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 moles of IF for every 1 mole of \(\mathrm{BN}\), 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?

Step by step solution

01

Write down the given enthalpy values.

The following enthalpy values are given: Enthalpy of reaction: \(\Delta H_{r} = -307\,\mathrm{kJ}\) Enthalpy of formation for \(\mathrm{BN}(s): \Delta H_{f}(\mathrm{BN}) = -254\,\mathrm{kJ/mol}\) Enthalpy of formation for \(\mathrm{IF}(g): \Delta H_{f}(\mathrm{IF}) = -96\,\mathrm{kJ/mol}\) Enthalpy of formation for \(\mathrm{BF}_{3}(g): \Delta H_{f}(\mathrm{BF}_{3}) = -1136\,\mathrm{kJ/mol}\)

02

Use the enthalpy of reaction formula to find the enthalpy of formation for \(\mathrm{NI}_{3}(s)\).

The enthalpy of reaction can be expressed as: \[\Delta H_{r} = \sum \Delta H_{f}(\mathrm{products}) - \sum \Delta H_{f}(\mathrm{reactants})\] Substituting the given values, we get: \[-307\,\mathrm{kJ} = [\Delta H_{f}(\mathrm{BF}_{3}) + \Delta H_{f}(\mathrm{NI}_{3})] - [\Delta H_{f}(\mathrm{BN}) + 3\Delta H_{f}(\mathrm{IF})]\]

03

Solve for \(\Delta H_{f}(\mathrm{NI}_{3})\).

Now, we can solve for the enthalpy of formation of \(\mathrm{NI}_{3}(s)\): \[\Delta H_{f}(\mathrm{NI}_{3}) = -307 + 254 + 3 \times 96 - (-1136)\] \[\Delta H_{f}(\mathrm{NI}_{3}) = -53 + 288 + 1136\] \[\Delta H_{f}(\mathrm{NI}_{3}) = 1371\,\mathrm{kJ/mol}\] So, the enthalpy of formation for \(\mathrm{NI}_{3}(s)\) is \(1371\,\mathrm{kJ/mol}\). b. Find the molecular geometries and hybridizations of the central atoms in the by-product \(\left[\mathrm{IF}_{2}\right]^{+}\left[\mathrm{BF}_{4}\right]_{-}^{-}\):

04

Determine the electron domain geometries of the central atoms in the by-product species.

For \(\left[\mathrm{IF}_{2}\right]^{+}\), there are 2 bonded atoms (F) and 1 lone pair on the central I atom, which results in a trigonal planar electron domain geometry. For \(\left[\mathrm{BF}_{4}\right]^{-}\), there are 4 bonded atoms (F) to the central B atom, which results in a tetrahedral electron domain geometry.

05

Determine the molecular geometries of the by-product species.

For \(\left[\mathrm{IF}_{2}\right]^{+}\), the molecular geometry is bent (V-shaped) due to the presence of the lone pair on the central I atom. For \(\left[\mathrm{BF}_{4}\right]^{-}\), the molecular geometry is tetrahedral, as there are no lone pairs on the central B atom.

06

Determine the hybridizations of the central atoms in the by-product species.

For \(\left[\mathrm{IF}_{2}\right]^{+}\), as the I atom has a trigonal planar electron domain geometry, its hybridization is sp2. For \(\left[\mathrm{BF}_{4}\right]^{-}\), as the B atom has a tetrahedral electron domain geometry, its hybridization is sp3. In conclusion, the molecular geometries and hybridizations of the central atoms in the by-product \(\left[\mathrm{IF}_{2}\right]^{+}\left[\mathrm{BF}_{4}\right]_{-}^{-}\) are: - \(\left[\mathrm{IF}_{2}\right]^{+}\): Molecular geometry is bent (V-shaped), and the hybridization of the central I atom is sp2. - \(\left[\mathrm{BF}_{4}\right]^{-}\): Molecular geometry is tetrahedral, and the hybridization of the central B atom is sp3.

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