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 need to analyze the number of unpaired electrons in each ion and use the formula for the magnetic moment: \[ \mu = \sqrt{n(n + 2)} \] where \( n \) is 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).