Understanding Spin-Only Magnetic Moment

The spin-only magnetic moment measures the magnetism arising solely from the spin motion of unpaired electrons in an ion. It is calculated using the formula:

$ \mu_s = \sqrt{n(n+2)} \text{ BM} $

where 'n' represents the number of unpaired electrons.

A higher number of unpaired electrons results in a higher spin-only magnetic moment. Therefore, to find the ion with the highest magnetic moment, we need to find the ion with the maximum number of unpaired electrons.

Determining Unpaired Electrons for Each Ion

We examine the electronic configuration of each ion, focusing on the d-orbitals:

• Cr$^{2+}$: Neutral Cr (Z=24) is $[Ar] 4s^1 3d^5$. Cr$^{2+}$ loses two electrons, resulting in $[Ar] 3d^4$. The configuration of $3d^4$ has 4 unpaired electrons.
• Mn$^{2+}$: Neutral Mn (Z=25) is $[Ar] 4s^2 3d^5$. Mn$^{2+}$ loses two electrons, resulting in $[Ar] 3d^5$. The configuration of $3d^5$ has 5 unpaired electrons.
• Fe$^{2+}$: Neutral Fe (Z=26) is $[Ar] 4s^2 3d^6$. Fe$^{2+}$ loses two electrons, resulting in $[Ar] 3d^6$. The configuration of $3d^6$ has 4 unpaired electrons (following Hund's rule).
• Ni$^{2+}$: Neutral Ni (Z=28) is $[Ar] 4s^2 3d^8$. Ni$^{2+}$ loses two electrons, resulting in $[Ar] 3d^8$. The configuration of $3d^8$ has 2 unpaired electrons.

Comparing Ions and Identifying the Highest Moment

Comparing the number of unpaired electrons:

• Cr$^{2+}$: $n=4$
• Mn$^{2+}$: $n=5$
• Fe$^{2+}$: $n=4$
• Ni$^{2+}$: $n=2$

Mn$^{2+}$ has the maximum number of unpaired electrons ($n=5$).

Conclusion

Since the spin-only magnetic moment is directly proportional to the number of unpaired electrons, the ion with the most unpaired electrons will have the highest magnetic moment.

Therefore, Mn$^{2+}$ has the highest spin-only magnetic moment.