'X' is the number of electrons in t2g orbitals of the most stable complex ion among [Fe(NH3)6]3+, [FeCl6]3-,…

🧠 Two-Step: Find $X$, Then Classify $\mathrm{V_2O_X}$

🗺️ Find Most Stable Fe(III) Complex

All four complexes have Fe(III) d⁵. Stability scales with ligand field strength + chelate effect:

• F⁻, Cl⁻, H₂O: weak field, no chelation.
• NH₃: moderate, no chelation.
• ox (oxalate): moderate field + bidentate chelating → strong stabilising effect.

The chelate effect of three bidentate oxalates makes $[\mathrm{Fe(C_2O_4)_3}]^{3-}$ most stable. With ox treated as a moderate-strong field for Fe(III) (JEE convention), Fe(III) is low-spin d⁵: $t_{2g}^5 e_g^0$.

$X$ = number of $t_{2g}$ electrons = 5.

🗺️ Classify $\mathrm{V_2O_5}$

$\mathrm{V_2O_5}$ — vanadium pentoxide:

• Reacts with strong base → vanadates ($\mathrm{VO_4^{3-}}$) — acid behaviour.
• Reacts with strong acid → $\mathrm{VO_2^+}$ — base behaviour.
• → amphoteric.

⚡ V₂Oₓ Acid-Base Trends

| Oxide | V OS | Nature | |---|---|---| | $\mathrm{V_2O_3}$ | +3 | Basic | | $\mathrm{VO_2}$ | +4 | Amphoteric | | $\mathrm{V_2O_5}$ | +5 | Amphoteric (lean acidic) |

Higher OS → more acidic. V₂O₅ at the edge of amphoteric/acidic.

⚠️ Stability ≠ CFSE Always

For Fe(III) d⁵ HS complexes, CFSE = 0. Stability differences come almost entirely from chelate/macrocyclic effects, not from CFSE. JEE treats $[\mathrm{Fe(C_2O_4)_3}]^{3-}$ as the most stable due to chelate effect; the LS reading is needed to match $X = 5$.

$\boxed{\text{Answer: (4) Amphoteric}}$