Question

Give a reason for the statement. ${}^{\prime }\left[Ni\right(CN{)}_4{]}^{2-}$ is diamagnetic while $[NiC{l}_4{]}^{2-}$ is paramagnetic in nature.

• A. In $[NiC{l}_4{]}^{2-}$, no unpaired electrons are present while in $\left[Ni\right(CN{)}_4{]}^{2-}$ two unpaired electrons are present.
• B. In $\left[Ni\right(CN{)}_4{]}^{2-}$, no unpaired electrons are present while in $[NiC{l}_4{]}^{2-}$ two unpaired electons are present.
• C. $[NiC{l}_4{]}^{2-}$ shows $ds{p}^2$ hybridisation hence it is paramagnetic.
• D. $\left[Ni\right(CN{)}_4{]}^{2-}$, shows $s{p}^3$ hybridisation hence it is diamagnetic.

Answer 1

The correct option is B In $\left[Ni\right(CN{)}_4{]}^{2-}$, no unpaired electrons are present while in $[NiC{l}_4{]}^{2-}$ two unpaired electons are present.

$\left[Ni\right(CN{)}_4{]}^{2-}$, have 0 unpaired electrons are present and $[NiC{l}_4{]}^{2-}$ having two unpaired electrons are present.