Question

What is the shape of cuprammonium ions?
A.Octahedral
B.Tetrahedral
C.Trigonal
D.Square planar

Answer 1

Cuprammonium ion is a coordination complex that has copper ($Cu$) as a central metal atom and there are four amine groups attached as a ligand with the central metal atom. The central metal atom i.e. $Cu$ is present in $+ 2$ oxidation state. Also, copper is a d- block element.

Complete answer:
The IUPAC name of the coordination complex cuprammonium ion is tetraamminecopper (II) ion. Here copper ($Cu$) is present as a central metal atom and there are four amine groups attached as a ligand with the central metal atom. The oxidation state of copper ($Cu$) is $+ 2$
The chemical formula of cuprammonium ion is ${[Cu{(N{H_3})_4}]^{2 + }}$
Here, $Cu$is present in $+ 2$ oxidation state i.e. $C{u^{2 + }}$. The atomic number of Copper ($Cu$) and its electronic configuration of $Cu$ is $[Ar]\,3{d^{10}}\,4{s^1}$. When Copper ($Cu$) is in $+ 2$ oxidation state, then its electronic configuration is $[Ar]\,3{d^9}\,4{s^0}$. The d- orbital has nine electrons out of which one electron gets transferred to p- orbital. This is referred to as ‘$d$’ to ‘$p$’ transference.
$\boxed{ \uparrow \downarrow }\boxed{ \uparrow \downarrow }\boxed{ \uparrow \downarrow }\boxed{ \uparrow \downarrow }\boxed \uparrow \,\,\,\,\,\,\,\boxed{}\,\,\,\,\,\,\boxed{}\boxed{}\boxed{}$
$\,\,\,\,\,\,\,\,\,\,3d\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4s\,\,\,\,\,\,\,\,\,\,\,4p$
The unpaired electron in d- orbital is transferred to p- orbital
$\boxed{ \uparrow \downarrow }\boxed{ \uparrow \downarrow }\boxed{ \uparrow \downarrow }\boxed{ \uparrow \downarrow }\boxed{}\,\,\,\,\,\,\,\boxed{}\,\,\,\,\,\,\boxed{}\boxed{}\boxed \uparrow$
$\,\,\,\,\,\,\,\,\,\,3d\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4s\,\,\,\,\,\,\,\,\,\,\,4p$
Since, $N{H_3}$ is a strong field ligand hence, here hybridization will be $ds{p^2}$which indicates that the structure of the compound will be square planar.
Hence, the shape of the cuprammonium ion ${[Cu{(N{H_3})_4}]^{2 + }}$ is Square Planar.
Hence, the correct answer is an option (D).

Note:
However, $C{u^{2 + }}$ usually adopts a distorted octahedral geometry. This distortion is due to the Jahn- Teller distortion. All the cupric complexes irrespective of the kind of ligands are always $ds{p^2}$ square planar hybridization. Remember, copper is a d- block metal which is also known as a transition metal. Transition metal shows a magnetic moment due to the presence of unpaired electrons in d- orbitals. If there are no unpaired electrons, then it is diamagnetic in nature.