Question

Explain on the basis of valence bond theory that \([Ni(CN)_{4} ]^{2−}\) ion with square planar structure is diamagnetic and the \([NiCl_{4}]^{2−}\) ion with tetrahedral geometry is paramagnetic.

Answer 1

Magnetic nature

If the coordination compound have unpaired electrons in their d-orbitals then that will be paramagnetic in nature and if compound have zero unpaired electrons then that will be diamagnetic in nature.

Effect of ligand field strength on pairing of d-electrons

In both the compounds, the oxidation state of Nickel is \(+2.\) So, it has \(d^8\) configuration. Since \(CN^{−}\) is a strong ligand so pairing will occur but in case of \(Cl^{−}\) pairing will not take place as it is a weak ligand.

The configuration of both the compounds are given below:

\([Ni(CN)_{4}]^{2-}\)


Where, \(4s , 4p\) and one of \(3d\) are occupied by electrons of \(C𝑁^{−}\) ligand.

Thus \([Ni(CN)_{4} ]^{2−}\) has \(dsp^{2}\) hybridization and square planer structure.

Here all the electrons are paired. So, it has zero unpaired electrons and is diamagnetic in nature.

\([NiCl_{4}]^{2-}\)



Where, \(4s\) and \(4p\) are occupied by
electrons of \(C𝑙^{−}\) ligand.
\([NiCl_{4} ]^{2−}\) has \(sp^{3}\) hybridization and tetrahedral geometry.

There are \(2\) unpaired electrons so the nature will be paramagnetic.