Question

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.

Short answer

(a) \text{In: [Kr] 4d^{10} 5s^{2} 5p^{1}}; \text{In}^{+}: [Kr] 4d^{10} 5s^{2}; \text{In}^{2+}: [Kr] 4d^{10} 5s^{1}; \text{In}^{3+}: [Kr] 4d^{10}. (b) \text{In}, \text{In}^{+}, and \text{In}^{2+} are paramagnetic, \text{In}^{3+} is diamagnetic. (c) Oxidation state of In in \text{InCl}_{2} is +2. (d) InCl_{2} is diamagnetic due to its paired electrons.

Step-by-step solution

- Write the electron configuration for In

Find the electron configuration for neutral indium (In). Indium has an atomic number of 49, so it has 49 electrons. The electron configuration is \[\text{[Kr]} 4d^{10} 5s^{2} 5p^{1}\]

- Write the electron configuration for \(\text{In}^{+}\)

In the \(\text{In}^{+}\) ion, indium has lost one electron. The electron configuration is \[\text{[Kr]} 4d^{10} 5s^{2}\]

- Write the electron configuration for \(\text{In}^{2+}\)

In the \(\text{In}^{2+}\) ion, indium has lost two electrons. The electron configuration is \[\text{[Kr]} 4d^{10} 5s^{1}\]

- Write the electron configuration for \(\text{In}^{3+}\)

In the \(\text{In}^{3+}\) ion, indium has lost three electrons. The electron configuration is \[\text{[Kr]} 4d^{10}\]

- Determine diamagnetic and paramagnetic species

Diamagnetic species have all paired electrons, while paramagnetic species have at least one unpaired electron. \(\text{In}\), \(\text{In}^{+}\), and \(\text{In}^{2+}\) all have unpaired electrons, so they are paramagnetic. \(\text{In}^{3+}\), with all paired electrons, is diamagnetic.

- Determine the oxidation state of In in \(\text{InCl}_{2}\)

Since the formula is \(\text{InCl}_{2}\), In must balance the \(-2\) charge from two chlorine atoms. The oxidation state of In in \(\text{InCl}_{2}\) is \(+2\).

- Explain the diamagnetic nature of InCl2

Given that \(\text{In}^{2+}\) is paramagnetic, for \(\text{InCl}_{2}\) to be diamagnetic, it must form a compound in such a way that paired electrons result. In forming \(\text{InCl}_{2}\), the compound structure leads to all electrons being paired, thereby making it diamagnetic.

Key concepts

Electron Configuration

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

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

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

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.