Chapter 2, Problem 2B.23E

(a)

Interpretation Introduction

Interpretation:

The formal charge on each atom has to be determined and structure of lower energy has to be identified.

Concept Introduction:

The formal charge on each atom in the Lewis structure can be calculated from the equation written as follows:

    $\text{Formal charge}=\left[V-L-\frac{1}{2}B\right]$

Here,

$V$ denotes valence electrons in free atom.

$L$ denotes electrons present as lone pairs.

$B$ denotes electrons present as bond pairs.

Lowest energy structure is the one that has zero or nearly zero formal charge.

Explanation of Solution

The formal charge on each atom in the Lewis structure can be calculated from the equation as follows:

  $\text{Formal charge}=\left[V-L-\frac{1}{2}B\right]$        (1)

For first structure:

Substitute 6 for $V$, 4 for $L$ and 4 for $B$ in equation (1) to calculate formal charge on doubly bonded $\text{O}$.

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

Substitute 6 for $V$, 4 for $L$ and 4 for $B$ in equation (1) to calculate formal charge on $\text{O}$ attached to $\text{H}$.

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

Substitute 7for $V$, 2 for $L$ and 10 for $B$ in equation (1).

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

Substitute 1 for $V$, 0 for $L$ and 2 for $B$ in equation (1).

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

For first structure:

Similarly for other molecule substitute 6 for $V$, 6 for $L$ and 2 for $B$ in equation (1) to $\text{O}$.

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

Substitute 6 for $V$, 4 for $L$ and 4 for $B$ in equation (1) to calculate formal charge on $\text{O}$ attached to $\text{H}$.

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

Substitute 7 for $V$, 2 for $L$ and 4 for $B$ in equation (1).

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

Substitute 1 for $V$, 0 for $L$ and 2 for $B$ in equation (1).

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

Hence formal charge in the right structure is illustrated below:

Since the former hasformal charges close to zero for maximum number of atoms I is lower in energy than II.

(b)

Interpretation Introduction

Interpretation:

The formal charge on each atom has to be determined and structure of lower energy has to be identified.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The formal charge on each atom in the Lewis structure can be calculated from the equation as follows:

    $\text{Formal charge}=\left[V-L-\frac{1}{2}B\right]$        (1)

Substitute 6 for $V$, 4 for $L$ and 4 for $B$ in equation (1) to calculate formal charge on doubly bonded $\text{O}$.

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

Substitute 6 for $V$, 4 for $L$ and 4 for $B$ in equation (1) to calculate formal charge on $\text{S}$.

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

Substitute 4for $V$, 0 for $L$ and 8 for $B$ in equation (1).

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

Similarly for other Lewis structure, substitute 6 for $V$, 6 for $L$ and 2 for $B$ in equation (1) to $\text{O}$.

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

Substitute 6 for $V$, 4 for $L$ and 4 for $B$ in equation (1) to calculate formal charge on $\text{S}$.

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

Substitute 4 for $V$, 0 for $L$ and 8 for $B$ in equation (1).

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

Hence formal charge in the two structures is illustrated below:

Since the former has all formal charges equal to zero I is lower in energy than II hence I represents structure of lower energy.

(b)

Interpretation Introduction

Interpretation:

The formal charge on each atom has to be determined and structure of lower energy has to be identified.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The formal charge on each atom in the Lewis structure can be calculated from the equation as follows:

    $\text{Formal charge}=\left[V-L-\frac{1}{2}B\right]$        (1)

Substitute 1 for $V$, 0 for $L$ and 2 for $B$ in equation (1).

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

Substitute 4 for $V$, 0 for $L$ and 8 for $B$ in equation (1).

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

Substitute 5for $V$, 2 for $L$ and 6  for $B$ in equation (1).

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

Similarly for other Lewis structure, substitute 1 for $V$, 0 for $L$ and 2 for $B$ in equation (1).

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

Substitute 4 for $V$, 2 for $L$ and 6 for $B$ in equation (1).

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

Substitute 5 for $V$, 2 for $L$ and 4  for $B$ in equation (1).

    $\begin{array}{c}\text{Formal charge}\left(\text{N}\right)=5-2-\frac{1}{2}\left(4\right)\\ =+1\end{array}$

Hence formal charge in the two structures is illustrated below:

Since the former has all formal charges equal to zero I is lower in energy than II hence I represents structure of lower energy.