Chapter 8, Problem 45QP

Interpretation Introduction

Interpretation:

The Lewis structures of the given molecules and ions are to be represented.

Concept introduction:

A Lewis structure is the representation of bonding and non-bonding electron pairs present in the outermost shell of all atoms present in the molecule.

The number of bonds formed by an atom in the molecule is determined by the valance electron pairs.

Dots are placed above and below as well as to the left and right of symbol.

Number of dots is important in Lewis dot symbol but not the order in which the dots are placed around the symbol.

In writing symbol pairing is not done until absolutely necessary.

For metals, the number of dots represents the number of electrons that are lost when the atom forms a cation.

For second period non metals, the number of unpaired dots is the number of bonds the atom can form.

Atomic ions can also be represented by dot symbols, by simply adding (for anions) and subtracting (for cations) the appropriate number of dots from Lewis dot symbol.

The octet rule states that every atom reacts to form bonds till its octet of electrons gets completely filled.

Answer to Problem 45QP

Solution: The Lewis structures of the given molecules and ions are shown below.

a)

b)

c)

d)

e)

f)

Explanation of Solution

a) ${\text{COBr}}_2$

The electronic configuration of carbon, oxygen, and bromine in ${\text{COBr}}_2$ is as follows:

$\begin{array}{c}\text{C}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^2\\ \text{O}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^4\\ \text{Br}=\left[\text{Ar}\right]{\text{3d}}^{10}{\text{4s}}^2{\text{4p}}^5\end{array}$

A carbon atom has the tendency to form four bonds because of the presence of four electrons in its outermost shell, bromine has the tendency to form one bond because of the presence of five electrons in its $\text{4p}$ subshell, and oxygen has the tendency to form two bonds due to the presence of two electrons in its outermost shell. On the basis of bond forming tendency of carbon, bromine, and oxygen atoms, the Lewis structure of ${\text{COBr}}_2$ is represented below.

b) ${\text{H}}_2\text{Se}$

The electronic configuration of hydrogen and selenium in ${\text{H}}_2\text{Se}$ is as follows:

$\begin{array}{c}\text{Se}=\left[\text{Ar}\right]{\text{3d}}^{10}{\text{4s}}^2{\text{4p}}^4\\ \text{H}=1{\text{s}}^1\end{array}$

A selenium atom contains four valance electrons in its $\text{4p}$ subshell and a hydrogen atom contains one valance electrons in its $\text{1s}$ subshell. Therefore, selenium has the tendency to accept two electrons and hydrogen has the tendency to donate one electron in order to complete their outermost shell. Therefore, the Lewis structure of ${\text{H}}_2\text{Se}$ contains two $\text{Se}-\text{H}$ single bonds.

The Lewis structure of ${\text{H}}_2\text{Se}$ is as follows:

c) ${\text{NH}}_2\text{OH}$

The electronic configuration of nitrogen, oxygen, and hydrogen in ${\text{NH}}_2\text{OH}$ is as follows:

$\begin{array}{c}\text{O}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^4\\ \text{N}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^3\\ \text{H}=1{\text{s}}^1\end{array}$

A nitrogen atom has the tendency to form three bonds because of the presence of three electrons in its outermost shell, an oxygen atom has the tendency to form two bonds because of the presence of four electrons in its $2\text{p}$ subshell, and hydrogen has the tendency to form one bond because of the presence of one electron in its outermost shell. On the basis of bond forming tendency of nitrogen, oxygen, and hydrogen atoms, the Lewis structure of ${\text{NH}}_2\text{OH}$ is represented below:

d) ${\text{CH}}_3{\text{NH}}_2$

The electronic configuration of carbon, nitrogen, and hydrogen in ${\text{CH}}_3{\text{NH}}_2$ is as follows:

$\begin{array}{c}\text{N}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^3\\ \text{C}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^2\\ \text{H}=1{\text{s}}^1\end{array}$

A carbon atom has the tendency to form four bonds because of the presence of four electrons in its outermost shell, hydrogen has the tendency to form one bond because of the presence of one electron in its outermost shell, and nitrogen has the tendency to form three bonds due to the presence of three valance electrons in its $2\text{p}$ subshell. On the basis of bond forming tendency of carbon, hydrogen, and nitrogen atoms, the Lewis structure of ${\text{CH}}_3{\text{NH}}_2$ is represented below:

e) ${\text{CH}}_3{\text{CH}}_2\text{Br}$

The electronic configuration of oxygen, carbon, chlorine, and hydrogen in ${\text{CH}}_3{\text{CH}}_2\text{Br}$ is as follows:

$\begin{array}{c}\text{C}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^2\\ \text{Br}=\left[\text{Ar}\right]{\text{3d}}^{10}{\text{4s}}^2{\text{4p}}^5\\ \text{H}=1{\text{s}}^1\end{array}$

A carbon atom has the tendency to form four bonds because of the presence of four electrons in its outermost shell, hydrogen has the tendency to form one bond because of the presence of one electron in its outermost shell, and bromine has the tendency to form one bond because of the presence of five valance electrons in its $\text{4p}$ subshell. On the basis of bond forming tendency of carbon, hydrogen, and bromine atoms, the Lewis structure of ${\text{CH}}_3{\text{CH}}_2\text{Br}$ is represented below:

f) ${\text{NCl}}_3$

The electronic configuration of nitrogen and chlorine in ${\text{NCl}}_3$ is as follows:

$\begin{array}{c}\text{N}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^3\\ \text{Cl}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^2{\text{3p}}^5\end{array}$

A nitrogen atom contains three valance electrons in its $2\text{p}$ subshell and a chlorine atom contains five valance electrons in its $3\text{p}$ subshell. Therefore, nitrogen has the tendency to donate three electrons and chlorine has the tendency to accept one electron to complete their outermost shell. Therefore, the Lewis structure of ${\text{NCl}}_3$ contains three $\text{N}-\text{Cl}$ bonds and one lone pair on nitrogen atom.

The Lewis structure of ${\text{NCl}}_3$ is as follows: