Determine the oxidation state of manganese in $$\left[\mathrm{MnCl}_6\right]^{3-}:$$

Chloride (Cl⁻) has a charge of -1.

With six chloride ions, the total charge contributed by the ligands is $$6(-1) = -6.$$

Let the oxidation state of Mn be $$x.$$ Then:

$$x - 6 = -3 \quad \Longrightarrow \quad x = +3.$$

So, manganese is in the +3 oxidation state.

Find the d-electron count for Mn(III):

The neutral manganese atom has the electronic configuration $$[\mathrm{Ar}]\, 3d^5 4s^2.$$

Removing three electrons (first from the 4s, then from the 3d) gives:

$$\mathrm{Mn}^{3+} : [\mathrm{Ar}]\, 3d^4.$$

Therefore, Mn(III) is a $$d^4$$ system.

Predict the spin state in an octahedral complex:

Chloride (Cl⁻) is a weak-field ligand, meaning it produces a relatively small crystal field splitting ($$\Delta_o$$).

When $$\Delta_o$$ is small, the pairing energy is higher than $$\Delta_o,$$ so electrons prefer to remain unpaired.

For a $$d^4$$ configuration in a high-spin octahedral complex, the electrons will be distributed as:

Three electrons in the three $$t_{2g}$$ orbitals (one in each)

One electron in one of the $$e_g$$ orbitals

This results in a total of 4 unpaired electrons, making the complex paramagnetic.

Determine the type of hybridization:

In octahedral complexes, the two common hybridizations are:

$$d^2sp^3$$ (inner orbital complex), typically seen in low-spin complexes where inner 3d orbitals are available because the electrons are paired.

$$sp^3d^2$$ (outer orbital complex), seen in high-spin complexes where the inner $$3d$$ orbitals are occupied by unpaired electrons.

Since $$\left[\mathrm{MnCl}_6\right]^{3-}$$ is high-spin (because of Cl⁻ being a weak-field ligand), the inner 3d orbitals are not available for hybridization.

Therefore, the complex uses the outer orbitals (the 4s, 4p, and 4d orbitals), leading to an $$sp^3d^2$$ hybridization.

Conclusion:

The hybridization of $$\left[\mathrm{MnCl}_6\right]^{3-}$$ is $$sp^3d^2.$$

It is paramagnetic with four unpaired electrons.

Thus, the correct answer is:

Option A: $$sp^3d^2$$, paramagnetic with four unpaired electrons.