Question

Consider following complexes :

I: $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$
II :$\left[Fe\right(CN{)}_6{]}^{4-}$
III :$\left[Ni\right(CO{)}_4]$
IV :$\left[Ni\right(H_{2}O{)}_4{]}^{2+}$
V :$\left[Ni\right(CN{)}_4{]}^{2-}$

Out of this, select the complex with equal magnetic moment :

• A. $\left[Fe\right(CN{)}_6{]}^{4-}$ and $\left[Ni\right(CN{)}_4{]}^{2-}$
• B. $\left[Ni\right(CO{)}_4]$ and $\left[Fe\right(CN{)}_6{]}^{4-}$
• C. $\left[Ni\right(H_{2}O{)}_4{]}^{2+}$ and $\left[Ni\right(CO{)}_4]$
• D. None of these

Answer 1

The correct option is D None of these

$\left[Fe\right(H_{2}O{)}_6{]}^{2+}$

$Fe$ is in $+2$ state in above compound as $H_{2}O$ don't have any charge on it.

$F{e}^{2+}-\left[Ar\right]3d^6$

There are four unpaired electrons in 3d orbital as $H_{2}O$ is a weak ligand.

Magnetic Moment- $\sqrt{n(n+2)}$- $\sqrt{24}$ BM

$\left[Fe\right(CN{)}_6{]}^{4-}$
$x-6=-4$
$x=+2$

Fe is in +2 state in above compound.

$F{e}^{2+}-\left[Ar\right]3d^6$

There is no unpaired electron because $C{N}^{-}$ is a strong field ligand and paired the electrons due to high energy difference between $t_{2g}$ level and $e_g$ level.

Magnetic Moment = 0

$\left[Ni\right(H_{2}O{)}_4{]}^{2+}$

$Ni$ is in $+2$ state in above compound as $H_{2}O$ is neutral.

$N{i}^{2+}-\left[Ar\right]3d^8$

There are two unpaired electron in $e_g$ level.

Magnetic Moment- $\sqrt{n(n+2)}$- $\sqrt{8}$ BM

$\left[Ni\right(CN{)}_4{]}^{2-}$
$x-4=-2$
$x=+2$

$Ni$ is in $+2$ state in above compound.

$N{i}^{2+}-\left[Ar\right]3d^8$

There are two unpaired electron in $e_g$ level.

Magnetic Moment- $\sqrt{n(n+2)}$- $\sqrt{8}$ BM

Therefore, option D is correct.