Chapter 1, Problem 63P

Interpretation Introduction

Interpretation:

The equation that shows the reaction of the given acid with water by considering the Bronsted-Lowry acid-base theory is to be written. All the electron pairs, formal chargers and curved arrows that represent the electron movement in the respective reaction are to be shown.

Concept introduction:

An acid is a chemical substance that readily donates protons and a base is a chemical substance that can easily accept a proton. During an acid-base reaction, the interaction between an acid and a base is taken place because of the transfer of a proton. The stronger the acid, the smaller its $\text{pKa}$ value

Answer to Problem 63P

Solution:

a)

The formal charge on the the oxygen atom is $+1$

b)

The formal charge on both the the oxygen atom and the nitrogen atom is $+1$

c)

The formal charge on the oxygen atom is $+1$

Explanation of Solution

a) The reaction of an acid with water.

In the respective reaction, the water acts as a base. The unshared electron pair of oxygen atom present in water is used to remove the proton from the acid. Water, after accepting the proton, is converted to its conjugate acid, that is, hydronium ion and the acid is converted to its conjugate base.

The curved arrows showing the electron movement is given below:

The formula that is used to calculate the electron count on the conjugate base is as follows:

$\text{Electron}\text{count}\text{=}\frac{\text{1}}{\text{2}}\text{(Number}\text{of}\text{shared}\text{electrons)}\text{+}\left(\text{Number}\text{of}\text{unshared}\text{electrons}\right)$

Substitute $6$ for the number of shared electrons and $2$ for number of unshared electrons in the above expression

$\begin{array}{c}\text{Electron}\text{count}=\frac{1}{2}(6)+(2)\\ =5\end{array}$

The formula that is used to calculate the formal charge on the oxygen atom is as follows:

$\text{Formal}\text{charge}=\text{number}\text{of}\text{valence}\text{electrons}-\text{electron}\text{count}$

Substitute $6$ for a number of valence electrons in boron atom and $5$ for electron count in the above expression:

$\begin{array}{c}\text{Formal}\text{charge}=6-5\\ =+1\end{array}$

b) The reaction of an acid with water.

The given acid reacts with water. So water acts as a base. The unshared electron pair of the oxygen atom in water is used to remove the proton from the acid. Water, after accepting the proton is converted to its conjugate acid, that is, hydronium ion and the acid is converted to its conjugate base.

The curved arrows showing the electron movement are shown below:

The formula that is used to calculate the electron count on the nitrogen atom is as follows:

$\text{Electron}\text{count}\text{=}\frac{\text{1}}{\text{2}}\text{(Number}\text{of}\text{shared}\text{electrons)}\text{+}\left(\text{Number}\text{of}\text{unshared}\text{electrons}\right)$

Substitute $8$ for the number of shared electrons and $0$ for number of unshared electrons in the above expression

$\begin{array}{c}\text{Electron}\text{count}=\frac{1}{2}(8)+(0)\\ =4\end{array}$

The formula that is used to calculate the formal charge on the nitrogen atom is as follows:

$\text{Formal}\text{charge}=\text{number}\text{of}\text{valence}\text{electrons}-\text{electron}\text{count}$

Substitute $5$ for a number of valence electrons in boron atom and $4$ for electron count in the above expression:

$\begin{array}{c}\text{Formal}\text{charge}=5-4\\ =+1\end{array}$

For calculating formal charge on the oxygen atom, recall the electron count formula:

$\text{Electron}\text{count}\text{=}\frac{\text{1}}{\text{2}}\text{(Number}\text{of}\text{shared}\text{electrons)}\text{+}\left(\text{Number}\text{of}\text{unshared}\text{electrons}\right)$

Substitute $6$ for the number of shared electrons and $2$ for number of unshared electrons in the above expression

$\begin{array}{c}\text{Electron}\text{count}=\frac{1}{2}(6)+(2)\\ =5\end{array}$

Recall the formula for formal charge:

$\text{Formal}\text{charge}=\text{number}\text{of}\text{valence}\text{electrons}-\text{electron}\text{count}$

Substitute $6$ for a number of valence electrons in boron atom and $5$ for electron count in the above expression:

$\begin{array}{c}\text{Formal}\text{charge}=6-5\\ =+1\end{array}$

c) The given acid reacts with water.

So water acts as a base. The unshared electron pair of the oxygen atom in water is used to remove the proton from the acid. Water, after accepting the proton, is converted to its conjugate acid, that is, hydronium ion and the acid is converted to its conjugate base.

The curved arrows showing the electron movement are shown below:

The formula that is used to calculate the electron count on the oxygen is as follows:

$\text{Electron}\text{count}\text{=}\frac{\text{1}}{\text{2}}\text{(Number}\text{of}\text{shared}\text{electrons)}\text{+}\left(\text{Number}\text{of}\text{unshared}\text{electrons}\right)$

Substitute $6$ for the number of shared electrons and $2$ for number of unshared electrons in the above expression

$\begin{array}{c}\text{Electron}\text{count}=\frac{1}{2}(6)+(2)\\ =5\end{array}$

The formula that is used to calculate the formal charge on the oxygen atom is as follows:

$\text{Formal}\text{charge}=\text{number}\text{of}\text{valence}\text{electrons}-\text{electron}\text{count}$

Substitute $6$ for a number of valence electrons in boron atom and $5$ for electron count in the above expression:

$\begin{array}{c}\text{Formal}\text{charge}=6-5\\ =+1\end{array}$