'X' is the number of electrons in $$t_{2g}$$ orbitals of the most stable complex ion among $$[Fe(NH_3)_6]^{3+}$$, $$[Fe(Cl)_6]^{3-}$$, $$[Fe(C_2O_4)_3]^{3-}$$ and $$[Fe(H_2O)_6]^{3+}$$. The nature of oxide of vanadium of the type $$V_2O_x$$ is:

All the four complexes contain iron in the $$+3$$ oxidation state, so the metal ion is $$Fe^{3+}$$ having the electronic configuration $$[Ar]\,3d^5$$.

Case 1: $$[Fe(NH_3)_6]^{3+}$$  and  $$[Fe(H_2O)_6]^{3+}$$    Both $$NH_3$$ and $$H_2O$$ are monodentate ligands of only moderate field strength.    For a $$d^5$$ ion with such ligands, the complex is high-spin:    $$t_{2g}^3\,e_g^2$$  —  CFSE $$=0$$.

Case 2: $$[Fe(Cl)_6]^{3-}$$    $$Cl^-$$ is a very weak-field ligand, so the complex is again high-spin:    $$t_{2g}^3\,e_g^2$$  —  CFSE $$=0$$.

Case 3: $$[Fe(C_2O_4)_3]^{3-}$$    Oxalate $$(C_2O_4^{2-})$$ is a bidentate ligand.    Although its crystal-field strength is similar to $$H_2O$$, the chelate effect gives this complex a much larger overall (stepwise) stability constant.    With a moderate field ligand, $$Fe^{3+}$$ again remains high-spin:    $$t_{2g}^3\,e_g^2$$.

Because of the pronounced chelate effect, $$[Fe(C_2O_4)_3]^{3-}$$ possesses the greatest thermodynamic stability among the four. Hence the most stable complex ion is $$[Fe(C_2O_4)_3]^{3-}$$.

The number of electrons occupying the $$t_{2g}$$ set in this ion is therefore $$X = 3$$.

The oxide of vanadium corresponding to $$V_2O_X$$ with $$X = 3$$ is $$V_2O_3$$. Vanadium(III) oxide behaves like aluminium(III) oxide and chromium(III) oxide: it dissolves both in acids (forming $$V^{3+}$$ salts) and in strong bases (forming vanadates of lower oxidation state after air-oxidation). Such behaviour is termed amphoteric.

Hence, the nature of $$V_2O_3$$ is amphoteric  ⇒  Option D.