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

 

Question:

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

1. Cr²⁺ > Cr³⁺ > Cu²⁺ > Cu⁺

2. Cu²⁺ > Cu⁺ > Cr²⁺ > Cr³⁺

3. Cu⁺ > Cu²⁺ > Cr³⁺ > Cr²⁺

4. Cr³⁺ > Cr²⁺ > Cu⁺ > Cu²⁺

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Short Solution (Text):

Step 1: Write electronic configurations of ions

• Cu (Z = 29): [Ar] 3d¹⁰ 4s¹

  • Cu⁺ → [Ar] 3d¹⁰ → all paired, no unpaired e⁻ → μ = 0 BM

  • Cu²⁺ → [Ar] 3d⁹ → 1 unpaired e⁻ → μ = √1(1+2) = √3 ≈ 1.73 BM

• Cr (Z = 24): [Ar] 3d⁵ 4s¹

  • Cr²⁺ → [Ar] 3d⁴ → 4 unpaired e⁻ → μ = √4(4+2) = √24 ≈ 4.90 BM

  • Cr³⁺ → [Ar] 3d³ → 3 unpaired e⁻ → μ = √3(3+2) = √15 ≈ 3.87 BM

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Step 2: Compare values

• Cr²⁺ ≈ 4.90 BM

• Cr³⁺ ≈ 3.87 BM

• Cu²⁺ ≈ 1.73 BM

• Cu⁺ = 0 BM

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✅ Final Answer:
Correct order = Cr²⁺ > Cr³⁺ > Cu²⁺ > Cu⁺

(Option 1)

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Conclusion – Video Solution:

In this question, we compared the number of unpaired electrons in each ion, applied the spin-only formula μ = √[n(n+2)], and arranged them in decreasing order. The highest magnetic moment comes from Cr²⁺ (d⁴), followed by Cr³⁺ (d³), Cu²⁺ (d⁹), and finally Cu⁺ (d¹⁰, diamagnetic).