Given below are two statements:
Statement I: $$[Mn(CN)_6]^{3-}$$, $$[Fe(CN)_6]^{3-}$$ and $$[Co(C_2O_4)_3]^{3-}$$ are d$$^2$$sp$$^3$$ hybridised.
Statement II: $$[MnCl_6]^{3-}$$ and $$[FeF_6]^{3-}$$ are paramagnetic and have 4 and 5 unpaired electrons, respectively.
In the light of the above statements, choose the correct answer from the options given below:

First we examine the oxidation state and the d-electron count of the central metal ions.

For every complex the overall charge is known, so we write

$$\text{Charge of complex}= \text{Oxidation number of metal}+ \sum (\text{Charges of ligands}).$$

1. In $$[Mn(CN)_6]^{3-}$$ the six cyanide ligands are each $$-1$$.  Let the oxidation number of Mn be $$x$$:

$$x+6(-1)=-3 \;\Rightarrow\; x=+3.$$

Mn has atomic number 25, ground configuration $$[Ar]\,3d^5\,4s^2$$, so

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

2. In $$[Fe(CN)_6]^{3-}$$, likewise,

$$x+6(-1)=-3 \;\Rightarrow\; x=+3,$$

and with $$Fe : [Ar]\,3d^6\,4s^2$$ we obtain

$$Fe^{3+} : [Ar]\,3d^5.$$

3. In $$[Co(C_2O_4)_3]^{3-}$$, each oxalate ion is $$-2$$, so

$$x+3(-2)=-3 \;\Rightarrow\; x=+3,$$

thus $$Co^{3+} : [Ar]\,3d^6.$$

Now we decide whether these ions are low-spin or high-spin. The ligands present are $$CN^-$$ and $$C_2O_4^{2-}$$. Cyanide is a very strong-field ligand, and for a highly charged ion such as $$Co^{3+}$$ even the moderately strong oxalate produces a large splitting. Hence for all three complexes the crystal-field splitting energy $$\Delta_o$$ exceeds the pairing energy $$P$$, giving LOW-SPIN arrangements.

Low-spin electronic distributions in an octahedral field are:

$$ \begin{aligned} d^4 &: t_{2g}^4 e_g^0 \;(2\; \text{unpaired}),\\ d^5 &: t_{2g}^5 e_g^0 \;(1\; \text{unpaired}),\\ d^6 &: t_{2g}^6 e_g^0 \;(\text{all paired}). \end{aligned} $$

Because the two $$e_g$$ orbitals (which are 3d in character) remain completely empty, they can be donated to the hybrid set. The metal can therefore use two 3d (the vacant $$e_g$$ pair), one 4s and three 4p orbitals to form the hybrid set

$$d^2sp^3,$$

yielding inner-orbital (low-spin) octahedral complexes. So Statement I is correct.

Next we analyse the second set of complexes.

4. For $$[MnCl_6]^{3-}$$ again $$Mn^{3+}$$ is $$d^4$$, but chloride $$Cl^-$$ is a weak-field ligand. Now $$\Delta_o<P$$, giving a HIGH-SPIN distribution

$$d^4_{\text{high-spin}}: t_{2g}^3 e_g^1,$$

which clearly shows one unpaired electron in each of the four orbitals → 4 unpaired electrons.

5. In $$[FeF_6]^{3-}$$ we have $$Fe^{3+}\;(d^5)$$ and the very weak-field ligand $$F^-$$, so the HIGH-SPIN pattern is obtained:

$$d^5_{\text{high-spin}}: t_{2g}^3 e_g^2,$$

giving five unpaired electrons.

Since both complexes possess the predicted numbers of unpaired electrons, they are paramagnetic exactly as stated. Therefore Statement II is also correct.

Both statements are true, so we must choose Option D.

Hence, the correct answer is Option D.