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 magnetic moment :

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

Answer 1

Answer: (A)

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

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

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

Electronic configuration of $F{e}^{2+}$ is, $\left[Ar\right]3d^6$

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 the 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 the above compound as $H_{2}O$ is neutral.

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

There are two unpaired electrons 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 the above compound

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

There is two unpaired electrons in $e_g$ level.

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

So, the correct answer is (A)