JEE Main 2024 — Coordination Compounds Question with Solution

First, let us determine the $d$-electron count of the cobalt center in $[\mathrm{Co}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{3+}$.

1. Oxidation state and $d$-electron count

Neutral cobalt (Co) has an atomic number of 27 and an electronic configuration:

$\mathrm{Co}\left[\mathrm{Ar}\right]3d^{7}4s^2.$

In the +3 oxidation state, cobalt has lost a total of 3 electrons:

First two electrons are removed from the $4s$ orbital.

The third electron is removed from the $3d$ orbitals.

Thus, ${\mathrm{Co}}^{3+}$ has

$3d^{7-1}=3d^6.$

So in $[\mathrm{Co}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{3+}$, the cobalt center is $d^6$.

2. High-spin vs Low-spin for ${\mathrm{Co}}^{3+}$

Even though ${\mathrm{H}}_2\mathrm{O}$ is generally considered a weak field ligand, the ${\mathrm{Co}}^{3+}$ ion has a relatively high charge (+3). A higher charge on the metal center usually increases the ligand-field splitting (${\Delta }_{\mathrm{o}}$) significantly compared to lower oxidation states of the same metal.

In most typical octahedral complexes of ${\mathrm{Co}}^{3+}$, the splitting ${\Delta }_{\mathrm{o}}$ is large enough that the complex ends up being low spin.

Electronic configuration in an octahedral field

For a $d^6$ octahedral low-spin complex, all six electrons pair up in the lower-energy ${\mathrm{t}}_{2g}$ orbitals:

${\mathrm{t}}_{2g}^6{\mathrm{e}}_{\mathrm{g}}^0$

That leaves 0 unpaired electrons.

3. Final answer

Therefore, the number of unpaired $d$-electrons in

$[\mathrm{Co}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{3+}$ is $\boxed{0}.$

Correct Option: A (0)