1. Check Statement I

We examine the complexes:

• $[Mn(CN)_6]^{3-}$
• $[Fe(CN)_6]^{3-}$
• $[Co(C_2O_4)_3]^{3-}$

(a) $[Mn(CN)_6]^{3-}$

Let oxidation state of Mn be $x$: $x+6(-1)=-3 \Rightarrow x=+3$ So, $Mn^{3+}$ has configuration: $Mn: [Ar]3d^54s^2 \Rightarrow Mn^{3+}:[Ar]3d^4$ Since $CN^-$ is a strong field ligand, pairing occurs in $3d$ orbitals, giving a low-spin octahedral complex. Hence inner orbital complex with hybridisation: $d^2sp^3$

(b) $[Fe(CN)_6]^{3-}$

Oxidation state of Fe: $x+6(-1)=-3 \Rightarrow x=+3$ So, $Fe^{3+}$ is: $Fe:[Ar]3d^64s^2 \Rightarrow Fe^{3+}:[Ar]3d^5$ Again, $CN^-$ is strong field, so electrons pair up to form a low-spin octahedral complex. Thus hybridisation is: $d^2sp^3$

(c) $[Co(C_2O_4)_3]^{3-}$

Each oxalate ligand $(C_2O_4)^{2-}$ has charge $-2$. Let oxidation state of Co be $x$: $x+3(-2)=-3 \Rightarrow x=+3$ So, $Co^{3+}$ is: $Co:[Ar]3d^74s^2 \Rightarrow Co^{3+}:[Ar]3d^6$ For $Co^{3+}$, even with oxalate (which gives sufficiently large splitting with $Co^{3+}$), pairing occurs and the octahedral complex is low spin. Hence it is an inner orbital complex with hybridisation: $d^2sp^3$

So, Statement I is correct.

2. Check Statement II

The statement refers to:

• $[MnCl_6]^{3-}$
• $[FeF_6]^{3-}$

(a) $[MnCl_6]^{3-}$

Oxidation state of Mn: $x+6(-1)=-3 \Rightarrow x=+3$ Thus $Mn^{3+}$ is $3d^4$.

$Cl^-$ is a weak field ligand, so no pairing occurs in octahedral field (high spin). For high-spin $d^4$ octahedral configuration: $t_{2g}^3 e_g^1$ Number of unpaired electrons $=4$. Hence it is paramagnetic.

(b) $[FeF_6]^{3-}$

Oxidation state of Fe: $x+6(-1)=-3 \Rightarrow x=+3$ Thus $Fe^{3+}$ is $3d^5$.

$F^-$ is a weak field ligand, so high-spin octahedral complex forms. For high-spin $d^5$: $t_{2g}^3 e_g^2$ Number of unpaired electrons $=5$. Hence it is paramagnetic.

So, Statement II is also correct.

3. Final conclusion

• Statement I: True
• Statement II: True

Therefore, the correct option is: $\boxed{D}$

4. Comparison with stored answer

Stored correct answer: $D$

My derived answer also is $D$, so they agree.