Question

Which of the following complex has square planar structure?

• A. ${\left[Ni(CN{)}_4\right]}^{2-}$
• B. $Ni(CO{)}_4$
• C. ${\left[Zn{\left(N{H}_3\right)}_4\right]}^{2+}$
• D. ${\left[NiC{l}_4\right]}^{2-}$

Answer 1

Answer: (A)

$\left[Ni(CO{)}_4\right]$ - has a tetrahedral structure.
${\left[Ni(Cl{)}_4\right]}^{2-}$ - has a tetrahedral structure.
${\left[Ni(CN{)}_4\right]}^{2-}$ - has a square planar structure.
In ${\left[Ni(CN{)}_4\right]}^{2-},Ni$ is in $+2$ state
$N{i}^{2+}\Rightarrow [Ar]4s^{0}3d^8\Rightarrow 4s^{0}3d_{xy}^{2}3d_{yz}^{2}3d_{zz}^{2}3d_{z^2}^{2}3d_{x^2-y^2}^0[\because$
The ligand is strong it will pair up single electrons in the $d$-orbital] Hence the hybridization would be $ds{p}^2$ and the structure is square planar structure.