Chapter 9, Problem 9.57QE

(a)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{NO}}_{\text{2}}{}^{-}$ 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 9.57QE

The Lewis structure of ${\text{NO}}_{\text{2}}{}^{-}$ that shows formal charge is as follows:

Explanation of Solution

The given compound is made up of oxygen, 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.

$\text{O}$ 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.

The rules applied to obtain the Lewis structure of ${\text{NO}}_{\text{2}}{}^{-}$ are as follows:

1. Write the skeleton structure.

In the skeleton structure, two bonds are formed.

2. Calculate the total number of valence electrons.

The valence electron of oxygen is calculated as follows:

    $\begin{array}{c}2\left(\text{O}\right)=2\times 6\\ =12\end{array}$

The valence electron of nitrogen is calculated as follows:

    $\begin{array}{c}1\left(\text{N}\right)=\text{1}\times 5\\ =5\end{array}$

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:

  $\begin{array}{c}\text{Total valence electrons}=12+5+1\\ =18\end{array}$

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

The skeleton structure has two bonds. Therefore four electrons are used in bonds.

The remaining electrons are calculated as follows:

  $\begin{array}{c}\text{Remaining electrons}=18-4\\ =14\end{array}$

4 To obey the octet rule, the oxygen atom needs six electrons and nitrogen atom needs 6 electrons.

5. Satisfy the octet rule.

There are ten 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 on nitrogen and oxygen atom to satisfy octet.

The Lewis structure of ${\text{NO}}_{\text{2}}{}^{-}$ 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 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).

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

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

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

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

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

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

    $\begin{array}{c}\text{Formal charge}\left({\text{O}}_2\right)=\left(6\right)-\left(4\right)-\frac{1}{2}\left(4\right)\\ =0\end{array}$

In this Lewis structure, nitrogen has formal charge 0. First oxygen atom has formal charge $-1$ and second oxygen atom has formal charge 0.

The Lewis structure made from ${\text{NO}}_{\text{2}}{}^{-}$ that shows formal charge is as follows:

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a)

Answer to Problem 9.57QE

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

Explanation of Solution

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

$\text{C}$ 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.

$\text{S}$ 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.

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

1. Write the skeleton structure.

There is one sulfur atom, one oxygen atom and carbon is place as central atom. Therefore, two bonds are formed between carbon, sulfur and oxygen atom.

2. Calculate the total number of valence electrons.

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

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 total number of valence electrons is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=6+6+4\\ =14\end{array}$

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

The skeleton structure has two bonds. Therefore four electrons are used in bonds. The remaining electrons are calculated as follows:

  $\begin{array}{c}\text{Remaining electrons}=14-2\\ =12\end{array}$

4 To obey the octet rule, the oxygen atom needs six electrons, carbon atom needs 4electrons and sulfur atom needs 6 electrons.

5. Satisfy the octet rule.

There are 12 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 on sulfur and oxygen atom to satisfy octet.

The Lewis structure of $\text{OCS}$ 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 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).

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

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 sulfur 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).

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

In this Lewis structure, nitrogen has formal charge 0. Oxygen atom has formal charge0 and sulfur atom has formal charge 0.

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

(c)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a)

Answer to Problem 9.57QE

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

Explanation of Solution

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

$\text{S}$ is the symbol for Sulphur. 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.

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

1. Write the skeleton structure.

There are one sulfur atom and three oxygen atoms. Therefore, three bonds are formed between sulfur and each oxygen atoms.

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 sulfur is calculated as follows:

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

The total number of valence electrons is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=18+6\\ =24\end{array}$

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:

  $\begin{array}{c}\text{Remaining electrons}=24-6\\ =18\end{array}$

4 To obey the octet rule, the oxygen atom needs six electrons, carbon atom needs 4el and sulfur atom needs 6 electrons.

5. Satisfy the octet rule.

There are 18 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 on oxygen atom to satisfy octet.

The Lewis structure of ${\text{SO}}_3$ 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 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).

    $\begin{array}{c}\text{Formal charge}\left(\text{S}\right)=\left(6\right)-\left(0\right)-\frac{1}{2}\left(12\right)\\ =0\end{array}$

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).

    $\begin{array}{c}\text{Formal charge}\left({\text{O}}_2\right)=\left(6\right)-\left(4\right)-\frac{1}{2}\left(4\right)\\ =0\end{array}$

The formal charge on the 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).

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

In this Lewis structure, sulfur has formal charge 0. All oxygen atoms have formal charge 0.

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