Although nitrogen trifluoride \(\left(\mathrm{NF}_{3}\right)\) is a thermally stable compound, nitrogen triiodide (NI \(_{3} )\) 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 enthalpies of formation

Begin by writing down the given enthalpies of formation for each compound: \(\Delta H^{o_f}(\mathrm{BN}, s) = -254 \,\mathrm{kJ/mol}\) \(\Delta H^{o_f}(\mathrm{IF}, g) = -96 \,\mathrm{kJ/mol}\) \(\Delta H^{o_f}(\mathrm{BF}_3, g) = -1136 \,\mathrm{kJ/mol}\)

02

Apply Hess's Law to calculate the enthalpy of formation for NI3(s)

To find the enthalpy of formation for \(\mathrm{NI}_{3}(s)\), use Hess's Law, which states that the overall enthalpy change for a reaction is the sum of the enthalpy changes for each reaction step: \(\Delta H^{o} = \sum{(\Delta H_{final})}- \sum{(\Delta H_{initial})}\) We have the equation: \(\mathrm{BN}(s) + 3 \, \mathrm{IF}(g) \longrightarrow \mathrm{BF}_{3}(g) + \mathrm{NI}_{3}(g)\) \(\Delta H^{o} = -307 \, \mathrm{kJ}\) Apply Hess's Law: \(-307 = [\Delta H^{o_f}(\mathrm{BF}_{3},g)+\Delta H^{o_f}(\mathrm{NI}_{3},s)]-[\Delta H^{o_f}(\mathrm{BN},s)+3\Delta H^{o_f}(\mathrm{IF},g)]\)

03

Solve for the enthalpy of formation for NI3(s)

Now, solve for the enthalpy of formation for \(\mathrm{NI}_{3}(s)\): \(\Delta H^{o_f}(\mathrm{NI}_{3},s) = \Delta H^{o} + \Delta H^{o_f}(\mathrm{BN},s) + 3\Delta H^{o_f}(\mathrm{IF},g) - \Delta H^{o_f}(\mathrm{BF}_{3},g)\) \(\Delta H^{o_f}(\mathrm{NI}_{3},s) = -307 + (-254) + 3(-96) - (-1136)\) \(\Delta H^{o_f}(\mathrm{NI}_{3},s) = -171 \, \mathrm{kJ/mol}\) So, the enthalpy of formation for \(\mathrm{NI}_{3}(s)\) is \(-171 \, \mathrm{kJ/mol}\). #b. Molecular geometries and hybridizations of the species in the by-product#

04

Determine the molecular geometry of IF2+

The central atom in \(\mathrm{IF}_{2}^+\) is Iodine (I) and has 3 bonding pairs (with two F atoms) and 2 lone pairs. According to VSEPR theory, this would result in a T-shaped molecular geometry.

05

Determine the molecular geometry of BF4-

The central atom in \(\mathrm{BF}_{4}^-\) is Boron (B) with 4 bonding pairs (with four F atoms), and no lone pairs. According to VSEPR theory, this would result in a tetrahedral molecular geometry.

06

Determine the hybridizations of the central atoms for IF2+ and BF4-

The hybridization of the central atom in \(\mathrm{IF}_{2}^+\) is determined by the number of electron domains. With 3 bonding pairs and 2 lone pairs, there are 5 electron domains, corresponding to an sp\(^3\)d hybridization for Iodine. For \(\mathrm{BF}_{4}^-\), the central atom (B) has 4 electron domains (4 bonding pairs), corresponding to an sp\(^3\) hybridization. In summary, the molecular geometries and hybridizations of the species in the by-product are as follows: 1. \(\mathrm{IF}_{2}^+\): Molecular geometry – T-shaped; Hybridization – sp\(^3\)d 2. \(\mathrm{BF}_{4}^-\): Molecular geometry – Tetrahedral; Hybridization – sp\(^3\).

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