The d- electronic configuration of an octahedral Co(II) complex having magnetic moment of 3.95 BM is:

Co(II) has the electronic configuration $$[Ar]3d^7$$.

The magnetic moment is given by $$ \mu = \sqrt{n(n+2)} \text{ BM} $$, where $$n$$ is the number of unpaired electrons.

Substituting the observed magnetic moment $$\mu = 3.95$$ BM into this formula gives

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

and hence

$$ 15.6 \approx n(n+2). $$

Testing integer values of $$n$$ shows that for $$n = 3$$, we have $$3 \times 5 = 15$$ and therefore

$$ \mu = \sqrt{15} = 3.87 \text{ BM}, $$

which is very close to the observed value of 3.95 BM. Thus, there are three unpaired electrons in the complex.

For a $$d^7$$ octahedral complex with three unpaired electrons, the high-spin configuration must be

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

In this arrangement, the $$t_{2g}$$ orbitals contain five electrons (↑↓ ↑↓ ↑), giving one unpaired electron, and the $$e_g$$ orbitals contain two electrons (↑ ↑), giving two unpaired electrons, for a total of three unpaired electrons.

This matches the given magnetic moment.

The correct answer is Option 3: $$t_{2g}^5 e_g^2$$.