The spin only magnetic moment ( μ ) value (B.M.) of the compound with strongest oxidising power among Mn₂ O₃, TiO and VO is ____________ B.M. (Nearest integer).

Solution

Chemistry ma’am · 4 steps

Find the ion that most wants to gain an electron, then count its unpaired d electrons.

1. Oxidation states. Oxygen is -2, so the metals are Mn³⁺ in Mn₂O₃, Ti²⁺ in TiO and V²⁺ in VO.
2. Strongest oxidant. Mn³⁺ (3d⁴) readily gains one electron to become Mn²⁺ (3d⁵), which is extra stable because its d subshell is half-filled (E°_(Mn³⁺/Mn²⁺) = +1.57 V). Ti²⁺ and V²⁺ lose electrons easily instead, so they are reducing agents. The answer must come from Mn₂O₃.
3. Unpaired electrons. Mn is [Ar] 3d⁵ 4s². Removing three electrons gives 3d⁴, with all four d electrons unpaired, so n = 4.

   μ = √n(n+2) = √(4 × 6) = √24 ≈ 4.90 B.M.

4. So. The spin-only magnetic moment of Mn₂O₃ is about 4.90 B.M., which rounds to 5.

The common slip: 6. Treating Mn in Mn₂O₃ as d⁵ (the electron count of Mn²⁺) gives √35 = 5.92, which rounds to 6. Mn³⁺ is d⁴.

The idea to keep. An ion that is one electron away from d⁵ or d¹⁰ gains it eagerly, so it is a strong oxidant. Then count unpaired electrons and use μ = √n(n+2).

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