The correct decreasing order of spin only magnetic moment values (BM) of Cu⁺, Cu²⁺, Cr²⁺ and Cr³⁺ ions is :

Solution

Chemistry ma’am · 2 steps

Write each ion's d-electron count, count the unpaired ones, and the order follows.

1. Remove 4s first. Cr is [Ar] 3d⁵ 4s¹ and Cu is [Ar] 3d¹⁰ 4s¹. When these atoms form ions, the 4s electron leaves before any 3d electron.

   μ = √n(n+2) BM

   Ion	d configuration	Unpaired electrons n	μ (BM)
   Cr²⁺	3d⁴	4	√24 = 4.90
   Cr³⁺	3d³	3	√15 = 3.87
   Cu²⁺	3d⁹	1	√3 = 1.73
   Cu⁺	3d¹⁰	0	0

2. So. The order is Cr²⁺ > Cr³⁺ > Cu²⁺ > Cu⁺, which is option (c).

The idea to keep. Spin-only moment depends only on the number of unpaired electrons. Write the d configuration, count the singles, and the ion with more unpaired electrons always has the larger moment.

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