Chemistry - Coordination Chemistry Question with Solution | TestHub

To determine if a species is diamagnetic, we need to check if all its electrons are paired. This involves analyzing the electron configuration or molecular orbital diagram.

A. ${\mathrm{NO}}^{\ominus }$: Total electrons = 7 (N) + 8 (O) + 1 (charge) = 16 electrons. The molecular orbital configuration for ${\mathrm{NO}}^{\ominus }$ shows unpaired electrons in the ${\pi }^{\ast }$ antibonding orbitals, making it paramagnetic.

B. ${\left[{\mathrm{NiCl}}_4\right]}^{2-}$: Ni is in the +2 oxidation state ($3d^8$). ${\mathrm{Cl}}^{-}$ is a weak field ligand. In a tetrahedral field, the $d$ orbitals split into $e$ and $t_2$. The 8 electrons will occupy these orbitals, resulting in unpaired electrons. Thus, it is paramagnetic.

C. ${\left[{\mathrm{CuCl}}_4\right]}^{3\ominus }$: Cu is in the +1 oxidation state ($3d^{10}$). All $d$ electrons are paired. This species is diamagnetic.

D. ${\mathrm{O}}_2\left[{\mathrm{AsF}}_6\right]$: This compound contains the ${\mathrm{O}}_2^{+}$ ion. ${\mathrm{O}}_2^{+}$ has 15 electrons. The molecular orbital configuration for ${\mathrm{O}}_2^{+}$ shows one unpaired electron in the ${\pi }^{\ast }$ antibonding orbital, making it paramagnetic.

Therefore, ${\left[{\mathrm{CuCl}}_4\right]}^{3\ominus }$ is the diamagnetic species.