Question

Although nitrogen trifluoride \(\left(\mathrm{NF}_{3}\right)\) is a thermally stable compound, nitrogen triiodide \(\left(\mathrm{NI}_{3}\right)\) is known to be a highly explosive material. \(\mathrm{NI}_{3}\) can be synthesized according to the equation $$\mathrm{BN}(s)+3 \mathrm{IF}(g) \longrightarrow \mathrm{BF}_{3}(g)+\mathrm{NI}_{3}(g)$$ a. What is the enthalpy of formation for \(\mathrm{NI}_{3}(s)\) given the enthalpy of reaction ( \(-307 \mathrm{~kJ}\) ) and the enthalpies of formation for \(\mathrm{BN}(s)(-254 \mathrm{~kJ} / \mathrm{mol}), \mathrm{IF}(g)(-96 \mathrm{~kJ} / \mathrm{mol})\), and \(\mathrm{BF}_{3}(g)\) \((-1136 \mathrm{~kJ} / \mathrm{mol}) ?\) b. It is reported that when the synthesis of \(\mathrm{NI}_{3}\) is conducted using \(4 \mathrm{~mol} \mathrm{IF}\) for every \(1 \mathrm{~mol} \mathrm{BN}\), one of the by-products isolated is \(\left[\mathrm{IF}_{2}\right]^{+}\left[\mathrm{BF}_{4}\right]^{-}\). What are the molecular geometries of the species in this by-product? What are the hybridizations of the central atoms in each species in the by-product?

Short answer

The enthalpy of formation for \(\mathrm{NI}_{3}(s)\) is \(1371 \mathrm{~kJ/mol}\). In the by-product \(\left[\mathrm{IF}_{2}\right]^{+}\left[\mathrm{BF}_{4}\right]^{-}\), the molecular geometry of \(\left[\mathrm{IF}_{2}\right]^{+}\) is bent with sp² hybridization on the central iodine atom, and the molecular geometry of \(\left[\mathrm{BF}_{4}\right]^{-}\) is tetrahedral with sp³ hybridization on the central boron atom.

Step-by-step solution

Write down the given enthalpy values

We have the following enthalpy values given in the exercise: - Enthalpy of reaction: \(-307 \mathrm{~kJ}\) - Enthalpy of formation of \(\mathrm{BN}(s)\): \(−254 \mathrm{~kJ/mol}\) - Enthalpy of formation of \(\mathrm{IF}(g)\): \(-96 \mathrm{~kJ/mol}\) - Enthalpy of formation of \(\mathrm{BF}_{3}(g)\): \(-1136 \mathrm{~kJ/mol}\)

Use Hess's Law

Hess's Law states that the enthalpy change of a reaction is equal to the sum of the enthalpy changes for the individual steps of the reaction. In this case, we have the enthalpy of reaction and the enthalpies of formation for all compounds in the equation except for \(\mathrm{NI}_{3}(s)\). We can find the enthalpy of formation for \(\mathrm{NI}_{3}(s)\) using the following equation derived from Hess's Law: \(\Delta H_{r} = \Delta H_{f}\left(\mathrm{BF}_{3}\right) + \Delta H_{f}\left(\mathrm{NI}_{3}\right) - \Delta H_{f}\left(\mathrm{BN}\right) - 3 \cdot \Delta H_{f}\left(\mathrm{IF}\right)\)

Solve for the enthalpy of formation for \(\mathrm{NI}_{3}(s)\)

Replacing the given values into the equation, we get: \({-307 \mathrm{~kJ}= -1136 \mathrm{~kJ/mol} + \Delta H_{f}\left(\mathrm{NI}_{3}\right) - {-254 \mathrm{~kJ/mol}} - 3 \cdot (-96 \mathrm{~kJ/mol})}\) Simplifying and solving for \(\Delta H_{f}\left(\mathrm{NI}_{3}\right)\), we find: \(\Delta H_{f}\left(\mathrm{NI}_{3}\right) = -307 + 1136 + 254 + (3 \cdot 96)\) \(\Delta H_{f}\left(\mathrm{NI}_{3}\right) = -307 + 1136 + 254 + 288\) \(\Delta H_{f}\left(\mathrm{NI}_{3}\right) = 1371 \mathrm{~kJ/mol}\) So, the enthalpy of formation for \(\mathrm{NI}_{3}(s)\) is \(1371 \mathrm{~kJ/mol}\). #Part b: Molecular geometries and hybridizations of the central atoms in the by-product#

Identify the central atoms and valence electrons

The central atoms in \(\left[\mathrm{IF}_{2}\right]^{+}\) is Iodine (I) and in \(\left[\mathrm{BF}_{4}\right]^{-}\) is Boron (B). We first need to determine the number of valence electrons in each central atom: - Iodine (I) has 7 valence electrons. - Boron (B) has 3 valence electrons.

