1. Find oxidation state of Mn in $[\mathrm{MnCl}_6]^{3-}$

Let oxidation state of Mn be $x$.

$x+6(-1)=-3$ $x-6=-3$ $x=+3$

So, the metal ion is $\mathrm{Mn}^{3+}$.

2. Write electronic configuration of $\mathrm{Mn}^{3+}$

Atomic number of Mn = 25.

Neutral Mn: $\mathrm{Mn}: [\mathrm{Ar}]\,3d^5 4s^2$

For $\mathrm{Mn}^{3+}$, remove two electrons from $4s$ and one from $3d$: $\mathrm{Mn}^{3+}: [\mathrm{Ar}]\,3d^4$

3. Nature of ligand

$\mathrm{Cl}^-$ is a weak field ligand, so it does not cause pairing of electrons in $3d$ orbitals.

Thus, the complex is a high-spin octahedral complex.

For $d^4$ high spin octahedral arrangement: $t_{2g}^3 e_g^1$

So, number of unpaired electrons = 4.

Hence, the complex is paramagnetic with four unpaired electrons.

4. Determine hybridization

Since $\mathrm{Cl}^-$ is weak field, no pairing occurs in inner $3d$ orbitals. Therefore, two vacant inner $3d$ orbitals are not available for $d^2sp^3$ hybridization.

So, the complex uses outer orbitals $4s, 4p,$ and $4d$ to form octahedral hybrid orbitals:

$sp^3d^2$

This is an outer orbital octahedral complex.

5. Match with options

• A: $sp^3d^2$, paramagnetic with four unpaired electrons. ✅
• B: $d^2sp^3$, paramagnetic with four unpaired electrons. ❌ hybridization incorrect
• C: $sp^3d^2$, paramagnetic with two unpaired electrons. ❌ unpaired electrons incorrect
• D: $d^2sp^3$, paramagnetic with two unpaired electrons. ❌ both incorrect

6. Final answer

The correct option is:

$\boxed{\text{A}}$