Question:medium

The calculated spin only magnetic moment of Cr\(^{2+}\) ion is

Show Hint

You can quickly estimate the magnetic moment. The value \(\sqrt{n(n+2)}\) is always slightly less than n+1. For n=4, the value is just under 5. For n=3, it's just under 4 (3.87). This can help you select the correct option without a calculator.
Updated On: Apr 28, 2026
  • 3.87 BM
  • 4.90 BM
  • 5.92 BM
  • 2.84 BM
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The magnetic moment of transition metal ions is primarily determined by the number of unpaired electrons in their d-orbitals.
Step 2: Key Formula or Approach:
Write the electronic configuration of the ion, count the number of unpaired electrons ($n$), and use the spin-only formula: $\mu = \sqrt{n(n+2)}$ BM.
Step 3: Detailed Explanation:
1. The atomic number of Chromium (Cr) is 2
4. Its neutral electronic configuration is an exception: $[Ar] 3d^{5} 4s^{1}$.

2. To form the $Cr^{2+}$ cation, 2 electrons are removed. Electrons are removed from the outermost shell first (one from $4s$), then from the inner shell (one from $3d$).
Configuration of $Cr^{2+}$: $[Ar] 3d^{4}$.

3. In a $3d^{4}$ configuration, applying Hund's rule, there are 4 unpaired electrons ($n = 4$).

4. Calculation using the formula:
\[ \mu = \sqrt{4(4+2)} \]
\[ \mu = \sqrt{4 \times 6} = \sqrt{24} \]
\[ \mu \approx 4.898 \approx 4.90 \text{ BM} \]
Step 4: Final Answer:
The calculated spin-only magnetic moment is 4.90 BM.
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