Question

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

Answer 1

Hint: 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.