Question

Give reasons:
(I) \([ \text{Ni(CO)}_4 ]\) is diamagnetic whereas \([ \text{NiCl}_4 ]^{2-}\) is paramagnetic. [Atomic number: Ni = 28]
(II) CO is a stronger complexing agent than NH\(_3\).
(III) The trans isomer of complex \([ \text{Co(en)}_2\text{Cl}_2 ]^{+}\) is optically inactive.

Hint:

In square planar and octahedral complexes, the arrangement of ligands and the presence of strong or weak field ligands can determine magnetic properties and optical activity.

Answer 1

(I) The electronic configuration of Ni in \([ \text{Ni(CO)}_4 ]\) is such that it forms a stable low-spin configuration, leading to no unpaired electrons and thus diamagnetic behavior. However, in \([ \text{NiCl}_4 ]^{2-}\), the higher oxidation state of Ni leads to unpaired electrons, making it paramagnetic.
(II) CO has a strong \(\pi\)-acceptor property, allowing it to form stronger bonds with metal centers compared to NH\(_3\), which is a weaker field ligand.
(III) The trans isomer of \([ \text{Co(en)}_2\text{Cl}_2 ]^{+}\) is optically inactive due to the symmetry of the complex, which prevents optical isomerism.