Among VO₂⁺, MnO₄⁻ and Cr₂ O₇²⁻, the spin-only magnetic moment value of the species with least oxidising ability is __________ BM (Nearest integer).
(Given atomic member V=23, Mn=25, Cr=24)

Solution

Chemistry ma’am · 2 steps

Find each metal's oxidation state first: all three turn out to be d⁰, so no unpaired electrons anywhere.

1. Least oxidising species. Compare standard reduction potentials in acid: MnO₄⁻ about 1.51 V, Cr₂O₇²⁻ about 1.33 V, VO₂⁺ about 1.0 V. The lowest value means the weakest oxidising agent, so the species is VO₂⁺.

   Species	Oxidation state	d-electron count	Unpaired electrons
   VO₂⁺	V: x - 4 = +1, so +5	23 - 18 - 5 = 0, 3d⁰	0
   MnO₄⁻	Mn: +7	3d⁰	0
   Cr₂O₇²⁻	Cr: +6	3d⁰	0

   μ = √n(n+2) BM = √(0 × 2) = 0

2. So. VO₂⁺ has no unpaired electrons, so its spin-only magnetic moment is 0 BM. In fact all three are d⁰ and diamagnetic.

The common slip: 2. Mixing up VO₂⁺ (V in +5, d⁰) with vanadyl VO²⁺ (V in +4, d¹) gives √3 = 1.73, which rounds to 2.

The idea to keep. Oxo-ions with the metal in its group oxidation state (V⁵⁺, Cr⁶⁺, Mn⁷⁺) are all d⁰. No unpaired electrons means a spin-only moment of zero.

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