(a) What does the term paramagnetism mean? (b) How can one determine experimentally whether a substance is paramagnetic? (c) Which of the following ions would you expect to be paramagnetic: \(\mathrm{O}_{2}^{+}, \mathrm{N}_{2}^{2-}, \mathrm{Li}_{2}^{+}, \mathrm{O}_{2}^{2-} ?\) For those ions that are paramagnetic, determine the number of unpaired electrons.

Paramagnetism is a type of magnetism exhibited by materials that have unpaired electrons in their atomic or molecular orbitals. These unpaired electrons give rise to a magnetic moment that interacts with an external magnetic field, causing the material to be attracted to the magnetic field.

One can determine if a substance is paramagnetic through an experiment called the Gouy balance method. In this method, a sample of a substance is placed between the poles of a strong electromagnet, and its mass change is measured upon applying the magnetic field. If the mass of the sample decreases, it means that the substance is attracted to the magnetic field and, therefore, is paramagnetic. If the mass remains the same or increases, the substance is not paramagnetic.

We will now check if each of the given ions is paramagnetic, and if so, calculate the number of unpaired electrons: (a) \(\mathrm{O}_{2}^{+}\): Molecular oxygen has 12 valence electrons. The \(\mathrm{O}_{2}^{+}\) ion has one less electron, so it has 11 valence electrons. According to Molecular Orbital (MO) theory, these electrons fill the molecular orbitals in the following order: \(1\sigma_g, 1\sigma_u^*, 2\sigma_g, 2\sigma_u^*, 1\pi_u(2), 3\sigma_g, 1\pi_g(3)\). Since the \(\mathrm{O}_{2}^{+}\) ion has one unpaired electron in the \(1\pi_g\) molecular orbital, it is paramagnetic. (b) \(\mathrm{N}_{2}^{2-}\): Molecular nitrogen has 10 valence electrons. The \(\mathrm{N}_{2}^{2-}\) ion has two additional electrons, so it has 12 valence electrons. These electrons fill the following molecular orbitals, similar to molecular oxygen: \(1\sigma_g, 1\sigma_u^*, 2\sigma_g, 2\sigma_u^*, 1\pi_u(2), 3\sigma_g, 1\pi_g(2)\). Since the electrons completely fill the molecular orbitals, there are no unpaired electrons, so \(\mathrm{N}_{2}^{2-}\) is not paramagnetic. (c) \(\mathrm{Li}_{2}^{+}\): A lithium molecule (\(\mathrm{Li}_{2}\)) has 4 valence electrons. It loses one electron to form \(\mathrm{Li}_{2}^{+}\) which has 3 valence electrons. According to MO theory, these electrons fill the following molecular orbitals: \(1\sigma_g, 1\sigma_u^*\). Since \(\mathrm{Li}_{2}^{+}\) has an unpaired electron in the \(1\sigma_u^*\) molecular orbital, it is paramagnetic. (d) \(\mathrm{O}_{2}^{2-}\): Molecular oxygen has 12 valence electrons. The \(\mathrm{O}_{2}^{2-}\) ion has two additional electrons, so it has 14 valence electrons. Similar to molecular oxygen, these electrons fill the following molecular orbitals: \(1\sigma_g, 1\sigma_u^*, 2\sigma_g, 2\sigma_u^*, 1\pi_u(2), 3\sigma_g, 1\pi_g(4)\). Since the \(\mathrm{O}_{2}^{2-}\) ion has two unpaired electrons in the \(1\pi_g\) molecular orbital, it is paramagnetic.

For the paramagnetic ions, the number of unpaired electrons are: - \(\mathrm{O}_{2}^{+}\): 1 unpaired electron - \(\mathrm{Li}_{2}^{+}\): 1 unpaired electron - \(\mathrm{O}_{2}^{2-}\): 2 unpaired electrons