JEE Chemistry 2025 Question with Solution

Given: An octahedral Co(II) complex has magnetic moment $\mu = 3.95 \, \text{BM}$.

Find: The d-electronic configuration in the octahedral field.

Use the spin-only formula:

$$\mu = \sqrt{n(n+2)}$$

where $n$ is the number of unpaired electrons.

From the given value,

$$\sqrt{n(n+2)} = 3.95$$

Squaring both sides,

$$(3.95)^2 \approx 15.6 \approx n(n+2)$$

So,

$$n(n+2) \approx 16$$

which gives approximately $n = 3$ unpaired electrons.

For Co(II), the ion is $\text{Co}^{2+}$ and it corresponds to $3d^7$.

In an octahedral complex, the d-orbitals split into $t_{2g}$ and $e_g$ levels. A $d^7$ octahedral arrangement with $3$ unpaired electrons is:

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

Therefore, the correct option is C, and the d-electronic configuration is $t_{2g}^5 e_g^2$.

Given: Co(II) in an octahedral field and magnetic moment $3.95 \, \text{BM}$.

Find: The correct split d-electron configuration.

Neutral cobalt has configuration $[\text{Ar}]\,3d^7 4s^2$. Therefore,

$$\text{Co}^{2+} = [\text{Ar}]\,3d^7$$

The magnetic moment value $3.95 \, \text{BM}$ corresponds to about $3$ unpaired electrons using

$$\mu = \sqrt{n(n+2)}$$

For an octahedral $d^7$ configuration, the distribution that gives $3$ unpaired electrons is

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

This matches the magnetic behavior stated in the question.

Hence, the correct option is C.