Find the oxidation state and d-electron count

The metal is Co(II).

Cobalt has atomic number $27$, so its ground-state electronic configuration is: $[Ar] \, 3d^7 4s^2$

For $\mathrm{Co^{2+}}$, two electrons are removed (from $4s$ first), so: $\mathrm{Co^{2+}} : [Ar] \, 3d^7$

Thus, the complex has a $d^7$ configuration.

Use the magnetic moment to find number of unpaired electrons

Given magnetic moment: $\mu = 3.95\ \text{BM}$

Using spin-only formula: $\mu = \sqrt{n(n+2)}$ where $n$ is the number of unpaired electrons.

Check for $n=3$: $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\ \text{BM}$

This is very close to $3.95$ BM, so the complex has 3 unpaired electrons.

Octahedral splitting for a $d^7$ ion

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

• lower: $t_{2g}$
• higher: $e_g$

For $d^7$, there are two possibilities:

• Low spin: $t_{2g}^6 e_g^1$ This has 1 unpaired electron.

• High spin: $t_{2g}^5 e_g^2$ This has 3 unpaired electrons.

Since the magnetic moment corresponds to 3 unpaired electrons, the complex must be high-spin octahedral: $t_{2g}^5 e_g^2$

Match with the options

• A: $e^4 t_2^3$ — not the standard octahedral notation and not correct here.
• B: $t_{2g}^6 e_g^1$ — low-spin $d^7$, only 1 unpaired electron.
• C: $t_{2g}^5 e_g^2$ — high-spin $d^7$, 3 unpaired electrons. Correct.
• D: $t_{2g}^3 e_g^0$ — corresponds to $d^3$, not $d^7$.

Final answer

The correct d-electronic configuration is: $\boxed{t_{2g}^5 e_g^2}$

So, Option C is correct.