Question

Which of the following transition metal ion has magnetic moment \(3.87\) BM?

• A. \(Mn^{2+}\)
• B. \(Co^{3+}\)
• C. \(Fe^{2+}\)
• D. \(Co^{2+}\)

Hint:

Remember spin-only values: \[ n=1 \Rightarrow 1.73,\quad n=2 \Rightarrow 2.83,\quad n=3 \Rightarrow 3.87 \] \[ n=4 \Rightarrow 4.90,\quad n=5 \Rightarrow 5.92 \]

Answer 1

The Correct Option is D

Concept:
The spin-only magnetic moment is calculated by: \[ \mu=\sqrt{n(n+2)}\ \text{BM} \] where: \[ n=\text{number of unpaired electrons} \] Given magnetic moment: \[ \mu=3.87\ \text{BM} \]

Step 1: Find the number of unpaired electrons.
Use: \[ \mu=\sqrt{n(n+2)} \] Try \(n=3\): \[ \mu=\sqrt{3(3+2)} \] \[ \mu=\sqrt{15} \] \[ \mu=3.87\ \text{BM} \] So the ion must have: \[ 3 \text{ unpaired electrons} \]

Step 2: Check \(Co^{2+}\).
Cobalt atomic number is: \[ 27 \] Electronic configuration of Co: \[ [Ar]3d^74s^2 \] For \(Co^{2+}\), remove two electrons from \(4s\): \[ Co^{2+}=[Ar]3d^7 \] A \(d^7\) ion commonly has 3 unpaired electrons in high-spin state. So: \[ \mu=\sqrt{3(3+2)}=\sqrt{15}=3.87\ \text{BM} \]

Step 3: Check other options.
\(Mn^{2+}\) has \(d^5\), generally 5 unpaired electrons: \[ \mu\approx5.92\ \text{BM} \] \(Fe^{2+}\) has \(d^6\), high-spin usually 4 unpaired electrons: \[ \mu\approx4.90\ \text{BM} \] \(Co^{3+}\) has \(d^6\), which may be low spin or high spin depending on ligand field, but it is not the standard answer for \(3.87\ \text{BM}\) in this option set.

Step 4: Final selection.
The ion matching 3 unpaired electrons is: \[ Co^{2+} \] Hence, the correct answer is: \[ \boxed{(D)\ Co^{2+}} \]