Question

Which of the following has square planar structure?

• A. $[NiC{l}_4{]}^{2\ominus }$
• B. $\left[Ni\right(CO{)}_4]$
• C. $\left[Ni\right(CN{)}_4{]}^{2\ominus }$
• D. $\left[Cd\right(CN{)}_4{]}^{2\ominus }$

Answer 1

The correct option is C $\left[Ni\right(CN{)}_4{]}^{2\ominus }$

As per VBT, cyanide is a strong field ligand. This means that it would force pairing of nickel electrons in the d subshells.
Remember, $Ni$ in its ground state has a configuration of $3d^8,4s^2$, hence $N{i}^{2+}$ would have a configuration of $3d^8$. Due to cyanide, 4 d orbitals would get completely filled up by the 8 d electrons which leaves the inner d oribtal of nickel open to hybdridisation.
Due to this, the given complex becomes an inner orbital complex i.e $\left[Ni\right(CN{)}_4{]}^{2-}$ has $ds{p}^2$ hybridisation and hence square planar geometry.