Chapter 9, Problem 9.65QP

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 9.65QP

Solution

The Lewis structure of ${\text{ClO}}_4{}^{-}$ is figure 1.

The resonating structure I has formal charge close to zero.

The shape of the molecule is tetrahedral and the hybridization of the chlorine is $s{p}^3{d}^3$

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 ${\text{ClO}}_4{}^{-}$ is,

$7+4\times 6+1=32$

The Lewis structure of ${\text{ClO}}_4{}^{-}$ 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 I resonating structure of the given compound from figure 2,

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

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

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{7}\right)-\left(\text{0}\right)-\frac{1}{2}\left(\text{14}\right)\right]\\ =0\end{array}$

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,

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

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

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

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

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

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

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{7}\right)-\left(\text{0}\right)-\frac{1}{2}\left(\text{12}\right)\right]\\ =+1\end{array}$

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

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

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

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

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

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

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

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

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

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{7}\right)-\left(\text{0}\right)-\frac{1}{2}\left(\text{14}\right)\right]\\ =0\end{array}$

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

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

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

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

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

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

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

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

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

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

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

$\begin{array}{c}\text{Formal}\text{charge}=\left[\left(\text{7}\right)-\left(\text{0}\right)-\frac{1}{2}\left(\text{10}\right)\right]\\ =+2\end{array}$

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

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

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

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

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

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

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

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

Resonating structure 1 has formal charge close to zero

The molecular shape of ${\text{ClO}}_{\text{2}}{}^{-}$ is tetrahedral

From the resonating structure I it is concluded that the chlorine atom forms three pi bonds this means it uses three $p$ orbitals, and the four sigma bonds are formed with three $d$ orbitals and one $s$ orbitals. Therefore, the hybridization of central atom chlorine in resonating structure I is $s{p}^3{d}^3$.

Conclusion

The Lewis structure of ${\text{ClO}}_4{}^{-}$ is figure 1.

The resonating structure I has formal charge close to zero.

The shape of the molecule is tetrahedral and the hybridization of the chlorine is $s{p}^3{d}^3$