Question

The complex ion $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ has:

• A. the tetrahedral configuration with one unpaired electron configuration
• B. square planar configuration with one unpaired electron
• C. tetrahedral configuration with all electrons paired
• D. square planar configuration with all electrons paired

Answer 1

The correct option is B square planar configuration with one unpaired electron

the hybridization should be $ds{p}^2$.

The Cu atom is in form of $C{u}^{+2}$ in the compound. In $Cu^{+2}, the electronic configuration would be

$1s^2,2s^2,2p^6,3s^2,3p^6,3d^9,4s^0$

So, there would be a rearrangement of electrons in $C{u}^{2+}$ because of the $N{H}_3$ ligand (which is a strong one). And the last electron in the d-orbital would be out waiting for the N's electrons to fill up first.

[because $N{H}_3$ is a strong ligand and the electrons donated from it should have got a place first--this is only punctuation]

So, the 4 electron pairs from N would be in one 3d, one 4s, & two 4p orbitals and in the third place of 4p orbital, the $e^{-}$ from $3d$ would take place. So, you can say the hybridisation here would be $ds{p}^2$. Square planar with one unpaired electron.