Chapter 3, Problem 3.25E

(a)

Interpretation Introduction

Interpretation:

Lewis structure of ${\text{NI}}_3$ has to be drawn. Shape of ${\text{NI}}_3$ has to be identified. Also, whether it can participate in dipole-dipole interactions or not has to be determined.

Concept Introduction:

Lewis structure represents covalent bonds and describes valence electrons configuration of atoms. The covalent bonds are depicted by lines and unshared electron pairs by pairs of dots. The sequence to write Lewis structure of some molecule is given as follows:

• The central atom is identified and various other atoms are arranged around it. This central atom so chosen is often the least electronegative.
• Total valence electrons are estimated for each atom.
• single bond is first placed between each atom pair.
• The electrons left can be allocated as unshared electron pairs or as multiple bonds around symbol of element to satisfy the octet (or duplet) for each atom.
• Add charge on overall structure in case of polyatomic cation or anion.

Hybridization is calculated by the hybrid orbitals and to calculate hybrid orbitals we need to know the steric number that is given by,

  $\text{Steric number}=\left[\begin{array}{c}\left(\text{number of atoms bonded to the central atom}\right)+\\ \left(\text{number of lone pairs on the central atom}\right)\end{array}\right]$        (1)

The relation of steric number with geometry or shape is depicted in the table below.

$\begin{array}{cccc}\text{Steric Number}& \text{Number of lone pairs}& \text{Molecular Geometry}& \text{Shape}\\ 2& 0& \text{linear}& \text{linear}\\ 3& \begin{array}{c}0\\ 1\end{array}& \begin{array}{c}\text{trigonal planar}\\ \text{trigonal planar}\end{array}& \begin{array}{c}\text{trigonal planar}\\ \text{bent}\end{array}\\ 4& \begin{array}{c}0\\ 1\\ 2\end{array}& \begin{array}{c}\text{tetrahedral}\\ \text{tetrahedral}\\ \text{tetrahedral}\end{array}& \begin{array}{c}\text{tetrahedral}\\ \text{trigonal pyramidal}\\ \text{bent}\end{array}\\ 5& \begin{array}{c}0\\ 1\\ 2\\ 3\end{array}& \begin{array}{c}\text{trigonal bipyramidal}\\ \text{trigonal bipyramidal}\\ \text{trigonal bipyramidal}\\ \text{trigonal bipyramidal}\end{array}& \begin{array}{c}\text{trigonal bipyramidal}\\ \text{see-saw}\\ \text{T-shaped}\\ \text{linear}\end{array}\\ 6& \begin{array}{c}0\\ 1\\ 2\end{array}& \begin{array}{c}\text{octahedral}\\ \text{octahedral}\\ \text{octahedral}\end{array}& \begin{array}{c}\text{octahedral}\\ \text{square pyramidal}\\ \text{square planar}\end{array}\end{array}$

Explanation of Solution

The molecule ${\text{NI}}_3$ consists of one $\text{N}$ and four $\text{I}$ atoms. Since $\text{N}$ is least electronegative and is drawn as central atom.

Electronic configuration of $\text{N}$ is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^3$. Thus, it possesses 5 valence electrons.

Electronic configuration of $\text{I}$ is $\left[\text{Kr}\right]{\text{ 4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{5}}$. Thus, it possesses 7 valence electrons.

Thus total valence electrons are sum of the valence electrons for each atom in ${\text{NI}}_3$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=5+3\left(7\right)\\ =26\left(13\text{ pairs}\right)\end{array}$

The skeleton structure ${\text{NI}}_3$ has three bonds that comprise 6 electrons. The electrons left to be allocated are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=26-6\\ =20\left(\text{10 pairs}\right)\end{array}$

Hence, 20 electrons are allocated as 3 lone pairs on each iodine atom and 1 lone pair on nitrogen. The Lewis structure of ${\text{NI}}_3$ is as follows:

As from the structure above, there are 3 bond pairs and 1 lone pair on central atom.

Substitute 3 for number of atoms bonded to central atom and 1 for number of lone pairs on central atom in equation (1) to determine steric number.

$\begin{array}{c}\text{Steric number}=3+1\\ =4\end{array}$

The shape of the molecule is defined by the lone pairs present on the central atom. Molecule ${\text{NI}}_3$ has steric number 4 with 1 lone pair so the electronic arrangement of the molecule and shape is trigonal pyramidal. Since molecule is polar thus it can show dipole-dipole interactions.

(b)

Interpretation Introduction

Interpretation:

Lewis structure of ${\text{BI}}_3$ has to be drawn. Shape of ${\text{BI}}_3$ has to be identified. Also, whether it can participate in dipole-dipole interactions or not has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The molecule ${\text{BI}}_3$ consists of one $\text{B}$ and four $\text{I}$ atoms. Since $\text{B}$ is least electronegative and is drawn as central atom.

Electronic configuration of $\text{B}$ is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^1$. Thus, it possesses 3 valence electrons.

Electronic configuration of $\text{I}$ is $\left[\text{Kr}\right]{\text{ 4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{5}}$. Thus, it possesses 7 valence electrons.

Thus total valence electrons is sum of the valence electrons for each atom in ${\text{BI}}_3$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=3+3\left(7\right)\\ =24\left(12\text{ pairs}\right)\end{array}$

The skeleton structure ${\text{BI}}_3$ has three bonds that comprise 6 electrons. The electrons left to be allocated are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=24-6\\ =18\left(\text{9 pairs}\right)\end{array}$

Hence, 18 electrons are allocated as 3 lone pairs on each iodine atom. The Lewis structure of ${\text{BI}}_3$ is as follows:

As from the structure above, there are 3 bond pairs and 0 lone pairs on central atom.

Substitute 3 for number of atoms bonded to central atom and 0 for number of lone pairs on central atom in equation (1) to determine steric number.

$\begin{array}{c}\text{Steric number}=3+0\\ =0\end{array}$

The shape of the molecule is defined by the lone pairs present on the central atom. Molecule ${\text{BI}}_3$ has steric number 3 with 0 lone pair so the electronic arrangement of the molecule and shape is trigonal planner. Since all dipoles of molecule cancel out each other thus molecule is nonpolar and cannot show dipole-dipole interactions.