Chapter 20, Problem 118IP

Although nitrogen trifluoride (NF3) is a thermally stable compound,  nitrogen triiodide (Nl3) is known to be a highly explosive material. NI3 can be synthesized according to the equation

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

a. What is the enthalpy of formation for NI₃(s) given the enthalpy of reaction ( − 307 kJ) and the enthalpies of formation for BN(s) (−254 kJ/mol), IF(g) ( −96 kJ/mol), and BF₃(g) ( − 1136 kJ/mol)?

b. It is reported that when the synthesis of NI₃ is conducted using 4 moles of IF for every l mole of BN. one of the by-products isolated is [IF₂]⁺ [BF₄]⁻. 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?

Interpretation Introduction

Interpretation: The reaction between $\text{BN}$ and $\text{IF}$ is given. The value of enthalpy of formation of ${\text{NI}}_{\text{3}}$ is to be calculated from the given values. The molecular structure and hybridization of ${\left[{\text{IF}}_{\text{2}}\right]}^{+}$ and ${\left[{\text{BF}}_{\text{4}}\right]}^{-}$ is to be stated.

Concept introduction: The following steps are to be followed to determine the molecular structure of a given compound,

• The central atom is identified.

• Its valence electrons are determined.

• The value $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.

• The total number obtained is divided by $2$ to find the number of electron pairs.

• This further gives the hybridization of the given compound.

To determine: The value of enthalpy of formation of ${\text{NI}}_{\text{3}}$.

Explanation of Solution

The value of enthalpy of formation of ${\text{NI}}_{\text{3}}$ is $\underset{\_}{287kJ/mol}$.

Given

Enthalpy of reaction is $-307\text{kJ}$.

Enthalpy of formation for $\text{BN}$ is $-254\text{kJ/mol}$.

Enthalpy of formation for $\text{IF}$ is $-96\text{kJ/mol}$.

Enthalpy of formation for ${\text{BF}}_3$ is $-1136\text{kJ/mol}$.

The reaction that takes place is,

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

The formula of $\Delta H°$ is,

$\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}$

Where,

• $\Delta H°$ standard enthalpy of reaction.

• $n_{\text{p}}$ is number of moles of each product.

• $n_{\text{r}}$ is number of moles each reactant.

• $\Delta H{°}_{\left(\text{product}\right)}$ is standard enthalpy of product at a pressure of $\text{1}\text{atm}$.

• $\Delta H{°}_{\left(\text{reactant}\right)}$ is standard enthalpy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}\\ \Delta H°=\left[\text{1}\left(\Delta H{°}_{{\text{BF}}_{\text{3}}}\right)+\text{1}\left(\Delta H{°}_{{\text{NI}}_{\text{3}}}\right)-\left\{\left(\Delta H{°}_{\text{BN}}\right)+\text{3}\left(\Delta H{°}_{\text{IF}}\right)\right\}\right]\\ -\text{307}\text{kJ/mol}=\left[\text{1}\left(-\text{1136}\right)+\text{1}\left(\Delta H{°}_{{\text{NI}}_{\text{3}}}\right)-\left\{\text{1}\left(-\text{254}\right)+\text{3}\left(-\text{96}\right)\right\}\right]\text{kJ/mol}\\ \Delta H{°}_{{\text{NI}}_{\text{3}}}=-\text{307}+\text{1136}+\left\{\text{1}\left(-\text{254}\right)+\text{3}\left(-\text{96}\right)\right\}\text{kJ/mol}\end{array}$

Simplify the above equation.

$\begin{array}{c}\Delta H{°}_{{\text{NI}}_{\text{3}}}=-\text{307}+\text{1136}+\left\{\text{1}\left(-\text{254}\right)+\text{3}\left(-\text{96}\right)\right\}\text{kJ/mol}\\ =\underset{\_}{287kJ/mol}\end{array}$

Interpretation Introduction

Interpretation: The reaction between $\text{BN}$ and $\text{IF}$ is given. The value of enthalpy of formation of ${\text{NI}}_{\text{3}}$ is to be calculated from the given values. The molecular structure and hybridization of ${\left[{\text{IF}}_{\text{2}}\right]}^{+}$ and ${\left[{\text{BF}}_{\text{4}}\right]}^{-}$ is to be stated.

Concept introduction: The following steps are to be followed to determine the molecular structure of a given compound,

• The central atom is identified.

• Its valence electrons are determined.

• The value $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.

• The total number obtained is divided by $2$ to find the number of electron pairs.

• This further gives the hybridization of the given compound.

To determine: The molecular structure and hybridization of ${\left[{\text{IF}}_{\text{2}}\right]}^{+}$ and ${\left[{\text{BF}}_{\text{4}}\right]}^{-}$.

Explanation of Solution

The molecular structure and hybridization of ${\left[{\text{IF}}_{\text{2}}\right]}^{+}$ is as follows.

The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of central atom iodine $\left(\text{I}\right)$ is $\text{53}$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{5}}$

The valence electron of iodine is $\text{7}$.

The atomic number of fluorine $\left(\text{F}\right)$ is $9$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{5}}$

The valence electron of fluorine is $7$.

The molecule ${\left[{\text{IF}}_{\text{2}}\right]}^{+}$ is made of two fluorine and single iodine atom. This ion also contains a positive charge. Hence, the total number of valence electrons is,

$\begin{array}{c}\text{I}\text{+}2\text{F}-1\text{=}\left(7\text{+}2\times 7-1\right)\\ \text{=}20\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

Where,

• $X$ is the number of electron pairs.

• $V$ is the valence electrons of central atom.

• $M$ is the number of monovalent atoms.

• $C$ is the charge on compound.

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{8}{2}\\ =4\end{array}$

This means that the central atom shows ${\text{sp}}^3$ hybridization and should have a tetrahedral geometry. But due to presence of repulsive force between lone pairs, the structure becomes V- shaped.

The Lewis structure of ${\left[{\text{IF}}_{\text{2}}\right]}^{+}$ is,

Figure 1

The molecular structure and hybridization of ${\left[{\text{BF}}_{\text{4}}\right]}^{-}$ is as follows.

The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of central atom boron $\left(\text{B}\right)$ is $\text{5}$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{1}}$

The valence electron of boron is $\text{3}$.

The atomic number of fluorine $\left(\text{F}\right)$ is $9$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{5}}$

The valence electron of fluorine is $7$.

The molecule ${\left[{\text{BF}}_{\text{4}}\right]}^{-}$ is made of four fluorine and single boron atom. This ion also contains a negative charge. Hence, the total number of valence electrons is,

$\begin{array}{c}\text{B}\text{+}4\text{F}+1\text{=}\left(3\text{+}4\times 7+1\right)\\ \text{=}32\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

Where,

• $X$ is the number of electron pairs.

• $V$ is the valence electrons of central atom.

• $M$ is the number of monovalent atoms.

• $C$ is the charge on compound.

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{8}{2}\\ =4\end{array}$

This means that the central atom shows ${\text{sp}}^3$ hybridization and should have a tetrahedral geometry.

The Lewis structure of ${\left[{\text{BF}}_{\text{4}}\right]}^{-}$ is,

Figure 2