Question

List the individual steps used in constructing a Born-Haber cycle for the formation of \(\mathrm{Bal}_{2}\) from the elements. Which of the steps would you expect to be exothermic?

Short answer

The steps for constructing a Born-Haber cycle for the formation of \(\mathrm{BaI}_{2}\) from its elements are as follows: 1. Atomization of elements (\(\mathrm{Ba(s) \to Ba(g)}\) and \(\mathrm{I_{2}(s) \to 2I(g)}\)) - endothermic. 2. Ionization of gaseous barium (\(\mathrm{Ba(g) \to Ba^{2+}(g) + 2e^{-}}\)) - endothermic. 3. Formation of gaseous iodine ions (\(2 \times \mathrm{I(g) + e^{-} \to I^{-}(g)}\)) - exothermic. 4. Formation of solid barium iodide (\(\mathrm{Ba^{2+}(g) + 2I^{-}(g) \to BaI_{2}(s)}\)) - exothermic. Steps 1 and 2 are endothermic processes, while steps 3 and 4 are exothermic processes.

Step-by-step solution

Formation of gaseous atoms

First, we need to form gaseous atoms from the elements. We start with solid barium (Ba) and diatomic iodine molecules (I2). The first step is to atomize these elements into gaseous atoms: \[\mathrm{Ba(s) \to Ba(g)}\] \[\mathrm{I_{2}(s) \to 2I(g)}\] Atomization of both Ba and I2 requires energy input to break the bonds. Therefore, these steps are endothermic.

Ionization of gaseous barium

Next, we need to ionize the gaseous barium atom so that it loses two electrons to form a Ba+2 ion: \[\mathrm{Ba(g) \to Ba^{2+}(g) + 2e^{-}}\] This ionization process requires energy input to remove the electrons from the atom. Therefore, this step is endothermic.

Formation of gaseous iodine ions

Now we need to form gaseous iodine ions by adding electrons to the gaseous iodine atoms: \[2 \times \mathrm{I(g) + e^{-} \to I^{-}(g)}\] This process is called electron affinity. It is the energy change that occurs when an electron is added to an atom. For iodine, the electron affinity is exothermic, as the ion gains stability by obtaining an extra electron to fill up the outer electron shell.

Formation of solid barium iodide

Finally, we combine the gaseous barium and iodine ions to form solid barium iodide: \[\mathrm{Ba^{2+}(g) + 2I^{-}(g) \to BaI_{2}(s)}\] This step is the lattice formation process, which is very exothermic. The ions in the solid-state are attracted to each other by electrostatic forces, releasing energy when they combine to form the ionic compound. So, to summarize: - Steps 1 and 2 (atomization of elements and ionization of gaseous barium) are endothermic processes. - Steps 3 and 4 (formation of gaseous iodine ions and formation of solid barium iodide) are exothermic processes.