1. Goal

We must count complexes that satisfy both conditions:

• they are paramagnetic, and
• they involve $d^2sp^3$ hybridization (inner-orbital octahedral complex).

2. Useful idea

For an octahedral complex:

• $d^2sp^3$ $ightarrow$ inner orbital complex
• $sp^3d^2$ $ightarrow$ outer orbital complex

Inner orbital octahedral complexes usually form when pairing occurs in the $(n-1)d$ orbitals, leaving two vacant $d$ orbitals for hybridization.

We now examine each complex one by one.

3. Complex 1: $[\mathrm{FeF}_6]^{3-}$

• Oxidation state of Fe: $x+6(-1)=-3 \Rightarrow x=+3$
• So, $\mathrm{Fe}^{3+}$ has configuration: $[\mathrm{Ar}]3d^5$
• $\mathrm{F^-}$ is a weak field ligand, so no pairing occurs.
• Octahedral high-spin $d^5$ uses outer orbitals: $sp^3d^2$
• It is paramagnetic, but not $d^2sp^3$.

✅ Paramagnetic

❌ Not $d^2sp^3$

So, not counted.

4. Complex 2: $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$

• Oxidation state of Fe: $x+6(-1)=-3 \Rightarrow x=+3$
• So, $\mathrm{Fe}^{3+}=3d^5$
• $\mathrm{CN^-}$ is a strong field ligand, so pairing occurs.
• Low-spin octahedral $d^5$ arrangement: $t_{2g}^5e_g^0$
• Two $3d$ orbitals become available, hence hybridization is: $d^2sp^3$
• Number of unpaired electrons in $t_{2g}^5$ is 1, so it is paramagnetic.

✅ Paramagnetic

✅ $d^2sp^3$

So, counted.

5. Complex 3: $[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$

• Oxidation state of Mn: $x+6(-1)=-3 \Rightarrow x=+3$
• So, $\mathrm{Mn}^{3+}=3d^4$
• $\mathrm{CN^-}$ is strong field, so low-spin octahedral complex forms.
• Low-spin $d^4$ arrangement: $t_{2g}^4e_g^0$
• Pairing occurs, so complex is inner orbital: $d^2sp^3$
• In $t_{2g}^4$, there are 2 unpaired electrons, so it is paramagnetic.

✅ Paramagnetic

✅ $d^2sp^3$

So, counted.

6. Complex 4: $[\mathrm{Co}(\mathrm{C}_2\mathrm{O}_4)_3]^{3-}$

• Oxidation state of Co: $x+3(-2)=-3 \Rightarrow x=+3$
• So, $\mathrm{Co}^{3+}=3d^6$
• $\mathrm{C_2O_4^{2-}}$ is generally treated as causing pairing with $\mathrm{Co}^{3+}$ because $\mathrm{Co}^{3+}$ has large $\Delta_o$.
• Thus low-spin octahedral $d^6$: $t_{2g}^6e_g^0$
• Hybridization: $d^2sp^3$
• But all electrons are paired, so it is diamagnetic.

❌ Not paramagnetic

✅ $d^2sp^3$

So, not counted.

7. Complex 5: $[\mathrm{MnCl}_6]^{3-}$

• Oxidation state of Mn: $x+6(-1)=-3 \Rightarrow x=+3$
• So, $\mathrm{Mn}^{3+}=3d^4$
• $\mathrm{Cl^-}$ is a weak field ligand, so high-spin complex forms.
• High-spin octahedral $d^4$ uses outer orbitals: $sp^3d^2$
• It is paramagnetic, but not $d^2sp^3$.

✅ Paramagnetic

❌ Not $d^2sp^3$

So, not counted.

8. Complex 6: $[\mathrm{CoF}_6]^{3-}$

• Oxidation state of Co: $x+6(-1)=-3 \Rightarrow x=+3$
• So, $\mathrm{Co}^{3+}=3d^6$
• Although $\mathrm{F^-}$ is weak field, for $\mathrm{Co}^{3+}$ the crystal field splitting is sufficiently large and $[\mathrm{CoF}_6]^{3-}$ is taken as high-spin outer orbital in standard JEE treatment.
• Hence hybridization: $sp^3d^2$
• High-spin $d^6$ is paramagnetic.

✅ Paramagnetic

❌ Not $d^2sp^3$

So, not counted.

9. Count the required complexes

The complexes that are both paramagnetic and $d^2sp^3$ are:

• $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$
• $[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$

Therefore, the number is

$\boxed{2}$

10. Comparison with stored answer

Stored correct answer = $2$

Our derived answer = $2$

Hence, the answer agrees with the stored answer.