Question

The correct decreasing order of spin only magnetic moment values (BM) of $ \text{Cu}^+ $, $ \text{Cu}^{2+} $, $ \text{Cr}^{2+} $ and $ \text{Cr}^{3+} $ ions is:

• A. \( \text{Cu}^+ > \text{Cu}^{2+} > \text{Cr}^{3+} > \text{Cr}^{2+} \)
• B. \( \text{Cr}^{3+} > \text{Cr}^{2+} > \text{Cu}^+ > \text{Cu}^{2+} \)
• C. \( \text{Cu}^{2+} > \text{Cu}^+ > \text{Cr}^{2+} > \text{Cr}^{3+} \)
• D. \( \text{Cr}^{2+} > \text{Cr}^{3+} > \text{Cu}^{2+} > \text{Cu}^+ \)

Hint:

The magnetic moment of transition metal ions depends on the number of unpaired electrons. A higher number of unpaired electrons results in a higher magnetic moment.

Answer 1

The Correct Option is D

To rank the spin-only magnetic moments of \( \text{Cu}^+ \), \( \text{Cu}^{2+} \), \( \text{Cr}^{2+} \), and \( \text{Cr}^{3+} \) in decreasing order, we must first determine the number of unpaired electrons for each ion using their electronic configurations.

The spin-only magnetic moment (\( \mu \)) is calculated using the formula:

\(\mu = \sqrt{n(n+2)} \, \text{BM}\)

where \( n \) represents the count of unpaired electrons.

1. For \( \text{Cu}^+ \):
   • Cu: [Ar] \(3d^{10} 4s^1\).
   • Cu⁺: [Ar] \(3d^{10}\).
   • \( n = 0 \).
   • \( \mu = \sqrt{0(0+2)} = 0 \, \text{BM}\).
2. For \( \text{Cu}^{2+} \):
   • Cu²⁺: [Ar] \(3d^{9}\).
   • \( n = 1 \).
   • \( \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \, \text{BM}\).
3. For \( \text{Cr}^{2+} \):
   • Cr: [Ar] \(3d^5 4s^1\).
   • Cr²⁺: [Ar] \(3d^{4}\).
   • \( n = 4 \).
   • \( \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \text{BM}\).
4. For \( \text{Cr}^{3+} \):
   • Cr³⁺: [Ar] \(3d^{3}\).
   • \( n = 3 \).
   • \( \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{BM}\).

The calculated spin-only magnetic moments in decreasing order are:

\( \text{Cr}^{2+} (4.90 \, \text{BM}) > \text{Cr}^{3+} (3.87 \, \text{BM}) > \text{Cu}^{2+} (1.73 \, \text{BM}) > \text{Cu}^+ (0 \, \text{BM}) \)

Therefore, the correct descending order is \( \text{Cr}^{2+} > \text{Cr}^{3+} > \text{Cu}^{2+} > \text{Cu}^+ \).