The spin-only magnetic moment value of the most basic oxide of vanadium among V₂O₃, V₂O₄ and V₂O₅ is _____________ B.M. (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Lowest oxidation state means most basic oxide, so pick V₂O₃ and count its d electrons.

1. Find oxidation states. Vanadium is +3 in V₂O₃, +4 in V₂O₄ and +5 in V₂O₅. A lower oxidation state makes the oxide more ionic and more basic. A higher one makes it more covalent and more acidic.
2. Most basic oxide. V₂O₃ is basic, V₂O₄ is amphoteric and V₂O₅ is amphoteric but mainly acidic. So the most basic oxide is V₂O₃, which contains V³⁺.
3. Count unpaired electrons. Neutral V is [Ar] 3d³ 4s². Losing three electrons gives V³⁺ as 3d², so n = 2 unpaired electrons.

   μ = √n(n+2) = √(2 × 4) = √8 ≈ 2.83 B.M.

4. So. The spin-only moment is 2.83 B.M., which rounds to 3.

The common slip: 0. Picking V₂O₅ as the most basic oxide gives V⁵⁺, which is 3d⁰ with no unpaired electrons, so the moment comes out as 0. In fact V₂O₅ is the most acidic of the three.

The idea to keep. For oxides of the same metal, the lowest oxidation state gives the most basic oxide and the highest gives the most acidic. Then use μ = √n(n+2) with the unpaired d electrons of that ion.

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