Question

Both $\left[Ni\right(CO{)}_4]$ and $\left[Ni\right(CN{)}_4{]}^{2-}$ are diamagnetic. The hybridizations of nickel in these complexes, respectively, are:

• A. $s{p}^3,s{p}^3$
• B. $s{p}^3,ds{p}^2$
• C. $ds{p}^2,s{p}^3$
• D. $ds{p}^2,ds{p}^2$

Answer 1

Answer: (B)

$s{p}^3,ds{p}^2$

In $\left[Ni\right(CO{)}_4]$, $Ni$ is in 0 oxidation state so there are 8 electrons in 3 d subshell and 2 electrons in 4s. $CO$ is the strong ligand, so it does pairing with the d subshell atom leaving one d subshell with unpaired electron.now the electron from s shell shifts to d creating vacant space in s subshell and thus the hybridization is $s{p}^3$

In $\left[Ni\right(CN{)}_4{]}^{2-}$, $Ni$ is in +2 oxidation state so it has 8 electrons in d subshell and $CN$ is also a strong ligand which pairs with d subshell atom and leaves one d orbital empty and thus its hybridization is $ds{p}^2$