Question

A d-block element forms an octahedral complex but its spin magnetic moment remains the same in the presence of both strong field and weak field ligands. Which of the following is/are correct?

• A. It always forms colourless compounds
• B. The number of electrons in $t_{2g}$ orbitals is more than those in $e_g$ orbitals
• C. It can have either $d^3$ or $d^8$ configuration
• D. It can have either $d^7$ or $d^8$configuration

Answer 1

Answer: (B), (C)

It can have either $d^3$ or $d^8$ configuration

The very first thing we need to understand here is that there is no change in magnetic moment when the nature of ligand is changed. This implies that the number of unpaired electrons is not changing with a change in the strength/nature of ligand.
It is possible for the said element to have zero unpaired electrons in case of coordination number six, at a certain oxidation state. This way, the number of unpaired electrons will not change with the strength of the ligand. However, this will not be the case with every oxidation state and hence, it is incorrect to say that the element will always form colorless compounds.
We have been told that the complex formed is octahedral. This means that the degeneracy of d-orbitals is lost due to splitting. There are 3 $t_{2g}$ and 2 $e_g$ orbitals. Due to the specific approach of ligands along the axes, we know that $t_{2g}$ orbitals have lower energy than $e_g$ orbitals. This means that at any time, the $t_{2g}$ orbitals will have more electrons than $e_g$ orbitals (Aufbau's principle).
Consider (in Octahedral geometry)
$d^3$ configuration-
Strong field- 3 unpaired electrons
Weak field- 3 unpaired electrons
$d^7$ configuration-
Strong field- 1 unpaired electron
Weak field- 3 unpaired electrons
$d^8$ configuration-
Strong field- 2 unpaired electrons
Weak field- 2 unpaired electrons

Hence, it can have either $d^3$ or $d^8$ configuration.