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 maximum number of unpaired electrons.

• A. $\left[Fe\right(CN{)}_6{]}^{4-}$
• B. $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$
• C. $\left[Ni\right(CO{)}_4]$
• D. $\left[Ni\right(H_{2}O{)}_4{]}^{2+}$

Answer 1

The correct option is B $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$

$\left[Fe\right(H_{2}O{)}_6{]}^{2+}$
$Fe$ 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.
$\left[Fe\right(CN{)}_6{]}^{4-}$
$x-6=-4$
$x=+2$
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
$\left[Ni\right(CO{)}_4]$
$Ni$ 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.
$\left[Ni\right(CN{)}_4{]}^{2-}$
$x-4=-2$
$x=+2$
$Ni$ is in +2 state in above compound
$N{i}^{2+}-\left[Ar\right]3d^8$
There are two unpaired electron in $e_g$ level.