1. Identify the metal ion configuration

All the given complexes contain $\mathrm{Fe}^{3+}$.

Atomic number of Fe $=26$

$\mathrm{Fe}:[\mathrm{Ar}]\,3d^6 4s^2$

So,

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

Thus, each complex is an octahedral $d^5$ system.

2. Find the most stable complex among the given options

Given complexes:

• $[\mathrm{Fe}(\mathrm{NH}_3)_6]^{3+}$
• $[\mathrm{FeCl}_6]^{3-}$
• $[\mathrm{Fe}(\mathrm{C}_2\mathrm{O}_4)_3]^{3-}$
• $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{3+}$

Stability depends strongly on ligand strength and chelation.

• $\mathrm{Cl}^-$ is a weak field ligand.
• $\mathrm{H_2O}$ is stronger than $\mathrm{Cl}^-$ but still not very strong.
• $\mathrm{NH_3}$ is stronger than $\mathrm{H_2O}$.
• $\mathrm{C_2O_4^{2-}}$ (oxalate) is a bidentate ligand, and due to the chelate effect, its complex is especially stable.

Hence, the most stable complex is:

$[\mathrm{Fe}(\mathrm{C}_2\mathrm{O}_4)_3]^{3-}$

3. Determine spin state and number of electrons in $t_{2g}$

For $\mathrm{Fe}^{3+}$, we have $d^5$.

Oxalate is not strong enough to force low spin in $\mathrm{Fe}^{3+}$ in the usual JEE-level treatment; thus it is taken as high spin octahedral.

High-spin octahedral $d^5$ configuration:

$t_{2g}^3 e_g^2$

Therefore, the number of electrons in $t_{2g}$ orbitals is

$X=3$

4. Find the oxide $\mathrm{V_2O_X}$

The question now asks for the nature of oxide of vanadium of type $\mathrm{V_2O_X}$, where $X$ is the value found above.

Since $X=3$, the oxide is:

$\mathrm{V_2O_3}$

Now, $\mathrm{V_2O_3}$ is an amphoteric oxide.

5. Match with options

• A: Amphoteric $\checkmark$
• B: Acidic
• C: Basic
• D: Neutral

So the correct option is:

$\boxed{\text{A: Amphoteric}}$

6. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.