${\left[NiC{l}_4\right]}^{2-}$ is diamagnetic in nature.

Q.

[NiCl₄]²⁻ is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why?

Q. $\left[NiC{l}_4\right]$ is paramagnetic while $\left[NiC{O}_4\right]$ is diamagnetic though both are tetrahedral. Explain.

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

Q. The complex ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$ is diamagnetic and the complex ${\left[NiC{l}_4\right]}^{2-}$ is paramagnetic. What can you conclude about their molecular geometries?

Q. The anion $[NiC{l}_4{]}^{2-}$ is paramagnetic but when $C{N}^{-}$ ions are added, the product $\left[Ni\right(CN{)}_4{]}^{2-}$ is diamagnetic because in $[NiC{l}_4{]}^{2-}$, $C{l}^{-}$ is a weak ligand and thus, electrons are unpaired in $N{i}^{2+}$ making it paramagnetic, $s{p}^3$ hybridised and tetrahedral.
If true enter 1, else enter 0.