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 number of unpaired electrons :

• A. $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$
• B. $\left[Ni\right(CO{)}_4]$
• C. $\left[Ni\right(H_{2}O{)}_4{]}^{2+}$
• 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.

$\left[Ni\right(CO{)}_4]$

Ni is in 0 state in above compound as CO is neutral.

$Ni=\left[Ar\right]4s^{2}3d^8$

There is no unpaired electron because $CO$ is a strong field ligand and it pushes the two electrons from 4s orbital to 3d orbital to pair the unpaired electron.

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

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.

Therefore, option D is correct.