Chapter 8, Problem 68QP

Interpretation Introduction

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

Concept introduction:

The Lewis structure is a representation of bonding and nonbonding electron pairs present in the outermost shells of the atoms present in a molecule.

The number of bonds formed by an atom in the molecule is determined by the valence electrons 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 68QP

Solution:

a)

b)

c)

d)

e)

Explanation of Solution

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

The electronic configurations of xenon and fluorine in ${\text{XeF}}_2$ are as follows:

$\begin{array}{c}\text{Xe}=\left[\text{Kr}\right]{\text{4d}}^{10}{\text{5s}}^2{\text{5p}}^6\\ \text{F}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^5\end{array}$

A fluorine atom contains five valence electrons in its $\text{2p}$ subshell due to which it requires only one electron to complete its outermost shell. Two electrons of xenon form bonds with two fluorine atoms. Hence, the Lewis structure of ${\text{XeF}}_2$ contains two $\text{Xe}-\text{F}$ bonds.

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

b) ${\text{XeF}}_4$

The electronic configurations of xenon and fluorine in ${\text{XeF}}_4$ are as follows:

$\begin{array}{c}\text{Xe}=\left[\text{Kr}\right]{\text{4d}}^{10}{\text{5s}}^2{\text{5p}}^6\\ \text{F}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^5\end{array}$

A fluorine atom contains five valence electrons in its $\text{2p}$ subshell due to which it requires only one electron to complete its outermost shell. Four electrons of Xenon form bonds with four fluorine atoms.

Hence, the Lewis structure of ${\text{XeF}}_4$ contains four $\text{Xe}-\text{F}$ bonds.

The Lewis structure of ${\text{XeF}}_4$ is as follows:

c) ${\text{XeF}}_6$

The electronic configurations of xenon and fluorine in ${\text{XeF}}_6$ are as follows:

$\begin{array}{c}\text{Xe}=\left[\text{Kr}\right]{\text{4d}}^{10}{\text{5s}}^2{\text{5p}}^6\\ \text{F}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^5\end{array}$

A fluorine atom contains five valence electrons in its $\text{2p}$ subshell due to which it requires only one electron to complete its outermost shell. Six electrons of Xenon form bonds with six fluorine atoms.

Hence, the Lewis structure of ${\text{XeF}}_6$ contains six $\text{Xe}-\text{F}$ bonds.

The Lewis structure of ${\text{XeF}}_6$ is as follows:

d) ${\text{XeOF}}_4$

The electronic configurations of xenon, oxygen, and fluorine in ${\text{XeOF}}_4$ are as follows:

$\begin{array}{c}\text{Xe}=\left[\text{Kr}\right]{\text{4d}}^{10}{\text{5s}}^2{\text{5p}}^6\\ \text{F}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^5\\ \text{O}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^4\end{array}$

A fluorine atom contains five valence electrons and oxygen contains four electrons in its $\text{2p}$ subshell, respectively, due to which fluorine requires one electron and oxygen requires two electrons to complete their outermost shells. Two electrons of xenon form two bonds with oxygen whereas four electrons of xenon form four bonds with four fluorine. Hence, the Lewis structure of ${\text{XeOF}}_4$ contains four $\text{Xe}-\text{F}$ single bonds and one $\text{Xe}-\text{O}$ double bond.

The Lewis structure of ${\text{XeOF}}_4$ is as follows:

e) ${\text{XeO}}_2{\text{F}}_2$

The electronic configurations of xenon, oxygen, and fluorine in ${\text{XeO}}_2{\text{F}}_2$ are as follows:

$\begin{array}{c}\text{Xe}=\left[\text{Kr}\right]{\text{4d}}^{10}{\text{5s}}^2{\text{5p}}^6\\ \text{F}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^5\\ \text{O}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^4\end{array}$

A fluorine atom contains five valence electrons and oxygen contains four electrons in its $\text{2p}$ subshell, respectively, due to which fluorine requires one electron and oxygen requires two electrons to complete their outermost shells. Four electrons of xenon form four bonds with two oxygen atoms whereas two electrons of xenon form two bonds with two fluorine. Hence, the Lewis structure of ${\text{XeO}}_2{\text{F}}_2$ contains two $\text{Xe}-\text{F}$ single bonds and two $\text{Xe}-\text{O}$ double bonds.

The Lewis structure of ${\text{XeO}}_2{\text{F}}_2$ is as follows: