Chapter 8, Problem 8.142AP

Interpretation Introduction

Interpretation: The Lewis structure of the given oxyacids of chlorine using the formal charge and the possible resonance structure is to be identified.

Concept introduction: The Lewis symbols of the atoms indicate the total number of valence electrons surrounding the atom or the ions.

Formal charge is calculated by the formula,

$\text{Formal}\text{charge}=\left(\begin{array}{l}\text{Valence}\text{electrons}\\ -\left(\text{Number}\text{of}\text{nonbonding}\text{electrons}+\frac{\text{Number}\text{of}\text{bonding}\text{electrons}}{2}\right)\end{array}\right)$

To determine: The Lewis structure of all the resonance forms of $\text{HClO}$, ${\text{HClO}}_2$ and ${\text{HClO}}_3$ using the formal charges.

Answer to Problem 8.142AP

Solution

There are two resonating structures of ${\text{HClO}}_2$ and three resonating structures of ${\text{HClO}}_3$.

Explanation of Solution

Explanation

The atomic number of oxygen is $8$.

Therefore, the electronic configuration of oxygen is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^4$.

Thus, the number of valence electrons of oxygen is $6$.

The atomic number of chlorine is $17$.

Therefore, the electronic configuration of chlorine is $\left[\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^5$.

Thus, the number of valence electrons of chlorine is $7$.

The atomic number of hydrogen is $1$.

Therefore, the electronic configuration of hydrogen is ${\text{1s}}^1$.

Thus, the number of valence electrons of hydrogen is $1$.

Thus, the total number of valence electrons of $\text{HClO}$ is,

$1+7+6=14$

Therefore, the possible Lewis structures of $\text{HClO}$ are,

Figure 1

Formal charge is calculated by the formula,

$\text{Formal}\text{charge}=\left(\begin{array}{l}\text{Valence}\text{electrons}\\ -\left(\text{Number}\text{of}\text{nonbonding}\text{electrons}+\frac{\text{Number}\text{of}\text{bonding}\text{electrons}}{2}\right)\end{array}\right)$

Therefore, the formal charge of chlorine on the right Lewis structure containing four nonbonding electrons and four bonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{7}-\left(4+\frac{\text{4}}{2}\right)\right)\\ =7-6\\ =+1\end{array}$

The formal charge of oxygen on the right Lewis structure containing six nonbonding electrons and two bonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{6}-\left(6+\frac{2}{2}\right)\right)\\ =6-7\\ =-1\end{array}$

The formal charge of chlorine on the left Lewis structure containing six nonbonding electrons and two bonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(7-\left(6+\frac{2}{2}\right)\right)\\ =7-7\\ =0\end{array}$

The formal charge of oxygen on the left Lewis structure containing four nonbonding electrons and four bonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{6}-\left(4+\frac{\text{4}}{2}\right)\right)\\ =6-6\\ =0\end{array}$

Therefore, the left Lewis structure is correct.

The total number of valence electrons of ${\text{HClO}}_2$ is,

$1+7+2\times 6=20$

Therefore, the possible Lewis structures of ${\text{HClO}}_2$ are,

Figure 2

The formal charge of chlorine on the right Lewis structure containing four nonbonding electrons and four bonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{7}-\left(4+\frac{\text{4}}{2}\right)\right)\\ =7-6\\ =+1\end{array}$

The formal charge of oxygen on the right Lewis structure containing six nonbonding electrons and two bonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{6}-\left(6+\frac{2}{2}\right)\right)\\ =6-7\\ =-1\end{array}$

The formal charge of chlorine on the left Lewis structure containing four nonbonding electrons and six bonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(7-\left(4+\frac{6}{2}\right)\right)\\ =7-7\\ =0\end{array}$

The formal charge of oxygen on the left Lewis structure containing four nonbonding electrons and four bonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{6}-\left(4+\frac{\text{4}}{2}\right)\right)\\ =6-6\\ =0\end{array}$

The left Lewis structure is correct and the two structures are the resonating structures of each other.

The total number of valence electrons of ${\text{HClO}}_3$ is,

$1+7+3\times 6=26$

Therefore, the possible Lewis structures of ${\text{HClO}}_3$ are,

Figure 3

The formal charge of chlorine on the Lewis structure containing two nonbonding electrons and eight bonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{7}-\left(2+\frac{\text{8}}{2}\right)\right)\\ =7-6\\ =+1\end{array}$

The formal charge of chlorine on the Lewis structure containing two nonbonding electrons and six bonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{7}-\left(2+\frac{6}{2}\right)\right)\\ =7-5\\ =+2\end{array}$

The formal charge of oxygen on the Lewis structure containing six nonbonding electrons and two bonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{6}-\left(6+\frac{2}{2}\right)\right)\\ =6-7\\ =-1\end{array}$

The formal charge of chlorine on the Lewis structure containing two nonbonding electrons and ten bonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(7-\left(2+\frac{5}{2}\right)\right)\\ =7-7\\ =0\end{array}$

The formal charge of oxygen on the left Lewis structure containing four nonbonding electrons and four bonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{6}-\left(4+\frac{\text{4}}{2}\right)\right)\\ =6-6\\ =0\end{array}$

Therefore, the extreme right Lewis structure is correct and the three structures are the resonating structures of each other.

Conclusion:

There are two resonating structures of ${\text{HClO}}_2$ and three resonating structures of ${\text{HClO}}_3$.