Among Co³⁺, Ti²⁺, V²⁺ and Cr²⁺ ions, one if used as a reagent cannot liberate H₂ from dilute mineral acid solution, its spin-only magnetic moment in gaseous state is ___________ B.M. (Nearest integer)

Solution

Chemistry ma’am · 3 steps

A reducing ion frees hydrogen, so pick out the oxidant Co³⁺ and count its unpaired electrons.

1. Who releases hydrogen. To give H₂, an ion must reduce H⁺, so it must lose an electron easily. Ti²⁺, V²⁺ and Cr²⁺ all have negative E°(M³⁺/M²⁺) values, so they are good reducing agents. Co³⁺ has E°(Co³⁺/Co²⁺) = +1.97 V, so it is a strong oxidising agent and cannot liberate H₂.

   Ion	Configuration	Behaviour with dilute acid
   Ti²⁺	3d²	reducing, gives H₂
   V²⁺	3d³	reducing, gives H₂
   Cr²⁺	3d⁴	reducing, gives H₂
   Co³⁺	3d⁶	oxidising, no H₂

2. Unpaired electrons. A gaseous ion is a free ion with no ligands, so no ligand field forces electrons to pair. The six 3d electrons of Co³⁺ fill all five orbitals singly and then pair once, which leaves n = 4 unpaired electrons.

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

3. So. The spin-only magnetic moment of Co³⁺ is 4.90 B.M., which rounds to 5.

The common slip: 0. Treating Co³⁺ as low-spin, as in its complexes, pairs all six electrons and gives 0. In the gaseous state there are no ligands, so 4 electrons stay unpaired.

The idea to keep. An ion that frees H₂ from acid must be a reducing agent, and a positive E° marks an oxidant. For a gaseous ion, apply Hund's rule with no ligand-field pairing.

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