JEE Chemistry 2023 Question with Solution

Given: The homoleptic octahedral complex is formed by $\mathrm{Co^{2+}}$ with $\mathrm{H_2O}$.

Find: Number of unpaired electrons in the $t_{2g}$ set of orbitals.

Step 1: Determine the electronic configuration of $\mathrm{Co^{2+}}$.

Atomic number of cobalt $\left(\mathrm{Co}\right) = 27$.

Electronic configuration of $\mathrm{Co}$ is

$$[\mathrm{Ar}]\,3d^7\,4s^2$$

For $\mathrm{Co^{2+}}$, two electrons are removed from the $4s$ orbital:

$$\mathrm{Co^{2+}} : [\mathrm{Ar}]\,3d^7$$

Step 2: Nature of the ligand and spin state.

Water $\left(\mathrm{H_2O}\right)$ is a weak field ligand.

Therefore, the octahedral complex $\left[\mathrm{Co(H_2O)_6}\right]^{2+}$ is a high-spin complex.

Step 3: Crystal field splitting and electron distribution.

In an octahedral field, the $3d$ orbitals split into

$$t_{2g} \; (\text{lower energy}) \quad \text{and} \quad e_g \; (\text{higher energy})$$

For a $d^7$ high-spin configuration, electrons are distributed as

$$t_{2g}^5\, e_g^2$$

Step 4: Count unpaired electrons in the $t_{2g}$ set.

The $t_{2g}$ set contains $5$ electrons, arranged as

$$\updownarrow \quad \updownarrow \quad \uparrow$$

Thus, there is $1$ unpaired electron in the $t_{2g}$ orbitals.

Therefore, the number of unpaired electrons in the $t_{2g}$ set is $1$.

Given: $\mathrm{Co^{2+}}$ in an octahedral complex with weak field ligand $\mathrm{H_2O}$.

Find: Unpaired electrons in $t_{2g}$.

Since $\mathrm{Co^{2+}}$ is $3d^7$ and $\mathrm{H_2O}$ is a weak field ligand, the complex is high spin. For high-spin octahedral $d^7$, the configuration is directly

$$t_{2g}^5 e_g^2$$

Now place $5$ electrons in the three $t_{2g}$ orbitals according to Hund's rule. Two orbitals become paired and one orbital contains a single electron, so the $t_{2g}$ set has exactly $1$ unpaired electron.

Therefore, the answer is $1$.