The spin-only magnetic moment value of the ion among Ti²⁺, V²⁺, Co³⁺ and Cr²⁺, that acts as strong oxidising agent in aqueous solution is _________ BM (Near integer).
(Given atomic numbers : Ti: 22, V: 23, Cr: 24, Co: 27)

Solution

Chemistry ma’am · 3 steps

First find which ion wants an electron. That is Co³⁺, so count its unpaired d electrons.

1. Pick the oxidiser. An oxidising agent gains electrons. Ti²⁺, V²⁺ and Cr²⁺ all lose an electron easily, so they are reducing agents. Co³⁺ readily takes an electron to become Co²⁺ (E° ≈ +1.8 V), so it is the strong oxidising agent.

   Ion	Configuration	Behaviour in water
   Ti²⁺	3d²	reducing, goes to Ti³⁺
   V²⁺	3d³	reducing, goes to V³⁺
   Cr²⁺	3d⁴	strong reducing, goes to stable 3d³ Cr³⁺
   Co³⁺	3d⁶	strong oxidising, goes to Co²⁺

2. Unpaired electrons. Co is [Ar]3d⁷ 4s², so Co³⁺ is 3d⁶. For the free ion, five electrons occupy the five d orbitals singly and the sixth pairs up, leaving n = 4 unpaired.

   μ = √n(n+2) = √(4 × 6) = √24 ≈ 4.90 BM

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

The common slip: 4. Reporting the number of unpaired electrons, 4, instead of the magnetic moment gives 4, not 5. Put n into √n(n+2).

The idea to keep. Read the redox behaviour first: M²⁺ ions of early 3d metals are reducing, while Co³⁺ and Mn³⁺ are oxidising. Then apply μ = √n(n+2) with n unpaired electrons.

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