Question

When the excess of ammonia is added to the copper sulphate solution, the deep blue coloured complex is formed. The complex is ______.
$A.$ Tetrahedral and paramagnetic
$B.$ Tetrahedral and diamagnetic
$C.$ Square planar and diamagnetic
$D.$ Square planar and paramagnetic

Answer 1

Before solving the given question we have an idea about the product formed. When excess of ammonia is added to the copper sulphate solution the deep blue coloured complex is formed. Here is the reaction for this process,
$CuS{O_4} + N{H_3} \to \left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]S{O_4} + 4{H_2}O$

Complete answer:
As we know the complex formed is $\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]S{O_4}$ , in the given question we have asked to determine the magnetic nature and geometry of this complex.
Magnetic nature of a complex depends on the number of unpaired electrons present in the $3d -$ orbital of metal atoms or ions. If an unpaired electron is present then it is paramagnetic and if an unpaired electron is not present then it is diamagnetic.
In this complex the central metal atom is copper. In this complex copper is in $+ 2$ oxidation state. The electronic configuration of $C{u^{2 + }}$ is $\left[ {Ar} \right]4{s^0}3{d^8}$, as $N{H_3}$ is a weak ligand so pairing of electrons do not take place. So there is one unpaired electron in $3d -$ orbital. So, the $\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]S{O_4}$ complex is paramagnetic in nature.
For geometry of $\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]S{O_4}$ complex, the central metal atom is copper. In this complex copper is in $+ 2$ oxidation state. The electronic configuration of $C{u^{2 + }}$ is $\left[ {Ar} \right]4{s^0}3{d^8}$, four $N{H_3}$ ligand occupies the one $3d$ , one $4s$ and two $4p$ orbital then the hybridization of this complex is $ds{p^2}$ . These types of hybridization show square planar complexes.
Therefore, the deep blue complex is formed. The complex square planar and paramagnetic

So, the correct option is $D$ .

Note:
In this question we can direct determine the magnetic nature of $\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]S{O_4}$ because it is a deep colour complex, we also know the colour of complex is also depend on the number of unpaired electrons present in $3d -$ orbital. So we can directly answer this question $\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]S{O_4}$ is paramagnetic.