Question

If which of the following orders the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment?
(i) $[Fe{F}_6{]}^{3-}$
(ii) $\left[Co\right(N{H}_3{)}_6{]}^{3+}$
(iii) $[NiC{l}_4{]}^{2-}$
(iv) $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$

• A. $\left(ii\right)>\left(i\right)>\left(iii\right)>\left(iv\right)$
• B. $\left(iii\right)>\left(iv\right)>\left(ii\right)>\left(i\right)$
• C. $\left(ii\right)>\left(iii\right)>\left(i\right)>\left(iv\right)$
• D. $\left(i\right)>\left(iii\right)>\left(iv\right)>\left(ii\right)$

Answer 1

The correct option is D $\left(i\right)>\left(iii\right)>\left(iv\right)>\left(ii\right)$

$[Fe{F}_6{]}^{3-}:F{e}^{3+}[Ar]3d^5\to 5$-unpaired electrons as $F^{-}$ is weak field ligand.

$\left[Co\right(N{H}_3{)}_6{]}^{3+}:C{o}^{3+}\left[Ar\right]3d^6\to$ No unpaired electron as $N{H}_3$ is strong field light and causes pairing.

$[NiC{l}_4{]}^{2-}:N{i}^{2+}[Ar]3d^8\to 2$-unpaired electrons

$\left[Cu\right(N{H}_3{)}_4{]}^{2+}:C{u}^{2+}\left[Ar\right]3d^9\to 1$ -unpaired electron.
We know that,
$\text{Spin only magnetic moment}\left(\mu \right)=\sqrt{n(n+2)}$
Where, $n=\text{number of unpaired electrons}$
$\Rightarrow \mu \propto n$
Hence, the correct order of decreasing spin only magnetic moment is $\left(i\right)>\left(iii\right)>\left(iv\right)>\left(ii\right)$.