Although nitrogen trifluoride (NF3) is a thermally stable compound, nitrogen triiodide (NI 3 ) i

Although nitrogen trifluoride (NF3) is a thermally stable...

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?

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?

Key Concepts

Hess's Law

Hess's Law states that the overall enthalpy change for a reaction is the same regardless of the pathway taken, allowing the use of known enthalpies of formation to compute the enthalpy change for a reaction or an unknown formation enthalpy via algebraic manipulation of the reaction equation. This principle underpins many thermochemical calculations and is central to understanding reaction energetics.

Enthalpy of Formation

The enthalpy of formation is defined as the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states. It serves as a fundamental baseline for calculating reaction heats and predicting the stability of compounds.

Reaction Stoichiometry in Thermochemistry

Reaction stoichiometry involves using the balanced chemical equation to relate the amounts of reactants and products, ensuring that the respective enthalpy changes are correctly accounted for per mole of substances. Accurate stoichiometric calculations are essential to derive correct thermodynamic values using Hess's Law.

Thermochemical Calculations

Thermochemical calculations combine stoichiometry, enthalpies of formation, and Hess's Law to determine the energy changes in chemical reactions. This approach is routinely used to compute unknown thermodynamic properties, such as the enthalpy of formation for a compound, by balancing energy inputs and outputs.

VSEPR Theory

VSEPR (Valence Shell Electron Pair Repulsion) theory helps predict the three-dimensional geometry of molecules based on the repulsions between electron pairs around a central atom. It is fundamental in understanding and rationalizing the molecular shapes of species, including polyatomic ions.

Hybridization in Molecular Structure

Hybridization theory explains the mixing of atomic orbitals to form new hybrid orbitals with specific geometries appropriate for bonding in molecules. This concept is crucial for determining the hybridization state of the central atoms in molecular species, which in turn helps predict molecular geometry and bonding properties.

Transcript

00:01 So in this video, we're going to talk about question 118 from chapter 20, which says, although nitrogen tripluride, nf3, is a thermally stable compound, nitrogen triiodide, ni3, is known to be a highly explosive material.

00:15 Ni3 can be synthesized according to the equation bn plus 3if yields bf3 plus ni3.

00:22 In part a, we're asked, what is the enthalpyope of formation for solid ni3 given the enthalpya of reaction and the enthalpyes of formation for bn, i3? and bf3.

00:34 So how are we going to solve this? well, we know that the enthalpy of reaction is equal to the enthalpy of formation of the products, minus the enthalpy of formation of the reactants, each multiplied by their stoichiometric coefficients.

00:48 So we have our delta h is equal to hf of bf 3 plus hf of ni3, minus hf of ni3, minus hf of bn.

00:57 So all we have to do is solve this equation for hf of n if.

01:02 And then plug in all the numbers that we were given in the problem.

01:07 So the equation we end up with is delta h minus hf of bf3 plus 3hf of if plus hf of bn.

01:14 And then once we plug everything in, this evaluates to 287 kilojoules per mole.

01:20 So that's the enthalpy of formation for ni3.

01:24 In part b, we have, it is reported that when the synthesis of ni3 is conducted using four moles of if for every one mole of bn, one of the byproducts isolated is if2 plus bf4 minus.

01:39 What are the molecular geometries of the species in this byproduct? what are the hybridizations of the central atoms in each species in the byproduct? so let's start with our cat ion, with our if2 plus.

01:51 So we know that each of our halogens is going to bring seven electrons.

01:57 So that would give us 21 electrons from our three halogens.

02:00 And then we have a plus one charge.

02:03 So we have one fewer electrons than that.

02:06 So we have 20 electrons total to work with...

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