Question

Why is [Ni(CN)4]2- square planar while [Ni(CO)4] is tetrahedral? [Atomic number : Ni = 28]

Hint:

Strong field ligands like CN\textsuperscript{−} induce pairing and lead to square planar geometry; weak field ligands like CO retain unpaired electrons, giving tetrahedral geometry.

Answer 1

The difference in geometry between [Ni(CN)4]\textsuperscript{2−} and [Ni(CO)4] arises due to the nature of ligands and hybridisation involved: \begin{itemize} \item [Ni(CN)4]\textsuperscript{2−}: The CN\textsuperscript{−} ligand is a strong field ligand. It causes pairing of electrons in the 3d orbitals of Ni2+. This leads to \textit{dsp\textsuperscript{2}} hybridisation and results in a square planar geometry. \item [Ni(CO)4]: The CO ligand is a weak field ligand and does not cause pairing of 3d electrons. Hence, Ni remains in its 0 oxidation state and uses \textit{sp\textsuperscript{3}} hybridisation, forming a tetrahedral structure. \end{itemize} Thus, the geometry difference arises due to ligand field strength and resulting hybridisation.