Chapter 9, Problem 9.60QE

Write the Lewis structures showing formal charge for the following species.

1. (a) ${\text{ClO}}_3^{-}$
2. (b) NCCN
3. (c) SOCl₂

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{ClO}}_3{}^{-}$ that shows formal charge has to be determined.

Concept Introduction:

A covalent bond is a bond that results from the mutual sharing of electrons between atoms. Lewis structures are representations of the covalent bond. In this, Lewis symbols show how the valence electrons are present in the molecule.

The steps to draw the Lewis structure of the molecule are as follows:

Step 1: Find the central atom and place the other atoms around it. The atom in a compound that has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Estimate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

The formula to calculate formal charge of the atom is as follows:

  $\text{Formal}\text{charge}=\left(\begin{array}{l}\text{number}\text{of}\\ \text{valence}\\ \text{electrons}\end{array}\right)-\left(\left(\begin{array}{l}\text{number}\text{of}\\ \text{lone pairs of}\\ \text{electrons}\end{array}\right)+\left(\frac{1}{2}\right)\left(\begin{array}{l}\text{number}\text{of}\\ \text{shared}\\ \text{electrons}\end{array}\right)\right)$

Answer to Problem 9.60QE

The Lewis structure of ${\text{ClO}}_3{}^{-}$ that shows formal charge is as follows:

Explanation of Solution

The given compound is made up of oxygen, chlorine and sulfur atoms.

$\text{S}$ is the symbol for sulfur. The electronic configuration of sulfur is $\left[\text{Ne}\right]3s^{2}3p^4$. It contains 6 valence electrons in its $3s$ and $3p$ orbital.

$\text{O}$ is the symbol for oxygen. The electronic configuration of oxygen is $\left[\text{He}\right]2s^{2}2p^4$. It contains 6 valence electrons in its $2s$ and $2p$ orbital.

$\text{Cl}$ is the symbol for chlorine. The electronic configuration of $\text{Cl}$ is $\left[\text{Ne}\right]3s^{2}3p^5$. It contains 7 valence electrons in its $3s$ and $3p$ orbital.

The rules applied to obtain the Lewis structure of ${\text{ClO}}_3{}^{-}$ are as follows:

1. Write the skeleton structure.

There are three oxygen atoms and one chlorine atom. Therefore, three bonds are formed.

2. Calculate the total number of valence electrons.

The valence electron of oxygen is calculated as follows:

    $\begin{array}{c}3\left(\text{O}\right)=3\times 6\\ =18\end{array}$

The valence electron of chlorine is calculated as follows:

    $\begin{array}{c}1\left(\text{Cl}\right)=1\times 7\\ =7\end{array}$

Also, the structure has charge of $-1$ that will be added to total number of valence electrons.

The total number of valence electrons is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=18+7+1\\ =26\end{array}$

3. Calculate the remaining electrons that are not used in skeleton structure.

The skeleton structure has three bonds. Therefore, six electrons are used in bonds.

The remaining electrons are calculated as follows:

  $\begin{array}{c}\text{Remaining electrons}=26-6\\ =20\end{array}$

4 To obey the octet rule, the oxygen atom needs 6 electron and chlorine atom needs 2 electrons

5. Satisfy the octet rule.

There are 20 remaining electrons. Multiple bonds can be formed. In this compound, additional bonds are needed to complete the structure. Also, remaining electrons are placed as lone pairs on atom to satisfy octet.

The Lewis structure of ${\text{ClO}}_3{}^{-}$ is as follows:

6. The Lewis structure is finished except for formal charges.

7. The formal charge on an atom in this Lewis structure can be calculated from the equation written as follows:

    $\text{Formal charge}=\left[\begin{array}{c}\left(\text{number of valence electrons in an atom}\right)\\ -\left(\text{number of lone pairs}\right)-\\ \frac{1}{2}\left(\text{number of shared electrons}\right)\end{array}\right]$        (1)

The formal charge on first oxygen atom is calculated as follows:

Substitute 6 for number of valence electrons, 4 for number of lone pairs and 4 for number of shared electrons in equation (1).

    $\begin{array}{c}\text{Formal charge}\left({\text{O}}_1\right)=\left(6\right)-\left(4\right)-\frac{1}{2}\left(4\right)\\ =0\end{array}$

The formal charge on second oxygen atom is calculated as follows:

Substitute 6 for number of valence electrons, 4 for number of lone pairs and 4 for number of shared electrons in equation (1).

