Question

What would be the geometrical shape of \([\mathrm{NiCl_4}]^{2-}\) and \([\mathrm{PtCl_4}]^{2-}\) compounds?

• A. Both \([\mathrm{NiCl_4}]^{2-}\) and \([\mathrm{PtCl_4}]^{2-}\) are square planar.
• B. Both \([\mathrm{NiCl_4}]^{2-}\) and \([\mathrm{PtCl_4}]^{2-}\) are tetrahedral.
• C. \([\mathrm{NiCl_4}]^{2-}\) is square planar while \([\mathrm{PtCl_4}]^{2-}\) is tetrahedral.
• D. \([\mathrm{NiCl_4}]^{2-}\) is tetrahedral while \([\mathrm{PtCl_4}]^{2-}\) is square planar.

Hint:

For \(d^8\) systems: \(Ni^{2+}\) with weak ligands usually forms tetrahedral complexes, whereas \(Pd^{2+}\) and \(Pt^{2+}\) commonly form square planar complexes.

Answer 1

The Correct Option is D

Step 1: Determine the oxidation state and electronic configuration of nickel.
For \([\mathrm{NiCl_4}]^{2-}\), \[ x+4(-1)=-2 \] \[ x=+2 \] Therefore, \[ Ni^{2+}=[Ar]\,3d^8 \] Since \(Cl^-\) is a weak field ligand, pairing of electrons does not occur. The complex uses outer orbitals and forms a tetrahedral geometry. \[ {[\mathrm{NiCl_4}]^{2-}\ \text{is tetrahedral}} \]

Step 2: Determine the oxidation state and electronic configuration of platinum.
For \([\mathrm{PtCl_4}]^{2-}\), \[ Pt^{2+}=5d^8 \] In \(5d\)-series metals, crystal field splitting is much larger. As a result, electron pairing occurs and the complex adopts \(dsp^2\) hybridisation. Hence the geometry becomes square planar. \[ {[\mathrm{PtCl_4}]^{2-}\ \text{is square planar}} \]

Step 3: Select the correct option.
\[ [\mathrm{NiCl_4}]^{2-} \rightarrow \text{Tetrahedral} \] \[ [\mathrm{PtCl_4}]^{2-} \rightarrow \text{Square planar} \] Therefore, \[ {\text{Option (D)}} \]