1. Find the oxidation state of Mn in $[MnCl_6]^{3-}$.

   Let the oxidation state of Mn be $x$. Each $Cl^-$ ligand has charge $-1$.

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

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

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

   Atomic number of Mn = 25

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

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

3. Nature of ligand.

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

4. Determine inner-orbital or outer-orbital complex.

   Since no pairing occurs in $3d^4$, two vacant $3d$ orbitals are not available for $d^2sp^3$ hybridisation.

   Therefore, the complex uses outer orbitals: $sp^3d^2$

5. Magnetic property.

   For $Mn^{3+} = 3d^4$ with weak field ligand, the configuration remains high spin with 4 unpaired electrons.

   Hence, the complex is paramagnetic.

6. Evaluate options.

   • A: $sp^3d^2$ and diamagnetic $\rightarrow$ wrong
   • B: $d^2sp^3$ and diamagnetic $\rightarrow$ wrong
   • C: $d^2sp^3$ and paramagnetic $\rightarrow$ wrong
   • D: $sp^3d^2$ and paramagnetic $\rightarrow$ correct

Therefore, the correct answer is D.