(a) What does the term diamagnetism mean? (b) How does a diamagnetic substance respond to a magnetic field? (c) Which of the following ions would you expect to be diamagnetic: \(\mathrm{N}_{2}^{2-}, \mathrm{O}_{2}^{2-}, \mathrm{Be}_{2}^{2+}, \mathrm{C}_{2}^{-} ?\)

Diamagnetism refers to the property of a substance where it has no net internal magnetic field (or moment) because all its electrons are paired, which means they cancel each other's magnetic fields. Diamagnetic substances repel external magnetic fields and do not get attracted to them.

When an external magnetic field is applied to a diamagnetic substance, it induces a weak opposing magnetic field within the substance. This induced field opposes the applied field, causing the substance to be slightly repelled by the magnetic field. Diamagnetic substances do not retain any magnetization in the absence of an external magnetic field.

We will analyze the molecular orbital electron configurations of the ions \(\mathrm{N}_{2}^{2-}, \mathrm{O}_{2}^{2-}, \mathrm{Be}_{2}^{2+},\) and \(\mathrm{C}_{2}^{-}\) to determine whether they are diamagnetic or not. (a) \(\mathrm{N}_{2}^{2-}\): Since N has 7 valence electrons, \(\mathrm{N}_{2}\) has 14 valence electrons. With the 2 extra electrons in \(\mathrm{N}_{2}^{2-}\), there are 16 valence electrons, which means that all orbitals up to and including the \(\pi_{2p}(4)\) orbitals are completely filled. Thus, all electrons are paired, and \(\mathrm{N}_{2}^{2-}\) is diamagnetic. (b) \(\mathrm{O}_{2}^{2-}\): An O atom has six valence electrons, resulting in \(\mathrm{O}_{2}\) having 12 valence electrons. \(\mathrm{O}_{2}^{2-}\) has 14 valence electrons, with 8 electrons filling the \(\sigma_{1s}(2), \sigma_{1s}^{*}(2), \sigma_{2s}(2),\) and \(\sigma_{2s}^{*}(2)\) orbitals, and the remaining 6 electrons partially filling the \(\pi_{2p}(4)\) and \(\pi_{2p}^{*}(2)\) orbitals. Since there are unpaired electrons in the \(\pi_{2p}^{*}\) orbitals, \(\mathrm{O}_{2}^{2-}\) is not diamagnetic. (c) \(\mathrm{Be}_{2}^{2+}\): A Be atom has two valence electrons, so \(\mathrm{Be}_{2}\) has 4 valence electrons. \(\mathrm{Be}_{2}^{2+}\) has only 2 valence electrons, thus only filling the \(\sigma_{1s}(2)\) orbital. All electrons are paired, so \(\mathrm{Be}_{2}^{2+}\) is diamagnetic. (d) \(\mathrm{C}_{2}^{-}\): A C atom has 4 valence electrons, which results in \(\mathrm{C}_{2}\) having 8 valence electrons. \(\mathrm{C}_{2}^{-}\) has 9 valence electrons, with 8 electrons filling the \(\sigma_{1s}(2), \sigma_{1s}^{*}(2), \sigma_{2s}(2),\) and \(\sigma_{2s}^{*}(2)\) orbitals, and one unpaired electron in the \(\pi_{2p}(1)\) orbital. Since there is one unpaired electron, \(\mathrm{C}_{2}^{-}\) is not diamagnetic.

Among the given ions \(\mathrm{N}_{2}^{2-}, \mathrm{O}_{2}^{2-}, \mathrm{Be}_{2}^{2+},\) and \(\mathrm{C}_{2}^{-}\), the following ions are diamagnetic: - \(\mathrm{N}_{2}^{2-}\) - \(\mathrm{Be}_{2}^{2+}\)