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 $B.M.$) is ___________.

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

Answer 1

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

$P=F{e}^{3+}$ (no. unpaired $e^{-}=5$)
$Q=V^{2+}$ (no. of unpaired $e^{-}=3$)
$R=F{e}^{2+}$ (no. unpaired $e^{-}=4$)

As all ligands are weak field, hence the no. of unpaired electrons remains the same in the complex ion.

$\mu =\sqrt{n(n+2)}B.M.$

where n is number of unpaired electrons.

Hence ($B$) is correct.