Question

Although nitrogen$\left({\text{NF}}_{\text{3}}\right)$is a thermally stable compound, nitrogen triiodide$\left({\text{NF}}_{\text{3}}\right)$is known to be a highly explosive material.${\text{NI}}_{\text{3}}$can be synthesized according to the equation

$\text{BN(s) + 3IF(g)}\to {\text{BF}}_{\text{3}}{\text{(g) + NI}}_{\text{3}}\text{(g)}$

1. What is the enthalpy of formation for ${\text{NI}}_{\text{3}}\text{(s)}$ given the enthalpy of $\left(\text{-307kJ}\right)$ and the enthalpies of formation for BN$\left(s\right)(-254kJ/mol),IF\left(g\right)$$(-96kJ/mol)$ , and ${\text{BF}}_3\left(\text{g}\right)\left(\text{-1136kJ/mol}\right)$ ?
2. It is reported that when the synthesis of ${\text{NI}}_{\text{3}}$is conducted using 4 moles of IF for every 1 mole of BN, one of the by-products isolated is ${\left[{\text{IF}}_{\text{2}}\right]}^{\text{+}}{\left[{\text{BF}}_{\text{4}}\right]}^{\text{-}}$

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

1. The ${\text{enthalpyofformationofNI}}_3\text{is287kJ/mol.}$
2. The structure of the hybridization is tetrahedral geometry.

Step-by-step solution

Concept Introduction

Hybridization is the concept of mixing atomic orbitals to generate new hybrid orbitals suited for electron pairing to form chemical bonds.

Calculate enthalpies of formation

a)

Subtracting the enthalpies of formation of reactants from the enthalpies of formation of products yields the enthalpy change for a specific reaction.

$\left.{\text{ΔH}}_{\text{reaction}}^{°}=\sum \Delta H_f^{°}\left(\text{products}\right)\text{-}\sum \Delta H_f^{°}(\text{reactants}\right)$

The reaction shown above has an enthalpy change of $\text{-307kJ}$, and the enthalpies of formation for:

$$\begin{gathered} \text{BN}\left(s\right)\text{is-254kJ/mol} \\ \text{-IF}\left(g\right)\text{is-96kJ/mol} \\ {\text{-BF}}_3\left(g\right)\text{is-1136kJ/mol} \end{gathered}$$

The equation above can be used to calculate the enthalpy of formation for the ${\text{Nl}}_{\text{3}}$${\text{ΔH}}_{\text{reaction}}^{°}\text{=1mol×Δ}{{\text{H}}^{°}}_f\left({\text{BF}}_3\right)\text{+1mol×Δ}{{\text{H}}^{°}}_f\left({\text{NI}}_3\right)\text{-1mol×Δ}{{\text{H}}^{°}}_f\left(\text{BN}\right)\text{-3mol×Δ}{{\text{H}}^{°}}_f\left(\text{IF}\right)$

$\text{1mol×Δ}{{\text{H}}^{°}}_f\left({\text{Nl}}_3\right)\text{=Δ}{{\text{H}}^{°}}_{\text{reaction}}\text{-1mol×Δ}{{\text{H}}^{°}}_f\left({\text{BF}}_3\right)\text{+1mol×Δ}{{\text{H}}^{°}}_f\left(\text{BN}\right)\text{+3mol×Δ}{{\text{H}}^{°}}_f\left(\text{IF}\right)$

$\text{1mol×Δ}{{\text{H}}^{°}}_f\left({\text{Nl}}_3\right)\text{=-307kJ-1mol×}\left(\text{-1136kJ/mol}\right)\text{+1mol×}\left(\text{-254kJ/mol}\right)\text{+3mol×}\left(\text{-96kJ/mol}\right)$

$\text{1mol×Δ}{{\text{H}}^{°}}_f\left({\text{Nl}}_3\right)\text{=287kJ}$

The final step is to divide the previous equation by 1 mol.

$\text{Δ}{{\text{H}}^{°}}_f\left({\text{NI}}_3\right)\text{=287kJ/mol}$

Therefore, the${\text{enthalpyofformationofNI}}_3\text{is287kJ/mol.}$

 Write the value of  IF2 + 

b)

The three-dimensional structure of an atom in a molecule is known as molecular geometry or molecular structure. It determines a substance's reactivity, polarity, phase of matter, colour, magnetism, and biological activity, among other features.

The molecular geometry is predicted as follows by the VSEPR theory.

The $\left[{\text{IF}}_{\text{2}}\right]\text{+}$total valence shell electrons can be calculated as:

${\left[{\text{IF}}_2\right]}^{+}{\text{=7e}}^{-}\left(\text{fromI}\right){\text{+2×7e}}^{-}\left(\text{fromF}\right)-\left(\text{netchargeofmolecule}\right)$

${\left[{\text{IF}}_{\text{2}}\right]}^{\text{+}}\text{= 7 + 14 - 1}$

${\left[{\text{IF}}_{\text{2}}\right]}^{\text{+}}{\text{= 20e}}^{\text{-}}$

As a result, it will have two lone pairs and two bond pairs, with a V-shaped geometry.

This image below depicts the representation of ${\left[{\text{IF}}_{\text{2}}\right]}^{\text{+}}$.

Electrons in the whole valence shell of the${\left[{\text{BF}}_{\text{4}}\right]}^{\text{-}}\text{can be calculated as:}$

${\left[{\text{IF}}_2\right]}^{+}{\text{=3e}}^{-}\left(\text{fromB}\right){\text{+4×7e}}^{-}(\text{from}\left.F\right)-(\text{netchargeofmolecule})$

${\left[{\text{IF}}_{\text{2}}\right]}^{\text{+}}\text{= 3 + 28 + 1}$

${\left[{\text{IF}}_{\text{2}}\right]}^{\text{+}}{\text{= 32e}}^{\text{-}}$

As a result, it will have four bond pairs and a tetrahedral geometry.

The image below depicts the depiction of ${\left[{\text{BF}}_{\text{4}}\right]}^{\text{-}}$.

Therefore, the structure of the hybridization is tetrahedral geometry.