    $\begin{array}{c}\text{Formal charge}\left({\text{O}}_2\right)=\left(6\right)-\left(6\right)-\frac{1}{2}\left(2\right)\\ =-1\end{array}$

The formal charge on third oxygen atom is calculated as follows:

Substitute 6 for number of valence electrons, 4 for number of lone pairs and 4 for number of shared electrons in equation (1).

    $\begin{array}{c}\text{Formal charge}\left({\text{O}}_3\right)=\left(6\right)-\left(4\right)-\frac{1}{2}\left(4\right)\\ =0\end{array}$

The formal charge on chlorine atom is calculated as follows:

Substitute 7 for number of valence electrons, 6 for number of lone pairs and 2 for number of shared electrons in equation (1).

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

In this Lewis structure, sulfur, chlorine and two oxygen atoms have formal charge 0. The second oxygen atom has formal charge $-1$.

The Lewis structure made from ${\text{ClO}}_3{}^{-}$ that shows formal charge is as follows:

Interpretation Introduction

Interpretation:

The Lewis structure of $\text{NCCN}$ that shows formal charge has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 9.60QE

The Lewis structure of $\text{NCCN}$ that shows formal charge is as follows:

Explanation of Solution

The given compound is made up of carbon and nitrogen atoms.

$\text{N}$ is the symbol for nitrogen atom. The electronic configuration of nitrogen is $\left[\text{He}\right]2s^{2}2p^3$. It contains five valence electrons in its $2s$ and $2p$ orbital.

$\text{C}$ is the symbol for carbon. The electronic configuration of $\text{C}$ is $\left[\text{He}\right]2s^{2}2p^4$. It contains four valence electrons in its $2s$ and $2p$ orbital.

The rules applied to obtain the Lewis structure of $\text{NCCN}$ are as follows:

1. Write the skeleton structure.

In the skeleton structure, three bonds are formed.

2. Calculate the total number of valence electrons.

The valence electron of carbon is calculated as follows:

    $\begin{array}{c}2\left(\text{C}\right)=1\times 4\\ =8\end{array}$

The valence electron of nitrogen is calculated as follows:

    $\begin{array}{c}2\left(\text{N}\right)=2\times 5\\ =10\end{array}$

The total number of valence electrons is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=10+8\\ =18\end{array}$

3. Calculate the remaining electrons that are not used in skeleton structure.

The skeleton structure has three bonds. Therefore six electrons are used in bonds.

The remaining electrons are calculated as follows:

  $\begin{array}{c}\text{Remaining electrons}=18-6\\ =12\end{array}$

4 To obey the octet rule, the carbon atoms need four electrons and nitrogen atoms need six electrons.

5. Satisfy the octet rule.

There are 12 remaining electrons. Multiple bonds can be formed. In this compound, additional bonds are needed to complete the structure. Also, remaining electrons are placed as lone pairs on nitrogen atom to satisfy octet.

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

6. The Lewis structure is finished except for formal charges.

7. The formal charge on an atom in this Lewis structure can be calculated from the equation written as follows:

    $\text{Formal charge}=\left[\begin{array}{c}\left(\text{number of valence electrons in an atom}\right)\\ -\left(\text{number of lone pairs}\right)-\\ \frac{1}{2}\left(\text{number of shared electrons}\right)\end{array}\right]$        (1)

The formal charge on first nitrogen atom is calculated as follows:

Substitute 5 for number of valence electrons, 2 for nmber of lone pairs and 6 for number of shared electrons in equation (1).

    $\begin{array}{c}\text{Formal charge}\left({\text{N}}_1\right)=\left(5\right)-\left(2\right)-\frac{1}{2}\left(6\right)\\ =0\end{array}$

The formal charge on second nitrogen atom is calculated as follows:

Substitute 5 for number of valence electrons, 2 for number of lone pairs and 6 for number of shared electrons in equation (1).

    $\begin{array}{c}\text{Formal charge}\left({\text{N}}_2\right)=\left(5\right)-\left(2\right)-\frac{1}{2}\left(6\right)\\ =0\end{array}$

The formal charge on first carbon atom is calculated as follows:

Substitute 4 for number of valence electrons, 0 for number of lone pairs and 8 for number of shared electrons in equation (1).

    $\begin{array}{c}\text{Formal charge}\left({\text{C}}_1\right)=\left(4\right)-\left(0\right)-\frac{1}{2}\left(8\right)\\ =0\end{array}$

The formal charge on second carbon atom is calculated as follows:

Substitute 4 for number of valence electrons, 0 for number of lone pairs and 8 for number of shared electrons in equation (1).

