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Understanding the electronic configuration of metal ions is crucial when analyzing their magnetic properties. Every element within the periodic table exhibits a unique distribution of electrons, described as its electronic configuration. This configuration tells us how the electrons are arranged within the atom's orbitals.

For metal ions, you must account for the removal of electrons when forming cations. Electron removal typically starts with the outermost `s` sublevel before advancing to the `d` sublevel. For example, consider the electron configuration of copper: \([Ar] 3d^{10} 4s^1\). To form the copper ion \(\mathrm{Cu}^{2+}\), two electrons are removed, resulting in \([Ar] 3d^9\).

This methodical change in electronic arrangement is key to understanding the behavior of ions. When analyzing ions like \(\mathrm{Ni}^{2+}\), \(\mathrm{Co}^{3+}\), and \(\mathrm{Fe}^{2+}\), it鈥檚 important to first determine the neutral atom鈥檚 electronic configuration before making ion-specific adjustments.

Unpaired electrons play a pivotal role in magnetic properties. Electrons occupy orbitals based on specific rules, like the Pauli Exclusion Principle and Hund's Rule, which help determine whether electrons pair up or remain unpaired. Electrons in unpaired states contribute to an atom's magnetic moment.

Let鈥檚 demystify this with an example. The ion \(\mathrm{Cu}^{2+}\), having an electron configuration of \([Ar] 3d^9\), has one unpaired electron, represented by a single electron not paired with a counterpart in its orbital. On the other hand, \(\mathrm{Ni}^{2+}\) with \([Ar] 3d^8\) has two unpaired electrons.

In both \(\mathrm{Co}^{3+}\) and \(\mathrm{Fe}^{2+}\), the electronic configuration \([Ar] 3d^6\) leads to four unpaired electrons due to the high-spin state of these ions. High-spin states occur when electrons occupy higher orbitals before pairing, resulting in more unpaired electrons.

To determine the magnetic moment of an ion, you can use the formula \( \mu = \sqrt{n(n+2)} \), where \( n \) is the number of unpaired electrons. The magnetic moment is measured in Bohr magnetons (\( \mu_B \)). By calculating the magnetic moment, we can understand how strong the magnetic field generated by the ion would be.

For instance, for the ion \(\mathrm{Cu}^{2+}\), with 1 unpaired electron, the formula gives us \( \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \mu_B \). Similarly, for \(\mathrm{Ni}^{2+}\) which has 2 unpaired electrons, the calculation is \( \mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \mu_B \).

As ions like \(\mathrm{Co}^{3+}\) and \(\mathrm{Fe}^{2+}\) possess 4 unpaired electrons, both their magnetic moments compute to \( \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \mu_B \). The formula helps compare magnetic moments across different ions, identifying the least magnetic to be \(\mathrm{Cu}^{2+}\) among the given choices.