Question

When excess ammonia is added to $CuS{O}_4$ solution, the deep blue coloured complex is formed. Complex is:

• A. square planar, paramagnetic
• B. tetrahedral, diamagnetic
• C. tetrahedral, paramagnetic
• D. square planar, diamagnetic

Answer 1

The correct option is A

square planar, paramagnetic

When excess ammonia is added to copper sulphate solution, a deep blue colour solution is formed due to formation of a complex $\left[Cu\right(N{H}_3{)}_4{]}^{2+}.$

$CuS{O}_4\left(aq\right)+4N{H}_3\left(aq\right)\to \left[Cu\right(N{H}_3{)}_4{]}^{2+}$ (aq)

Excess Deep blue solution

In $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ complex,

The name of $N{H}_3$ is amine and it is a monodentate neutral ligand. The charge on $N{H}_3$ is $0.$

Let the oxidation state of $Cu=x$

$x+4\times \left(0\right)=+2$

$x=+2$

The electronic configuration of

$Cu=\left[Ar\right]3d^{10}4s^1$

The electronic configuration of

$C{u}^{2+}=\left[Ar\right]3d^{9}4s^0$

$N{H}_3$ is a strong field ligand. So, due to the presence of strong field ligands, one of the $3d-$ electrons undergoes transference to an empty $4p-$ orbital.

IMAGE

Unpaired electron $=1$

Hybridization is $ds{p}^2$ so the geometry of the complex is square planar.

It is paramagnetic in nature because it contains one unpaired electron.

Hence the correct answer is option (D)