The spin only magnetic moment of the complex present in Fehling's reagent is __________ B.M. (Nearest integer).

Solution

Chemistry ma’am · 3 steps

Fehling's solution is deep blue because of copper(II), so start from Cu²⁺ and count its unpaired electrons.

1. Identify the complex. Fehling's reagent is made by mixing CuSO₄ solution with alkaline sodium potassium tartrate. The blue complex in it is a copper(II) tartrate complex, so the metal ion is Cu²⁺.
2. Count unpaired electrons. Copper is [Ar] 3d¹⁰4s¹, so Cu²⁺ is 3d⁹. Nine electrons in five d orbitals leave exactly one unpaired electron, so n = 1.

   μ = √n(n+2) = √(1 × 3) = √3 ≈ 1.73 B.M.

3. So. The spin-only magnetic moment is 1.73 B.M., which rounds to 2.

The common slip: 0. Taking copper as Cu⁺ from the red Cu₂O precipitate gives 3d¹⁰ and 0 B.M. That precipitate forms only after the reagent is reduced. The reagent itself contains Cu²⁺.

The idea to keep. For a spin-only moment, fix the oxidation state first, write the d count, then use √n(n+2). Any d⁹ ion, like Cu²⁺, always has one unpaired electron and gives 1.73 B.M.

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