Question

The coordination number of $N{i}^{2+}$ is 4.
$NiC{l}_2$ + KCN(excess) $\to$ A (Cyano complex)
$NiC{l}_2$ + conc HCl(excess) $\to$ B (Chloro complex)

Predict the magnetic nature of A and B.

• A. Both are diamagnetic
• B. A is diamagnetic and B is paramagnetic with one unpaired electron
• C. A is diamagnetic and B is paramagnetic with two unpaired electrons
• D. Both are paramagnetic

Answer 1

The correct option is C A is diamagnetic and B is paramagnetic with two unpaired electrons

$\left[Ni\right(CN{)}_4{]}^{2-}$
The oxidation number of Ni is +2.
Atomic number = 28
Ni = $\left[Ar\right]3d^{8}4s^2$
$N{i}^{2+}=\left[Ar\right]3d^8$
There are no unpaired electrons, so the complex is diamagnetic.
Spin magnetic moment = zero
$[NiC{l}_4{]}^{2-}$
The oxidation number of Ni is +2.
Atomic number = 28
$Ni=$ $\left[Ar\right]3d^{8}4s^2$
$N{i}^{2+}=\left[Ar\right]3d^8$
Chloride is a weak field ligand, no pairing.
$s{p}^3$ hybridization (tetrahedral)
There are two unpaired electrons, so the complex is paramagnetic.
Spin magnetic moment
$\mu =\sqrt{n(n+2)}BM=\sqrt{2(2+2)}BM=\sqrt{8}BM$