Chapter 18, Problem 108AE

(a)

Interpretation Introduction

Interpretation : The enthalpy of the formation of ${\text{NI}}_{\text{3}}\text{(s)}$ is to be calculated.

Concept Introduction :

The enthalpy is defined as the product of the pressure and volume added up with the system internal energy. It is the property of thermodynamic system. The heat can be absorbed or released is the change in enthalpy.

Answer to Problem 108AE

The enthalpy of formation of $N{I}_3$ is $287\text{kj}{\text{mol}}^{\text{-1}}$

Explanation of Solution

Nitrogen trifloride is colorless, inorganic thermally stable compound. It is non- flammable in nature. The stability is due to comparable size of nitrogen and fluoride. While the nitrogen iodide is an explosive compound due to the difference in the size of the atoms.

Consider the following reaction

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

Consider the enthalpy of formation of ${\text{NI}}_{\text{3}}$ as $x$

  $\begin{array}{l}\text{enthalpy}\text{of}\text{reaction}=\text{enthalpy of product-enthalpy of reactant}\\ \text{enthalpy}\text{of}\text{reaction}=\left[\text{ΔH}\text{of}{\text{BF}}_{\text{3}}\text{+ΔH}\text{of}{\text{NI}}_{\text{3}}\right]\text{-}\left[\text{ΔH}\text{of}\text{}\text{BN+3}\left(\text{ΔH}\text{of}\text{}{\text{IF}}_{\text{3}}\right)\right]\end{array}$

Subtitling $\text{ΔH}\text{of}{\text{BF}}_{\text{3}}=\text{-1136}\text{kj}{\text{mol}}^{\text{-1}}\text{,}\text{ΔH}\text{of}{\text{NI}}_{\text{3}}=x\text{,}\text{ΔH}\text{of}\text{BN}=\text{-254}\text{kj}{\text{Mol}}^{\text{-1}}$

  $\text{ΔH}\text{of}{\text{IF}}_{\text{3}}=\text{-96}\text{kj}{\text{mol}}^{\text{-1}}$ and the enthalpy of reaction as $\text{-307}\text{kj}{\text{mol}}^{\text{-1}}$ in the above reaction

  $\begin{array}{l}-307\text{kj}{\text{mol}}^{\text{-1}}=\left[-1136\text{kj}{\text{mol}}^{\text{-1}}+x\right]-\left[\left(-254\text{kj}{\text{mol}}^{\text{-1}}\right)+3\left(-96\text{kj}{\text{mol}}^{\text{-1}}\right)\right]\\ x=1136\text{kj}{\text{mol}}^{\text{-1}}-307\text{kj}{\text{mol}}^{\text{-1}}-354\text{kj}{\text{mol}}^{\text{-1}}-288\text{kj}{\text{mol}}^{\text{-1}}\\ -307\text{kj}{\text{mol}}^{\text{-1}}=287\text{kj}{\text{mol}}^{\text{-1}}\end{array}$

Thus, the enthalpy of formation of $N{I}_3$ is $287{\text{kJmol}}^{\text{-1}}$

(b)

Interpretation Introduction

Interpretation : The molecular geometries of the species are determined.

Concept Introduction :

The molecular geometry is the three dimensional shape of the atom that contains the molecule. It consists of the general shape which contains bond lengths, bond angles, transition angles and other parameters determining the position of atoms.

Answer to Problem 108AE

It will have 4 bond pairs and the geometry will become tetrahedral with $s{p}^3$ hybridization

  

Explanation of Solution

According to the vesper theory the molecular geometry are predicted as follows

Total valance shell electrons of the ${\left[{\text{IF}}_2\right]}^{+}$ can be calculated as

  $\begin{array}{l}{\left[{\text{IF}}_2\right]}^{+}=7e^{-}\text{of}I+2\left(7e^{-}\text{of}\text{F}\right)-\left(\text{net}\text{charge}\text{of}\text{molecule}\right)\\ =7+14-1\\ {\left[{\text{IF}}_2\right]}^{+}=20e^{-}\end{array}$

Thus it will have 2 lone pairs and 2 bond pairs and the geometry will become V-shaped geometry with $s{p}^3$ hybridization.

It is represented as follows:

  

Total valance shell electrons of the ${\left[B{F}_4\right]}^{-}$ can be calculated as follows:

  $\begin{array}{l}{\left[B{F}_4\right]}^{-}=3e^{-}\text{of}B+4\left(7e^{-}\text{of}F\right)-\left(\text{net charge of molecule}\right)\\ =3+28+1\\ {\left[B{F}_4\right]}^{-}=32e^{-}\end{array}$

Thus it will have 4 bond pairs and the geometry will become tetrahedral with $s{p}^3$ hybridization.