Question

A transition metal ion \(X^{3+}\) has a magnetic moment of \(\sqrt{15}\) BM. The atomic number of the metal \(X\) is

• A. 24
• B. 25
• C. 26
• D. 27

Hint:

Common spin-only magnetic moments: \[ d^1 \rightarrow \sqrt{3} \] \[ d^2 \rightarrow \sqrt{8} \] \[ d^3 \rightarrow \sqrt{15} \] \[ d^5 \rightarrow \sqrt{35} \] Memorizing these values saves a lot of time in exams.

Answer 1

The Correct Option is A

Concept: For most transition metal ions, the magnetic moment is calculated using the spin-only formula: \[ \mu = \sqrt{n(n+2)} \] where \[ n = \text{number of unpaired electrons} \] and \[ \mu = \text{magnetic moment in Bohr Magnetons (BM)} \] By comparing the given magnetic moment with the formula, we can determine the number of unpaired electrons and then identify the metal ion.

Step 1: Use the magnetic moment formula Given \[ \mu=\sqrt{15} \] Therefore, \[ \sqrt{n(n+2)} = \sqrt{15} \] Squaring both sides: \[ n(n+2)=15 \] \[ n^2+2n-15=0 \] \[ (n-3)(n+5)=0 \] \[ n=3 \] Thus the ion contains \[ 3 \text{ unpaired electrons} \]

Step 2: Examine the options Atomic number 24 corresponds to chromium. \[ Cr=[Ar]\,3d^5\,4s^1 \] \[ Cr^{3+} = [Ar]\,3d^3 \] A \(d^3\) configuration contains exactly three unpaired electrons.

Step 3: Verify magnetic moment For \(d^3\), \[ \mu = \sqrt{3(3+2)} = \sqrt{15} \] which exactly matches the given value.

Step 4: Final answer Therefore, \[ X = Cr \] whose atomic number is \[ \boxed{24} \] Hence the correct option is \[ \boxed{\text{(A)}} \]