The first transition series metal 'M' has the highest enthalpy of atomisation in its series. One of its aquated ion ($$M^{n+}$$) exists in green colour. The nature of the oxide formed by the above $$M^{n-}$$ ion is :

The metal that has the highest enthalpy of atomisation in the first transition (3d) series is vanadium, $$V$$. Its enthalpy of atomisation is about $$515\ \text{kJ mol}^{-1}$$, higher than that of neighbouring elements such as titanium and chromium.

In aqueous solution, the green-coloured ion of vanadium is $$[V(H_2O)_6]^{3+}$$, that is $$V^{3+}$$. Hence for the given statement $$M^{n+} = V^{3+}$$ and $$n = 3$$.

To obtain the corresponding oxide of this ion we write

$$2\,V^{3+} + 3\,O^{2-} \rightarrow V_2O_3$$

so the oxide formed is $$V_2O_3$$, in which vanadium is again in the $$+3$$ oxidation state.

The nature of transition-metal oxides depends on the oxidation state:
  • lower oxidation states (e.g. $$+2, +3$$) give basic oxides,
  • intermediate states often give amphoteric oxides,
  • very high oxidation states give acidic oxides.

Since $$V_2O_3$$ contains vanadium in the lower $$+3$$ state, it behaves as a basic oxide; it dissolves readily in mineral acids to regenerate the green $$V^{3+}$$ ion but does not react with alkali to any significant extent.

Therefore, the oxide of the given ion is basic.

Option C (basic) is correct.