Mn_2O_7, CrO and V_2O_5 respectively are:

The correct answer is Option (4) → Acidic, basic and amphoteric

For the oxides of a transition metal, the character changes with the oxidation state of the metal. In a low oxidation state the oxide is ionic and basic; in the highest oxidation state it is covalent and acidic; intermediate states give amphoteric oxides. (For manganese: MnO basic, \( \ce{Mn2O7} \) acidic.)

• Mn₂O₇: 2x + 7(−2) = 0, so x = +7, the highest state of Mn: acidic.

• CrO: x + (−2) = 0, so x = +2, the lowest common state of Cr: basic.

• V₂O₅: 2x + 5(−2) = 0, so x = +5, an intermediate case: amphoteric.

So the three oxides are acidic, basic and amphoteric respectively.

Why the other options are incorrect

• Option (1) Acidic, amphoteric and basic — Incorrect. CrO, with Cr in the +2 state, is basic, not amphoteric; and \( \ce{V2O5} \) is not basic.

• Option (2) Basic, acidic and amphoteric — Incorrect. \( \ce{Mn2O7} \) has Mn in its highest (+7) state, which makes it acidic, not basic.

• Option (3) Amphoteric, basic and acidic — Incorrect. \( \ce{Mn2O7} \) is acidic, not amphoteric, and \( \ce{V2O5} \) is amphoteric rather than only acidic.

NCERT (2026-27 edition):
Oxides and Oxoanions of Metals – “… As the oxidation number of a metal increases, ionic character decreases. In the case of Mn, Mn₂O₇ is a covalent green oil. Even CrO₃ and V₂O₅ have low melting points. In these higher oxides, the acidic character is predominant. Thus, Mn₂O₇ gives HMnO₄ and CrO₃ gives H₂CrO₄ and H₂Cr₂O₇. V₂O₅ is, however, amphoteric though mainly acidic and it gives VO₄^3– as well as VO₂⁺ salts. In vanadium there is gradual change from the basic V₂O₃ to less basic V₂O₄ and to amphoteric V₂O₅. … The well characterised CrO is basic but Cr₂O₃ is amphoteric.”