Determine the oxidation state of manganese in :

Chloride (Cl -1 ) has a charge of -1.

With six chloride ions, the total charge contributed by the ligands is .

Let the oxidation state of Mn be . Then:
.

So, manganese is in the +3 oxidation state.

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

The neutral manganese atom has the electronic configuration .

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

Therefore, is a system.
Predict the spin state in an octahedral complex:

Chloride (Cl -1 ) is a weak-field ligand, meaning it produces a relatively small crystal field splitting ( ).

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

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

Three electrons in the three orbitals (one in each)

One electron in one of the 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:
(inner orbital complex), typically seen in low-spin complexes where inner 3d orbitals are available because the electrons are paired.
(outer orbital complex), seen in highspin complexes where the inner orbitals are occupied by unpaired electrons.

Since is high-spin (because of Cl -1 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 hybridization.

Conclusion:
The hybridization of is .

It is paramagnetic with four unpaired electrons.

Thus, the correct answer is:

Option a: , paramagnetic with four unpaired electrons.