Question

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

Short answer

To construct a Born-Haber cycle for the formation of BaI\(_{2}\) from the elements, follow these steps: 1. Sublimation of barium: \(Ba_{(s)} \rightarrow Ba_{(g)}\) (endothermic) 2. Bond dissociation of iodine: \(I_{2 (s)} \rightarrow 2I_{(g)}\) (endothermic) 3. Ionization of barium: \(Ba_{(g)} \rightarrow Ba^{2+}_{(g)} + 2e^{-}\) (endothermic) 4. Electron affinity of iodine: \(I_{(g)} + e^{-} \rightarrow I^{-}_{(g)}\) (exothermic) 5. Formation of BaI\(_{2}\): \(Ba^{2+}_{(g)} + 2I^{-}_{(g)} \rightarrow BaI_{2 (s)}\) (exothermic) The exothermic steps in the Born-Haber cycle are the electron affinity of iodine (Step 4) and the formation of the ionic compound BaI\(_{2}\) (Step 5).

Step-by-step solution

Sublimation of barium

Starting from the solid elements, barium and iodine, we first consider the sublimation of barium. The sublimation process involves converting a solid into a gas. This involves the absorption of heat, making it an endothermic process. \[Ba_{(s)} \rightarrow Ba_{(g)}\]

Breaking the iodine molecule

Next, we need to break the diatomic iodine molecule (I2) into individual iodine atoms. This process is called bond dissociation. As the bond between two iodine atoms needs to be broken, this step is also an endothermic process. \[I_{2 (s)} \rightarrow 2I_{(g)}\]

Ionization of barium

The third step involves ionizing gaseous barium to form a positively charged ion with the removal of two electrons. In this process, we are providing energy to remove the electrons, making it an endothermic process. \[Ba_{(g)} \rightarrow Ba^{2+}_{(g)} + 2e^{-}\]

Electron affinity of iodine

The fourth step involves adding an electron to the gaseous iodine atoms to form negatively charged iodine ions. This process is known as electron affinity. As energy is released during electron addition, this step is exothermic. \[I_{(g)} + e^{-} \rightarrow I^{-}_{(g)}\] Please note that it happens twice because there are two iodine atoms.

Formation of the ionic compound BaI\(_{2}\)

Finally, the gaseous ions come together to form the ionic compound BaI2. As the positive and negative ions come together, there will be attractive forces between them. This process leads to a decrease in potential energy, releasing energy, making it an exothermic process. \[Ba^{2+}_{(g)} + 2I^{-}_{(g)} \rightarrow BaI_{2 (s)}\] Now, let's summarize the steps and their characteristics: 1. Sublimation of barium: endothermic 2. Bond dissociation of iodine: endothermic 3. Ionization of barium: endothermic 4. Electron affinity of iodine: exothermic 5. Formation of BaI2: exothermic The exothermic steps in constructing the Born-Haber cycle for BaI2 are the electron affinity of iodine (Step 4) and the formation of the ionic compound BaI2 (Step 5).

Key concepts

Sublimation

Sublimation involves the transformation of a substance from a solid to a gas without passing through the liquid state. For barium (Ba), this process is necessary in the Born-Haber cycle when forming barium iodide from the solid element. Sublimation requires the absorption of heat, making it an endothermic process. This means it takes in heat from the surroundings, which is necessary to break the bonds holding the solid lattice together.
During the sublimation step for barium, the equation is modeled as:

• \(Ba_{(s)} \rightarrow Ba_{(g)}\)

This step helps transform barium into the gaseous form needed for further reactions in the formation cycle.

Bond Dissociation

Bond dissociation is the process of breaking the bond between atoms in a molecule. Taking iodine as an example, this step involves splitting diatomic iodine (I_2) into individual iodine atoms. This process can require a significant amount of energy since the bond between the two iodine atoms has to be broken.
Because breaking bonds requires energy input, bond dissociation is considered endothermic. The energy absorbed during this process transitions iodine from a bonded diatomic state to free atoms:

• \(I_{2 (s)} \rightarrow 2I_{(g)}\)

Understanding bond dissociation is critical because it allows for the individual iodine atoms to later participate in forming an ionic bond with barium.

Ionization Energy

Ionization energy is the energy required to remove electrons from an atom or ion. When forming an ionic compound, it is important to consider the ionization energy of the involved elements, such as barium. This step in the Born-Haber cycle involves removing electrons from gaseous barium to create a positively charged ion (Ba^{2+}).
As energy is needed to overcome the attractive forces between the positively charged nucleus and the negatively charged electrons, ionization energy is inherently an endothermic process. The chemical reaction for this process is expressed as:

• \(Ba_{(g)} \rightarrow Ba^{2+}_{(g)} + 2e^{-}\)

The removed electrons will then be used in the subsequent electron affinity step.

Electron Affinity

Electron affinity refers to the energy change that occurs when an electron is added to a gaseous atom. For iodine, this involves the gain of an electron to form an iodide ion (I^-). Unlike ionization energy, electron affinity is usually exothermic. This means energy is released when an electron is added.
The process of adding an electron to iodine involves the release of energy because the negatively charged electron is attracted to the positively charged nucleus:

• \(I_{(g)} + e^{-} \rightarrow I^{-}_{(g)}\)

In the Born-Haber cycle, this process happens twice due to the requirement to form two iodide ions for every barium ion.

Ionic Compound Formation

Ionic compound formation is the final step where positive and negative ions like Ba^{2+} and two I^- ions come together to form a stable ionic compound. This process is exothermic since the attractive interactions between opposite charges reduce the potential energy of the system, releasing energy.
In the Born-Haber cycle, the formation of BaI(_{2}) from its constituent ions is an essential step demonstrating the stabilization through electrostatic forces:

• \(Ba^{2+}_{(g)} + 2I^{-}_{(g)} \rightarrow BaI_{2 (s)}\)

Understanding each phase of ionic compound formation helps explain how the thermodynamically favorable ionic lattice is created, highlighting the balance of energy changes in the Born-Haber cycle.