Indium (In) reacts with \(\mathrm{HCl}\) to form a diamagnetic solid with the formula \(\ln \mathrm{Cl}_{2}\). (a) Write condensed electron configurations for \(\mathrm{In}, \mathrm{In}^{+}, \mathrm{In}^{2+}\) and \(\mathrm{In}^{3+}\) (b) Which of these species is (are) diamagnetic and which paramagnetic? (c) What is the apparent oxidation state of In in \(\mathrm{InCl}_{2}\) ? (d) Given your answers to parts (b) and (c), explain how \(\mathrm{InCl}_{2}\) can be diamagnetic.

The electron configuration of an atom describes how the electrons are distributed in its atomic orbitals. For indium (In), which has an atomic number of 49, the electron configuration is \(\text{[Kr]} 4d^{10} 5s^{2} 5p^{1}\). This tells us that indium has its electrons filling up to the 5p orbital. For the ions, \(\text{In}^{+}\), \(\text{In}^{2+}\), and \(\text{In}^{3+}\), electrons are removed from the outermost orbitals in order. So, the configurations are: \(\text{In}^{+}\) : \(\text{[Kr]} 4d^{10} 5s^{2}\), \(\text{In}^{2+}\) : \(\text{[Kr]} 4d^{10} 5s^{1}\), and \(\text{In}^{3+}\) : \(\text{[Kr]} 4d^{10}\). Understanding this helps in determining the properties of each ion.

Diamagnetism is a property that occurs in materials where all electrons are paired. When exposed to a magnetic field, diamagnetic substances are repelled. For indium ions, only \(\text{In}^{3+}\) is diamagnetic because its electron configuration \(\text{[Kr]} 4d^{10}\) has no unpaired electrons. This is in contrast to \(\text{In}\), \(\text{In}^{+}\), and \(\text{In}^{2+}\), which contain unpaired electrons and are thus not diamagnetic. Because diamagnetic substances oppose external magnetic fields, they display weak magnetic susceptibility.

Paramagnetism is a property seen in materials with at least one unpaired electron. Paramagnetic materials are attracted to magnetic fields. In our case, indium \(\text{In}\), \(\text{In}^{+}\), and \(\text{In}^{2+}\) are paramagnetic. Their electron configurations - \(\text{[Kr]} 4d^{10} 5s^{2} 5p^{1}\), \(\text{[Kr]} 4d^{10} 5s^{2}\), and \(\text{[Kr]} 4d^{10} 5s^{1}\), respectively - all have unpaired electrons. These unpaired electrons align with external magnetic fields, causing the attraction seen in paramagnetic materials. Therefore, indium ions possessing these configurations will display paramagnetism.

Oxidation states refer to the charge of an ion when all its bonds are considered ionic. In the given problem, the formula \(\text{InCl}_{2}\) indicates indium's oxidation state in the compound. Since chlorine has a -1 oxidation state and there are two chlorine atoms, the total charge contributed by chlorine is \(-2\). To balance this, indium must have a +2 oxidation state in \(\text{InCl}_{2}\). Despite \(\text{In}^{2+}\) being paramagnetic normally, \(\text{InCl}_{2}\) is diamagnetic because the bonded electrons are paired, thus creating a diamagnetic compound.