Question

Correct decreasing order of spin-only magnetic moment values is:

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

Hint:

The magnetic moment increases with the number of unpaired electrons. For \( d \)-block elements, the higher the oxidation state, the fewer the unpaired electrons.

Answer 1

The Correct Option is C

To determine the decreasing order of spin-only magnetic moments, we analyze the number of unpaired electrons in each ion using the formula \( \mu = \sqrt{n(n + 2)} \), where \( n \) represents the number of unpaired electrons.

Step 1: Electron Configuration and Number of Unpaired Electrons:

1. 
   \( \text{Cr}^{3+} \):
   Electron configuration: \( [Ar] 3d^3 \)
   Number of unpaired electrons: 3
   
2. 
   \( \text{Cr}^{2+} \):
   Electron configuration: \( [Ar] 3d^4 \)
   Number of unpaired electrons: 4
   
3. 
   \( \text{Cu}^{2+} \):
   Electron configuration: \( [Ar] 3d^9 \)
   Number of unpaired electrons: 1
   
4. 
   \( \text{Cu}^{+} \):
   Electron configuration: \( [Ar] 3d^{10} \)
   Number of unpaired electrons: 0
   

Step 2: Magnetic Moment Calculation:
Using the formula \( \mu = \sqrt{n(n + 2)} \):

• For \( \text{Cr}^{3+} \), \( n = 3 \), so: \( \mu = \sqrt{3(3 + 2)} = \sqrt{15} \).
• For \( \text{Cr}^{2+} \), \( n = 4 \), so: \( \mu = \sqrt{4(4 + 2)} = \sqrt{24} \).
• For \( \text{Cu}^{2+} \), \( n = 1 \), so: \( \mu = \sqrt{1(1 + 2)} = \sqrt{3} \).
• For \( \text{Cu}^{+} \), \( n = 0 \), so: \( \mu = \sqrt{0} = 0 \).

Step 3: Decreasing Order of Magnetic Moments:
The decreasing order of the spin-only magnetic moments is:
\[\text{Cr}^{2+}>\text{Cr}^{3+}>\text{Cu}^{2+}>\text{Cu}^{+}\]

Final Answer: The correct option is (3).