Chapter 9, Problem 9.64QP

Interpretation Introduction

Interpretation: The Lewis structure and shape of the given compound, the hybridization of the central atom and the resonating structure that has formal charge close to zero is to be identified.

Concept introduction: The Lewis structure represents all bonding and non-bonding electrons surrounding the atoms involved in a molecule. A molecule exhibits a particular geometry in space considering the steric hindrances of the bond pairs and the lone pairs of electron.

Resonance structures are different forms of a molecule in which the chemical connectivity of the atoms in a molecule is same but there distribution of electrons is different.

The formal charge is calculated by the formula,

$\text{Formal}\text{charge}=\left[\begin{array}{r}\left(\text{Number}\text{of}\text{valence}\text{electrons}\right)-\\ \left(\begin{array}{l}\text{Number}\text{of}\text{non}\text{bonding}\text{electrons}\\ \text{around}\text{the}\text{atom}\end{array}\right)-\\ \frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{bonding}\text{electrons}\\ \text{around}\text{the}\text{atom}\end{array}\right)\end{array}\right]$

To determine: The Lewis structure and shape of the given compound, the hybridization of the central atom and the resonating structure that has formal charge close to zero.

Answer to Problem 9.64QP

Solution

The Lewis structure of ${\text{ClO}}_{\text{2}}{}^{-}$ is figure 1.

The resonating structure 1 has formal charge close to zero.

The shape of the molecule is bent and the hybridization of the central atom is $s{p}^3$

Explanation of Solution

Explanation

The atomic number of $\text{Cl}$ is $17$.

The electronic configuration of $\text{Cl}$ is,

$\left[\text{Ne}\right]{\text{3s}}^{2}3{\text{p}}^5$

The number of valence electrons of $\text{Cl}$ is $7$.

The atomic number of $\text{O}$ is $8$.

The electronic configuration of $\text{O}$ is,

${\text{1s}}^{\text{2}}{\text{2p}}^{\text{2}}{\text{2p}}^{\text{4}}$

The number of valence electrons of $\text{O}$ is $6$.

The total number of valence electrons of ${\text{ClO}}_{\text{2}}{}^{-}$ is,

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

The Lewis structure of ${\text{ClO}}_{\text{2}}{}^{-}$ is,

Figure 1

Resonance structures are different forms of a molecule in which the chemical connectivity of the atoms in a molecule is same but there distribution of electrons is different.

The resonance structure of ${\text{ClO}}_{\text{2}}{}^{-}$ is,

Figure 2

The formal charge is calculated by the formula,

$\text{Formal}\text{charge}=\left[\begin{array}{r}\left(\text{Number}\text{of}\text{valence}\text{electrons}\right)-\\ \left(\begin{array}{l}\text{Number}\text{of}\text{non}\text{bonding}\text{electrons}\\ \text{around}\text{the}\text{atom}\end{array}\right)-\\ \frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{bonding}\text{electrons}\\ \text{around}\text{the}\text{atom}\end{array}\right)\end{array}\right]$

According to the 1 resonating structure of the given compound from figure 2,

• Chlorine has six bonding and four nonbonding electrons.
• Oxygen (a) has four bonding electrons and four nonbonding electrons.
• Oxygen (b) has two bonding and six nonbonding electrons.

The formal charge of chlorine in resonating structure 1 is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{7}\right)-\left(\text{4}\right)-\frac{1}{2}\left(\text{3}\right)\right]\\ =0\end{array}$

The formal charge of oxygen (a) in resonating structure 1 is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{6}\right)-\left(\text{4}\right)-\frac{1}{2}\left(\text{4}\right)\right]\\ =0\end{array}$

The formal charge of oxygen (b) in resonating structure 1 is calculated as,

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

According to the 2 resonating structure of the given compound from figure 2,

• Chlorine has four bonding and four nonbonding electrons.
• Oxygen (c) has four bonding electrons and four nonbonding electrons.
• Oxygen (d) has four bonding and four nonbonding electron.

The formal charge of chlorine in resonating structure 2 is calculated as,

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

The formal charge of oxygen (a) in resonating structure 1 is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{6}\right)-\left(\text{4}\right)-\frac{1}{2}\left(\text{4}\right)\right]\\ =0\end{array}$

The formal charge of oxygen (b) in resonating structure 1 is calculated as,

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{6}\right)-\left(\text{4}\right)-\frac{1}{2}\left(\text{4}\right)\right]\\ =0\end{array}$

The central atom chlorine is zero in resonating structure 1.

The molecular shape of ${\text{ClO}}_{\text{2}}{}^{-}$ is bent, one $s$ and three $p$ orbitals are taking part in the hybridization of the ${\text{ClO}}_{\text{2}}{}^{-}$ ion. Therefore, the hybridization of central atom in the given ion is $s{p}^3$.

Conclusion

The Lewis structure of ${\text{ClO}}_{\text{2}}{}^{-}$ is figure 1.

The resonating structure 1 has formal charge close to zero.

The shape of the molecule is bent and the hybridization of the central atom is $s{p}^3$