    $\begin{array}{c}\text{Formal charge}\left({\text{C}}_2\right)=\left(4\right)-\left(0\right)-\frac{1}{2}\left(8\right)\\ =0\end{array}$

The Lewis structure made from $\text{NCCN}$ that shows formal charge is as follows:

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{SOCl}}_2$ that shows formal charge has to be determined.

Concept Introduction:

Refer to part (a)

Answer to Problem 9.60QE

The Lewis structure made from ${\text{SOCl}}_2$ that shows formal charge is as follows:

Explanation of Solution

The given compound is made up of oxygen, sulfur and chlorine atoms.

$\text{S}$ is the symbol for sulfur. The electronic configuration of sulfur is $\left[\text{Ne}\right]3s^{2}3p^4$. It contains 6 valence electrons in its $3s$ and $3p$ orbital.

$\text{O}$ is the symbol for oxygen. The electronic configuration of oxygen is $\left[\text{He}\right]2s^{2}2p^4$. It contains 6 valence electrons in its $2s$ and $2p$ orbital.

$\text{Cl}$ is the symbol for chlorine. The electronic configuration of $\text{Cl}$ is $\left[\text{Ne}\right]3s^{2}3p^5$. It contains 7 valence electrons in its $3s$ and $3p$ orbital.

The rules applied to obtain the Lewis structure of ${\text{SO}}_2{\text{Cl}}_2$ are as follows:

1. Write the skeleton structure.

There are two chlorine atoms, one oxygen atom and sulfur atom. Sulfur atom is placed as central atom. Therefore, three bonds are formed.

2. Calculate the total number of valence electrons.

The valence electron of oxygen is calculated as follows:

    $\begin{array}{c}1\left(\text{O}\right)=1\times 6\\ =6\end{array}$

The valence electron of sulfur is calculated as follows:

    $\begin{array}{c}1\left(\text{S}\right)=\text{1}\times 6\\ =6\end{array}$

The valence electron of chlorine is calculated as follows:

  $\begin{array}{c}2\left(\text{Cl}\right)=2\times 7\\ =14\end{array}$

The total number of valence electrons is calculated as follows:

    $\begin{array}{c}\text{Total valence electrons}=14+6+6\\ =26\end{array}$

3. Calculate the remaining electrons that are not used in skeleton structure.

The skeleton structure has 3 bonds. Therefore 6 electrons are used in bonds.

The remaining electrons are calculated as follows:

  $\begin{array}{c}\text{Remaining electrons}=26-6\\ =20\end{array}$

4 To obey the octet rule, oxygen atom needs 6 electrons, sulfur atom needs 2 electrons and chlorine atom needs 6 electrons.

5. Satisfy the octet rule.

There are 20 remaining electrons. Multiple bonds can be formed. In this compound, an additional bond is needed to complete the structure. Also, remaining electrons are placed as lone pairs to satisfy octet.

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

6. The Lewis structure is finished except for formal charges.

7. The formal charge on an atom in this Lewis structure can be calculated from the equation written as follows:

    $\text{Formal charge}=\left[\begin{array}{c}\left(\text{number of valence electrons in an atom}\right)\\ -\left(\text{number of lone pairs}\right)-\\ \frac{1}{2}\left(\text{number of shared electrons}\right)\end{array}\right]$        (1)

The formal charge on sulfur atom is calculated as follows:

Substitute 6 for number of valence electrons, 0 for number of lone pairs and 12 for number of shared electrons in equation (1)

    $\begin{array}{c}\text{Formal charge}\left(\text{S}\right)=\left(6\right)-\left(2\right)-\frac{1}{2}\left(8\right)\\ =0\end{array}$

The formal charge on oxygen atom is calculated as follows:

Substitute 6 for number of valence electrons, 4 for number of lone pairs and 4 for number of shared electrons in equation (1)

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

The formal charge on first chlorine atom is calculated as follows:

Substitute 7 for number of valence electrons, 6 for number of lone pairs and 2 for number of shared electrons in equation (1)

    $\begin{array}{c}\text{Formal charge}\left({\text{Cl}}_1\right)=\left(7\right)-\left(6\right)-\frac{1}{2}\left(2\right)\\ =0\end{array}$

The formal charge on second chlorine atom is calculated as follows:

Substitute 7 for number of valence electrons, 6 for number of lone pairs and 2 for number of shared electrons in equation (1)

    $\begin{array}{c}\text{Formal charge}\left({\text{Cl}}_2\right)=\left(7\right)-\left(6\right)-\frac{1}{2}\left(2\right)\\ =0\end{array}$

In this Lewis structure, all the atoms have formal charge 0.

The Lewis structure made from ${\text{SOCl}}_2$ that shows formal charge is as follows: