Question

What is the hybridization for $[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4{]}^{2-}$?

Answer 1

Hybridization for $[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4{]}^{2-}$:

• Nickel$\left(\mathrm{Ni}\right)$ has atomic number $28$ so its electronic configuration of Nickel is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{8}4{\mathrm{s}}^2$.
• In $[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4{]}^{2-}$, nickel exists in $+2$ oxidation state, Thus the electronic configuration of ${\mathrm{Ni}}^{2+}$ will be $3{\mathrm{d}}^{8}4{\mathrm{s}}^0$.
• Cyanide ion $\left({\mathrm{CN}}^{-}\right)$ is a strong field ligand, thus all electrons will be paired.
• The hybridization of $[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4{]}^{2-}$ is ${\mathrm{dsp}}^2$ and has square planar geometry.
• Since all electrons are paired thus this complex is diamagnetic in nature.

Hence, the hybridization for $[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4{]}^{2-}$ is ${\mathrm{dsp}}^2$.