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

Solution

Chemistry ma’am · 4 steps

Find M first: it is vanadium, and its green ion is V³⁺. Then judge the oxide V₂O₃.

1. Identify M. Enthalpy of atomisation rises across the first transition series towards the middle, where many unpaired d electrons make strong metallic bonds. The highest value belongs to vanadium, about 515 kJ mol⁻¹.
2. Find the green ion. Each aquated vanadium ion has its own colour. Only V³⁺, as [V(H₂O)₆]³⁺, is green, so n = 3.

   Oxidation state	Aquated ion and colour	Oxide	Nature of oxide
   +2	V²⁺, violet	VO	basic
   +3	V³⁺, green	V₂O₃	basic
   +4	VO²⁺, blue	VO₂ (V₂O₄)	amphoteric
   +5	VO₂⁺, yellow	V₂O₅	amphoteric, mainly acidic

3. Why basic. In a low oxidation state like +3, the metal behaves more like an ionic metal. V₂O₃ dissolves in acids to give V³⁺ salts but does not dissolve in alkalis, so it is a basic oxide.
4. So. M is vanadium, the green ion is V³⁺, and its oxide V₂O₃ is basic, option (a).

The idea to keep. For one transition metal, oxide character moves from basic to amphoteric to acidic as the oxidation state goes up. Low states give basic oxides, high states give acidic ones.

Answer from the NTA final answer key. Solution by PrepWiser.