Question

The complex ion $[Cu(NH_3​)4]^{2+}$ has:
(A) the tetrahedral configuration with one unpaired electron
(B) square planar configuration with one unpaired electron
(C) the tetrahedral configuration with all electrons paired
(D) square planar configuration with all electrons paired

Answer 1

Hint: This is an exceptional case which is explained correctly by ligand field theory, just assume that it is square planar and select your answer from the remaining options.

Complete step by step solution:
It is very convenient to use crystal field theory to discuss this.
It is usually assumed that in octahedral coordination the energy levels of the five d-orbitals are split, with two orbitals ($d_{z^2}$ and $d_{x^2−y^2}$) well above the other three. The splitting is assumed to be large enough to overcome electron pairing energy.
The first six electrons will populate three lower orbitals ($d_{xy}$, $d_{xz}$,$d_{yz}$). The next two electrons should occupy $d_{z^2}$ and $d_{x^2−y^2}$). However, if these two electrons occupy $d_{z^2}$ and, two corresponding opposite ligands become well shielded and leave. This is the common case of d8d8 complexes.
Adding one more electron makes the remaining ligand's bonding weaker. The charge of bivalent cation $Cu^{2+}$ is large enough to bond four charged ligands if available, such as in $[CuCl_4]^{2-}$, but the ligands are bound weakly and easily dissociate.
But, $Cu^{2+}$ ions usually adopt a distorted octahedral geometry, with two ligands having a longer bond length than the four others. These two are very weakly bound and exchange quickly. For this reason, the $NH_3$ complex is written with only four molecules; the two others are so weakly bound.
Copper has 9 electrons. The $d_{z^2}$ and $d_{x^2−y^2}$) are at a lower energy than the ($d_{xy}$, $d_{xz}$,$d_{yz}$ orbitals. Because there are 9 electrons, these three degenerate orbitals are not occupied equally (2 of them have 2 electrons but the other only has 1). This causes the complex to distort its geometry to rid itself of the degeneracy, which causes the formation of the square planar complex. This square planar complex is more energetically favourable.

So, the correct answer is option B: The complex-ion $[Cu(NH_3​)4]^{2+}$ has the square planar configuration with one unpaired electron.

Note: Although it can be explained at the 9th electron gets excited to p orbital and we have a $dsp^2$ hybridization, which makes the compound square planar, this is not the correct explanation or the complete picture.