)}=\sqrt{15}B.M.$

$R:\left[Fe\right(H_{2}O{)}_6{]}^{2+}$
$H_{2}O:$ (weak field ligand)
${\Delta }_0<P$

number of unpaired electron = 4
$\mu =\sqrt{4(4+2)}=\sqrt{24}B.M.$
Hence, Q < R < P