Question

Predict the geometry and number of unpaired spins in the following four-coordinate complexes:${\left[\mathbf{Ni}{\left(\mathbf{CN}\right)}_{\mathbf{4}}\right]}^{\mathbf{2}\mathbf{-}}$,${\left[{\mathbf{NiCl}}_{\mathbf{4}}\right]}^{\mathbf{2}\mathbf{-}}$. Explain your reasoning.

Short answer

A coordination compound is made up of a central metal atom and ligand which are usually organized in a coordination sphere. Ligands can be unidentate, bidentate, and ambidentate.

Step-by-step solution

Coordinate Complexes

A coordination compound is made up of a central metal atom and ligand which are usually organized in a coordination sphere. Ligands can be unidentate, bidentate, and ambidentate.

Explanation

As Ni is present in both compounds. The oxidation state of Ni is +2 in both compounds, and Ni has $d^8$which are arranged as follows:

In ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$, the ligand is $C{N}^{-}$and has four ions in the complex. These $C{N}^{-}$ions are strong filed ligands, which caused the pairing of unpaired 3d electrons. Thus, the configuration is as shown:

Step 3:

This compound undergoes $ds{p}^2$hybridization and there is no lone pair of electrons on the central Ni ion. Valence bond theory proposes the use of $ds{p}^2$hybridization for $N{i}^{+2}$and predicts the geometry to be square planar.

This compound undergoes $ds{p}^2$hybridization and there is no lone pair of electrons on the central Ni ion. Valence bond theory proposes the use of $ds{p}^2$hybridization for $N{i}^{+2}$and predicts the geometry to be square planar.

Square planar geometry of${\left[Ni{\left(CN\right)}_4\right]}^{2-}$

In the case of ${\left[NiC{l}_4\right]}^{2-}$, the ligand is and has four ions in the complex. These $C{l}^{-}$is a weak field ligand, which does not lead to the pairing of 3d electrons. Thus, the configuration is as shown:

As the pairing does not take place due to the presence of a weak field ligand, there are two unpaired 3d electrons present on Ni ion. The valence bond theory proposes the use of $s{p}^3$hybridization for $N{i}^{2+}$and predicts the geometry to be tetrahedral.