Question

The magnetic moment of \(Cr^{3+}\) ion is \(3.87\ \text{BM}\). The number of unpaired electrons present is:

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

Important spin-only magnetic moments:
• \(1\) unpaired electron \(\rightarrow 1.73\ \text{BM}\)
• \(2\) unpaired electrons \(\rightarrow 2.83\ \text{BM}\)
• \(3\) unpaired electrons \(\rightarrow 3.87\ \text{BM}\)
• \(4\) unpaired electrons \(\rightarrow 4.90\ \text{BM}\)
• \(5\) unpaired electrons \(\rightarrow 5.92\ \text{BM}\) These values are frequently used directly in competitive exams.

Answer 1

The Correct Option is C

Concept: The magnetic moment of transition metal ions can be calculated using the spin-only formula: \[ \mu = \sqrt{n(n+2)}\ \text{BM} \] where:
• \(\mu\) = magnetic moment in Bohr Magneton (BM)
• \(n\) = number of unpaired electrons This formula is commonly used for transition metal ions where orbital contribution is negligible.

Step 1: Writing the given magnetic moment.
Given: \[ \mu = 3.87\ \text{BM} \] Using the formula: \[ \mu = \sqrt{n(n+2)} \] Substitute the value of magnetic moment: \[ 3.87 = \sqrt{n(n+2)} \]

Step 2: Squaring both sides.
\[ (3.87)^2 = n(n+2) \] \[ 14.97 \approx n(n+2) \]

Step 3: Testing possible integer values of \(n\).
Check the options: For \(n=1\): \[ 1(1+2)=3 \] \[ \sqrt{3}=1.73\ \text{BM} \] Incorrect. For \(n=2\): \[ 2(2+2)=8 \] \[ \sqrt{8}=2.83\ \text{BM} \] Incorrect. For \(n=3\): \[ 3(3+2)=15 \] \[ \sqrt{15}=3.87\ \text{BM} \] This matches the given magnetic moment. For \(n=4\): \[ 4(4+2)=24 \] \[ \sqrt{24}=4.90\ \text{BM} \] Incorrect.

Step 4: Final conclusion.
Thus, the number of unpaired electrons present in \(Cr^{3+}\) ion is: \[ \boxed{3} \] Hence, the correct option is: \[ \boxed{(3)\ 3} \]