QJEEM24Ch8a65: JEE Main 2024 Chemistry - Class 12 JEEM24Ch8a65 Solution: The number of unpaired d-electrons in is ________.

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

1. Oxidation state and $d$ -electron count

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

$\mathrm{Co} \; \bigl[\mathrm{Ar}\bigr]\, 3d^7\, 4s^2.$

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 $\bigl[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6\bigr]^{3+}$ , 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:

$\mathrm{t_{2g}^6}\, \mathrm{e_g^0}$

That leaves 0 unpaired electrons.

3. Final answer

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

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

Correct Option: A (0)