Question

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?

Short answer

The enthalpy of formation for NI₃(s) is 287 kJ/mol. 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.

Step-by-step solution

Write down the reaction equation

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

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).

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)

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

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).

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.

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.

Key concepts

Enthalpy of Formation

To understand the enthalpy of formation, imagine it as the energy change when one mole of a compound is formed from its elements in their standard states. It's a crucial part of thermochemistry, helping us comprehend the energetic dynamics in chemical reactions. In our exercise, we're tasked to find the enthalpy of formation for nitrogen triiodide (\(\text{NI}_3\)s).
We apply Hess's Law, which states that the total enthalpy change in a reaction is the same, no matter the route taken. This allows us to combine known enthalpies to find unknown ones. For \(\text{NI}_3\), we rearrange the Hess's Law equation. We substitute known enthalpy values of reactants and products. A careful calculation following the equation \[\Delta H(\text{Reaction}) = \Delta H_f(\text{products}) - \Delta H_f(\text{reactants})\] reveals that the enthalpy of formation for \(\text{NI}_3\) is 287 kJ/mol.
Understanding this helps in predicting reaction feasibility and thermochemical analysis.

Molecular Geometry

Molecular geometry describes the shape of a molecule based on the arrangement of its atoms. Each molecule's geometry affects properties such as polarity and reactivity. We examined the geometries of IF \(_2^+\) and BF \(_4^-\), which are by-products from a reaction that used IF in an excess. The geometry depends on the electron domains surrounding the central atom.

• For IF \(_2^+\) with three electron domains (two bonding, one non-bonding), the geometry is bent.
• For BF \(_4^-\), which has four bonding domains, the geometry is tetrahedral.

Evaluating molecular geometries is important. It allows us to predict how a molecule will interact with others. This influences the overall reaction outcomes and stability.

Hybridization

Hybridization involves the mixing of atomic orbitals in an atom to form new hybrid orbitals. These explain atomic bonding in molecules. Knowing hybridization helps predict molecular structure and bond angles.
In the given scenario, we explored hybridization in IF \(_2^+\) and BF \(_4^-\).

• For IF \(_2^+\), since there's one non-bonding domain and two bonding ones, iodine undergoes \(\text{sp}^2\) hybridization.
• In BF \(_4^-\), boron is surrounded by four bonding domains, leading to \(\text{sp}^3\) hybridization.

Understanding how atoms hybridize is vital in chemistry as it attributes to the molecule's overall shape, which directly impacts reactivity and properties.

Chemical Reaction Equations

Chemical reaction equations serve as a symbolic representation of a chemical reaction. They illustrate the transformation of reactants into products, respecting the law of conservation of mass. Every atom in the reactant side must appear in the product side.
In this exercise, the equation \(\text{BN}(s)+3\text{IF}(g) \rightarrow \text{BF}_3(g)+\text{NI}_3(g)\) showcases this balance. Properly balancing equations is indispensable.
It ensures all interactions and changes happening between substances are precisely chronicled.
By understanding these reactions, we can predict the outcomes, control chemical processes, and interpret the energetic changes involved.

• Reactants such as BN and IF combine in specific ratios to produce products like BF \(_3\) and NI \(_3\)
• Mastering balancing equations is fundamental for deeper insights into chemical mechanisms.

Mastery of chemical equations is key to exploring complex chemical behavior in various systems.