Chapter 8, Problem 8.105QP

Interpretation Introduction

Interpretation: The Lewis structure for ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ is to be drawn. Either of the chlorine atoms in the structure having an expanded valence shell is to be stated.

Concept introduction: The Lewis structures are diagrams that give information about the bonding electron pairs and the lone pairs of electrons in a molecule. Similar to electron dot structure in Lewis diagram the lone pair electrons are represented as dots and they also contain lines which represent bonding electron pairs in a bond.

To determine: If either of the chlorine atoms in the structure has an expanded valence shell and the Lewis structures for ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$.

Answer to Problem 8.105QP

Solution

The Lewis structure for ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ molecule is shown in Figure 2. The central chlorine atom has an expanded valence shell.

Explanation of Solution

Explanation

The reaction of ${\text{FClO}}_{\text{2}}$ with aluminium chloride produces ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ and ${\text{AlFCl}}_{\text{2}}$.

${\text{FClO}}_{\text{2}}\left(g\right)+{\text{AlCl}}_{\text{3}}\left(s\right)\stackrel{}{\to }{\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}\left(g\right)+{\text{AlFCl}}_{\text{2}}\left(s\right)$

The given compound contains two chlorine and two oxygen atoms. The chlorine atom acts as a central metal atom which is attached with two oxygen atoms. Chlorine contains seven valence electrons and oxygen has six valence electrons. The total valence electrons in ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ are calculated as,

$\begin{array}{c}\text{2Cl}+\text{2O}=2\left(7\right)+2\left(6\right)\\ =14+12\\ =26\end{array}$

The skeleton structure is drawn in which $2$ oxygen and one chlorine atom is single bonded to another chlorine atom then the electron pairs around every atom are drawn so that each gets an octet. The formal charge over each atom is calculated by using the formula,

$\text{Formal}\text{charge}=\text{Valence}\text{electrons}-\left(\text{Lone}\text{pair}\text{electrons}+\frac{\text{1}}{\text{2}}\text{Bond}\text{pair}\text{electrons}\right)$

For oxygen atoms,

The number of valence electrons in oxygen atom is six, the lone pair electrons are six and the bonding electrons are two.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

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

For central chlorine atom,

The number of valence electrons in chlorine is seven, the lone pair electrons are two and the bonding electrons are six.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

$\begin{array}{c}\text{Formal}\text{charge}=7-\left(2+\frac{1}{\text{2}}\times 6\right)\\ =7-\left(2+3\right)\\ =7-5\\ =2\end{array}$

For bonded chlorine atom,

The number of valence electrons in chlorine is seven, the lone pair electrons are six and the bonding electrons are two.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

$\begin{array}{c}\text{Formal}\text{charge}=7-\left(6+\frac{1}{\text{2}}\times 2\right)\\ =7-\left(6+1\right)\\ =7-7\\ =0\end{array}$

The Lewis structure of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ is shown in Figure 1.

Figure 1

By assigning a formal charge over the atoms of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ molecule, the first and second chlorine atoms have $+2$ and a zero formal charge respectively. All the oxygen atoms carry a formal charge of $-1$. The atoms do not carry a minimal formal charge. Therefore, the given structure is not the most preferred Lewis structure of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ molecule. The chlorine atom belongs to group VII A and due to the presence of d orbital it expand its orbital and make another possible Lewis dot structure of the ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ molecule which is shown in Figure 2.

Figure 2

The formal charge over each atom is calculated by using the above formula.

For oxygen atoms,

The number of valence electrons in oxygen atom is six, the lone pair electrons are four and the bonding electrons are four.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

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

For central chlorine atom,

The number of valence electrons in chlorine is seven, the lone pair electrons are two and the bonding electrons are ten.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

$\begin{array}{c}\text{Formal}\text{charge}=7-\left(2+\frac{1}{\text{2}}\times 10\right)\\ =7-\left(2+5\right)\\ =7-7\\ =0\end{array}$

For bonded chlorine atom,

The number of valence electrons in chlorine is seven, the lone pair electrons are six and the bonding electrons are two.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

$\begin{array}{c}\text{Formal}\text{charge}=7-\left(6+\frac{1}{\text{2}}\times 2\right)\\ =7-\left(6+1\right)\\ =7-7\\ =0\end{array}$

The Lewis structure shown in Figure 2 shows all the atoms carries a zero formal charge. Therefore, it is the most preferred Lewis structure of the ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ molecule. The central chlorine atom has an expanded valence shell due to the presence of d orbital.

Conclusion

The Lewis structure for ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{2}}$ molecule is shown in Figure 2. The central chlorine atom has an expanded valence shell due to the presence of d orbital.