The type of hybridisation and magnetic property of the complex $$[MnCl_6]^{3-}$$, respectively, are:

We first identify the oxidation state of the central metal. Let the oxidation state of manganese be $$x$$. Each chloride ion carries a charge of $$-1$$ and there are six of them, so their total charge is $$6(-1)=-6$$. The overall charge on the complex ion is given to be $$-3$$. Hence, we write the simple equation

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

Solving, we get

$$x = -3 + 6 = +3$$

So manganese is in the $$+3$$ oxidation state in $$[MnCl_6]^{3-}$$.

Next we need the ground-state electronic configuration of a neutral manganese atom. Manganese has atomic number $$25$$, so

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

For $$Mn^{3+}$$ we remove first the two $$4s$$ electrons and then one $$3d$$ electron (because electrons are lost from the outer shell first). Thus

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

Now we must decide whether the complex is an inner-orbital ($$d^2sp^3$$) or outer-orbital ($$sp^3d^2$$) octahedral complex. For this we recall a qualitative result from crystal-field theory:

• Weak-field (high-spin) ligands do not cause pairing of $$d$$ electrons and therefore favour the use of the outer $$d$$ orbitals, producing $$sp^3d^2$$ hybridisation.
• Strong-field (low-spin) ligands do cause pairing and favour $$d^2sp^3$$ hybridisation.

Chloride ion ($$Cl^-$$) is a weak-field ligand. Hence no pairing of the four $$3d$$ electrons of $$Mn^{3+}$$ will occur, and the two $$d$$ orbitals needed for octahedral bonding will be taken from the next (fourth) shell. Therefore the hybridisation must be

$$sp^3d^2$$.

Because the four $$3d$$ electrons remain unpaired, the complex possesses four unpaired electrons. A substance with one or more unpaired electrons is paramagnetic.

So the complex $$[MnCl_6]^{3-}$$ is $$sp^3d^2$$ hybridised and paramagnetic.

Hence, the correct answer is Option D.