Question

Consider the following complex ions:
$P,Q,$ and $R$. $P=[Fe{F}_6{]}^{3-},Q=[V(H_{2}O{)}_6{]}^{2+}$ and $R=\left[Fe\right(H_{2}O{)}_6{]}^{2+}$
The correct order of the complex ions, according to their spin only magnetic moment values (in BM) is:

• A. $Q<R<P$
• B. $R<Q<P$
• C. $R<P<Q$
• D. $Q<P<R$

Answer 1

The correct option is A $Q<R<P$

${\mu }_{\text{spin only}}=\sqrt{n(n+2)}$
Magnetic moment arises due to the presence of unpaired electrons.
More the number of unpaired electrons, greater will be the magnetic moment.
$P$ has $Fe$ in $+3$ oxidation state.
$Fe(+3):3d^5$
$Q$ has $V$ in $+2$ oxidation state.
$V(+2):3d^3$
$R$ has $Fe$ in $+2$ oxidation state.
$Fe(+2):3d^6$
Hence the number of unpaired electrons in $P,Q$ and $R$ are $5,3$ and $4$ respectively.