Although nitrogen trifluoride (NF \(_{3}\) ) is a thermally stable compound, nitrogen triodide \(\left(\mathrm{NI}_{3}\right)\) 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 \(\mathrm{BN}(s)(-254 \mathrm{kJ} / \mathrm{mol}), \operatorname{IF}(g)(-96 \mathrm{kJ} / \mathrm{mol}),\) 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 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 reaction equation

Write down the given equation: \(BN(s)+3IF(g) \rightarrow BF_3(g)+NI_3(g)\)

02

Write down the known enthalpies and the enthalpy of the reaction

Given enthalpies of formation and enthalpy of reaction are: ∆H(Reaction) = -307 kJ ∆Hf(BN) = -254 kJ/mol ∆Hf(IF) = -96 kJ/mol ∆Hf(BF₃) = -1136 kJ/mol We need to find ∆Hf(NI₃).

03

Apply Hess's Law to find ∆Hf(NI₃)

According to Hess's Law: ∆H(Reaction) = ∆Hf(products) - ∆Hf(reactants) Substitute the known enthalpies into the equation: -307 kJ = [∆Hf(NI₃) + ∆Hf(BF₃)] - [∆Hf(BN) + 3 * ∆Hf(IF)] Now, rearrange the equation to solve for ∆Hf(NI₃): ∆Hf(NI₃) = -307 kJ - ∆Hf(BF₃) + ∆Hf(BN) + 3 * ∆Hf(IF)

04

Calculate the value of ∆Hf(NI₃)

Plug in the given values to the equation: ∆Hf(NI₃) = -307 kJ - (-1136 kJ/mol) + (-254 kJ/mol) + 3 * (-96 kJ/mol) Perform the calculation: ∆Hf(NI₃) = -307 + 1136 - 254 - 288 = 287 kJ/mol The enthalpy of formation for NI₃(s) is 287 kJ/mol. b. Molecular geometries and hybridizations of the by-product species

05

Determine the number of electron domains for each central atom

For the by-product, we have IF₂⁺ and BF₄⁻. For IF₂⁺, central atom is I, and it has 3 electron domains (2 bonding and 1 non-bonding). For BF₄⁻, central atom is B, and it has 4 electron domains (all 4 are bonding).

06

Assign molecular geometries based on electron domains

For IF₂⁺, there are 3 electron domains: Molecular geometry of IF₂⁺ is bent. For BF₄⁻, there are 4 electron domains: Molecular geometry of BF₄⁻ is tetrahedral.

07

Determine the hybridizations based on the electron domains

For IF₂⁺, hybridization is determined by the number of electron domains (3): Hybridization of Iodine (I) in IF₂⁺ is sp². For BF₄⁻, hybridization is determined by the number of electron domains (4): Hybridization of Boron (B) in BF₄⁻ is sp³. Thus, the molecular geometries of the species in the by-product are: IF₂⁺ has a bent molecular geometry, and BF₄⁻ has a tetrahedral molecular geometry. The hybridizations for the central atoms in each species are: Iodine (I) in IF₂⁺ is sp² hybridized, and Boron (B) in BF₄⁻ is sp³ hybridized.

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