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

Magnetic moment (\( \mu_{\text{spin}} \)) depends on the number of unpaired electrons in the ions.
The formula for the magnetic moment due to spin only is: \[ \mu_{\text{spin}} = \sqrt{n(n+2)} \, \text{BM} \] Where \( n \) is the number of unpaired electrons.

Step 1: Determine the number of unpaired electrons for each ion:
- \( \text{Cu}^+ \) (\( \text{Cu}^{1+} \)): The electron configuration of Cu is \( [Ar] 3d^{10} 4s^1 \).
For \( \text{Cu}^+ \), the electron configuration is \( [Ar] 3d^{10} \), so there are no unpaired electrons.
Thus, \( \mu_{\text{spin}} = 0 \, \text{BM} \).
- \( \text{Cu}^{2+} \): The electron configuration of \( \text{Cu}^{2+} \) is \( [Ar] 3d^9 \), which has 1 unpaired electron.
Thus, \( \mu_{\text{spin}} = \sqrt{1(1+2)} = \sqrt{3} \, \text{BM} \).
- \( \text{Cr}^{2+} \): The electron configuration of \( \text{Cr}^{2+} \) is \( [Ar] 3d^4 \), which has 4 unpaired electrons.
Thus, \( \mu_{\text{spin}} = \sqrt{4(4+2)} = \sqrt{24} = 2\sqrt{6} \, \text{BM} \). - \( \text{Cr}^{3+} \): The electron configuration of \( \text{Cr}^{3+} \) is \( [Ar] 3d^3 \), which has 3 unpaired electrons.
Thus, \( \mu_{\text{spin}} = \sqrt{3(3+2)} = \sqrt{15} \, \text{BM} \).
Step 2: Compare the magnetic moment values.
- \( \text{Cu}^+ \) has 0 unpaired electrons, so its magnetic moment is 0 BM.
- \( \text{Cu}^{2+} \) has 1 unpaired electron, so its magnetic moment is \( \sqrt{3} \, \text{BM} \).
- \( \text{Cr}^{3+} \) has 3 unpaired electrons, so its magnetic moment is \( \sqrt{15} \, \text{BM} \).
- \( \text{Cr}^{2+} \) has 4 unpaired electrons, so its magnetic moment is \( 2\sqrt{6} \, \text{BM} \).

Step 3: Final decreasing order.
Thus, the correct decreasing order of spin only magnetic moment values is: \[ \boxed{\text{Cr}^{2+} > \text{Cr}^{3+} > \text{Cu}^{2+} > \text{Cu}^+} \]