A transition metal 'M' among Sc, Ti, V, Cr, Mn and Fe has the highest second ionisation enthalpy. The spin-only magnetic moment value of M⁺ ion is _______ BM (Near integer)
(Given atomic number Sc: 21, Ti: 22, V: 23, Cr: 24, Mn: 25, Fe: 26)

Solution

Chemistry ma’am · 4 steps

The second electron comes out of M⁺, so find the M⁺ ion with a half-filled 3d⁵ shell.

1. Look at M+. The second ionisation enthalpy removes an electron from M⁺. It is highest when M⁺ has an extra-stable configuration.
2. Pick the metal. Chromium is [Ar] 3d⁵ 4s¹, so Cr⁺ is [Ar] 3d⁵, a half-filled d subshell. Taking an electron out of it costs the most, so M is Cr (about 1590 kJ mol⁻¹, the highest of the six).
3. Count unpaired electrons. In 3d⁵ each of the five d orbitals holds one electron, so n = 5.

   μ = √n(n+2) = √(5 × 7) = √35 ≈ 5.92 BM

4. So. The spin-only magnetic moment of Cr⁺ is 5.92 BM, which rounds to 6.

The common slip: 5. Using Cr²⁺ (3d⁴, 4 unpaired) instead of Cr⁺ gives √24 ≈ 4.9, so 5, not 6.

The idea to keep. The nth ionisation enthalpy depends on the ion left after n-1 removals. If that ion is d⁵ or d¹⁰, expect a big jump.

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