Determine electron domains and molecular geometries

Using VSEPR theory, we can determine the electron domains and molecular geometries of the central atoms: - For \(\left[\mathrm{IF}_{2}\right]^{+}\): Iodine (I) has 3 electron domains (2 bonded to F atoms and 1 lone pair). This gives us a bent molecular geometry. - For \(\left[\mathrm{BF}_{4}\right]^{-}\): Boron (B) has 4 electron domains (bonded to 4 F atoms). This gives us a tetrahedral molecular geometry.

Determine the hybridizations of the central atoms

Using the electron domains, we can determine the hybridizations of the central atoms: - For \(\left[\mathrm{IF}_{2}\right]^{+}\): Iodine (I) has 3 electron domains, so it is sp² hybridized. - For \(\left[\mathrm{BF}_{4}\right]^{-}\): Boron (B) has 4 electron domains, so it is sp³ hybridized. To summarize, in the by-product \(\left[\mathrm{IF}_{2}\right]^{+}\left[\mathrm{BF}_{4}\right]^{-}\), the molecular geometry of \(\left[\mathrm{IF}_{2}\right]^{+}\) is bent with sp² hybridization on the central iodine atom, and the molecular geometry of \(\left[\mathrm{BF}_{4}\right]^{-}\) is tetrahedral with sp³ hybridization on the central boron atom.

Key concepts

Hess's Law

Understanding Hess's Law is vital when calculating unknown enthalpies of formation. This fundamental principle of chemistry states that the total enthalpy change in a chemical reaction is the same, regardless of the number of steps taken to complete the reaction.

To make use of Hess's Law, we can imagine a reaction broken down into multiple steps for which enthalpy changes are known and then add those changes together to find the total enthalpy change for the reaction. This method becomes especially helpful when direct measurement of enthalpy change for a reaction is not possible.

In the context of the exercise provided, we used Hess's Law to calculate the enthalpy of formation for \(\mathrm{NI}_{3}(s)\). This was accomplished by rearranging the known enthalpy changes of the other compounds in the equation and solving for the unknown value. It allowed us to build upon the given information systematically to derive a solution without direct experimental data for the enthalpy of formation of \(\mathrm{NI}_{3}(s)\).

Molecular Geometries

The concept of molecular geometries arises from Valence Shell Electron Pair Repulsion (VSEPR) theory, which predicts the shape molecules will take based on the repulsion between electron pairs in the valence shell of the central atom. These geometries are crucial for understanding the physical and chemical properties of compounds.

Each molecule is made up of atoms with regions of electron density known as electron domains, which can be bonds or lone pairs of electrons. The electron domains will arrange themselves as far apart as possible to minimize repulsion, resulting in a definitive molecular geometry.

For instance, in a case where there are four electron domains around a central atom, like in \(\left[\mathrm{BF}_{4}\right]^{-}\), a tetrahedral geometry is formed. If there are three domains, like in the \(\left[\mathrm{IF}_{2}\right]^{+}\) molecule, and one of them is a lone pair, a bent molecular geometry is observed. These shapes are important for predicting many properties like polarity, reactivity, and intermolecular forces.

Hybridization

Hybridization is a concept in molecular chemistry that describes the mixing of atomic orbitals to form new hybrid orbitals suitable for the pairing of electrons to form chemical bonds. Hybridization helps explain the shapes of molecules that cannot be described by simple s-orbital or p-orbital overlap.

The type of hybridization depends on the number of electron domains around the central atom, as seen with molecular geometries. For example, if a central atom has three electron domains, it undergoes sp\(^{2}\) hybridization, resulting in three hybrid orbitals that are all in the same plane. This geometry corresponds to trigonal planar or bent shapes, depending on whether there are lone pairs involved.

In the specific exercise, the central boron atom in the \(\left[\mathrm{BF}_{4}\right]^{-}\) ion has four electron domains and is therefore sp\(^{3}\) hybridized, indicating tetrahedral geometry. In contrast, the central iodine in \(\left[\mathrm{IF}_{2}\right]^{+}\) has three electron domains and takes on sp\(^{2}\) hybridization, resulting in a bent shape.