Chemistry - Coordination Chemistry Question with Solution | TestHub

${\mathrm{O}}_2^{-}\to$ Superoxide ligand ion (Predicted on the basis of given ${\mathrm{t}}_{2\mathrm{g}}{}^6$ configuration)

Here's a detailed solution to the question, explaining each option:

The complex ion is ${\left[\mathrm{Fe}(\mathrm{Cl})(\mathrm{CN}{)}_4\left({\mathrm{O}}_2\right)\right]}^{4-}$, and its electronic configuration is ${\mathrm{t}}_{2\mathrm{g}}{}^6,{\mathrm{e}}_{\mathrm{g}}{}^0$. This indicates a low-spin $d^6$ configuration for the iron ion.

A. It is a paramagnetic complex.

Since the electronic configuration is ${\mathrm{t}}_{2\mathrm{g}}{}^6,{\mathrm{e}}_{\mathrm{g}}{}^0$, all the $t_{2g}$ orbitals are filled, and the $e_g$ orbitals are empty. This means there are no unpaired electrons. Therefore, the complex is DIAMAGNETIC, not paramagnetic. However, the presence of the superoxide ligand, ${\mathrm{O}}_2^{-}$, which has one unpaired electron, makes the overall complex paramagnetic. Thus, option A is CORRECT.

B. $\mathrm{O}-\mathrm{O}$ bond length will be more than found in ${\mathrm{O}}_2$ molecule.

The electronic configuration suggests the presence of a superoxide ligand (${\mathrm{O}}_2^{-}$). The bond order of ${\mathrm{O}}_2$ is 2, while the bond order of ${\mathrm{O}}_2^{-}$ is 1.5. A lower bond order means a weaker and longer bond. Therefore, the $\mathrm{O}-\mathrm{O}$ bond length in the complex will be longer than that in a neutral ${\mathrm{O}}_2$ molecule. Thus, option B is CORRECT.

C. Its IUPAC name will be chloridotetracyanidosuperoxidoferrate(II) ion.

To determine the oxidation state of iron, we consider the charges of the ligands:

- Cl: -1

- CN: -1 (there are 4, so -4 total)

- ${\mathrm{O}}_2$: We've established it's superoxide, so -1

Overall charge of the complex: -4

Let x be the oxidation state of Fe.

$x+(-1)+(-4)+(-1)=-4$

$x-6=-4$

$x=+2$

Thus, option C is CORRECT.

D. It will show geometrical but not optical isomerism.