Step 1: Formula for Magnetic Moment
The spin-only magnetic moment is calculated using the formula:
\[
\mu = \sqrt{n(n+2)} \text{ BM (Bohr Magneton)}
\]
Where \( n \) is the number of unpaired electrons. A higher value of \( n \) results in a higher magnetic moment.
Step 2: Electronic Configuration and Unpaired Electrons
We determine the number of unpaired electrons in the \(3d\) subshell for each ion.
\(Zn^{2+}\) (\(Z=30\)): Configuration is \( [Ar]\,3d^{10} \).
All electrons are paired.
\[
n = 0 \implies \mu = 0
\]
\(Fe^{2+}\) (\(Z=26\)): Configuration is \( [Ar]\,3d^{6} \).
d-orbital arrangement:
\( \uparrow\downarrow \)
\( \uparrow \)
\( \uparrow \)
\( \uparrow \)
\( \uparrow \)
\[
n = 4 \text{ unpaired electrons}
\]
\[
\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}
\]
\(Co^{2+}\) (\(Z=27\)): Configuration is \( [Ar]\,3d^{7} \).
d-orbital arrangement:
\( \uparrow\downarrow \)
\( \uparrow\downarrow \)
\( \uparrow \)
\( \uparrow \)
\( \uparrow \)
\[
n = 3 \text{ unpaired electrons}
\]
\[
\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}
\]
\(Ni^{2+}\) (\(Z=28\)): Configuration is \( [Ar]\,3d^{8} \).
d-orbital arrangement:
\( \uparrow\downarrow \)
\( \uparrow\downarrow \)
\( \uparrow\downarrow \)
\( \uparrow \)
\( \uparrow \)
\[
n = 2 \text{ unpaired electrons}
\]
\[
\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \text{ BM}
\]
Step 3: Conclusion
\(Fe^{2+}\) has the maximum number of unpaired electrons (4), so it has the highest magnetic moment.