Arrange the following oxides of manganese in decreasing order of thei…

The correct answer is Option (4) → (C), (D), (A), (B)

For the oxides of one metal, the ionic character decreases as the oxidation number of the metal increases. So the oxides are placed by the oxidation number of manganese in each, worked out from O = −2.

(C) MnO — first (most ionic). x + (−2) = 0, so Mn is +2, its lowest oxidation number here. The MO oxides of the 3d metals are ionic.

(D) Mn₂O₃ — second. 2x + 3(−2) = 0, so Mn is +3.

(A) MnO₂ — third. x + 2(−2) = 0, so Mn is +4; the oxide is less ionic still.

(B) Mn₂O₇ — fourth (least ionic). 2x + 7(−2) = 0, so Mn is +7, the highest. Mn₂O₇ is a covalent green oil, and its acidic character is predominant.

NCERT (2026-27 edition):
4.4.1 Oxides and Oxoanions of Metals – “…All the metals except scandium form MO oxides which are ionic.”

“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.”