Chapter 18, Problem 10A

(a)

Interpretation Introduction

Interpretation: The oxidation state of each atom in $\text{HBr}$ needs to be calculated.

Concept introduction: Oxidation state is the charge assigned to an atom in a compound. The oxidation-reduction reaction involves the gain and loss of electrons from atoms.

Answer to Problem 10A

The oxidation state of H is +1 and the oxidation state of Br is -1.

Explanation of Solution

In $\text{HBr}$ , the electronegativity of bromine is higher than hydrogen. The stable charge on bromine is -1. Thus, oxidation state of bromine is -1. $\text{HBr}$ is a neutral species and the sum of the oxidation state is equal to 0. Therefore, oxidation state of hydrogen is +1.

(b)

Interpretation Introduction

Interpretation: The oxidation state of each atom of $\text{HOBr}$ needs to be calculated.

Concept introduction: Oxidation state is the charge assigned to an atom in a compound. The oxidation-reduction reaction involves the gain and loss of electrons from atoms.

Answer to Problem 10A

In $\text{HOBr}$ , the oxidation state of bromine is +1 and hydrogen is +1 and oxygen is -2.

Explanation of Solution

In $\text{HOBr}$ , the electronegativity of oxygen is higher than bromine and hydrogen. The stable charge on the oxygen is -2. Thus, the oxidation state of oxygen is -2. Also, the oxidation state of hydrogen is +1. Here, $\text{HOBr}$ is a neutral species thus, the sum of the oxidation state is equal to 0. Calculation of the oxidation state of bromine is as follows:

  $\begin{array}{c}1+1\left(-2\right)+x=0\\ x=+1\end{array}$

Therefore oxidation state of bromine is +1.

(c)

Interpretation Introduction

Interpretation: The oxidation state of Br in ${\text{Br}}_{\text{2}}$ needs to be calculated.

Concept introduction: Oxidation state is the charge assigned to an atom in a compound. The oxidation-reduction reaction involves the gain and loss of electrons from atoms.

Answer to Problem 10A

The oxidation state of Br in ${\text{Br}}_{\text{2}}$ is 0.

Explanation of Solution

Given species is ${\text{Br}}_{\text{2}}$ . It is an elemental state of Br, the oxidation state of an element is 0. Thus, the oxidation state of oxygen is 0.

(d)

Interpretation Introduction

Interpretation: The oxidation state of each element in ${\text{HBrO}}_{\text{4}}$ needs to be calculated.

Concept introduction: Oxidation state is the charge assigned to an atom in a compound. The oxidation-reduction reaction involves the gain and loss of electrons from atoms.

Answer to Problem 10A

The oxidation state of bromine is +7, hydrogen is +1 and oxygen is -2.

Explanation of Solution

In ${\text{HBrO}}_{\text{4}}$ , the oxidation state of oxygen is higher than hydrogen and bromine. ${\text{HBrO}}_{\text{4}}$ is a neutral species. Thus, the sum of the oxidation state is equal to 0. As an anion charge of oxygen is -2. Calculation of the oxidation state of bromine is as follows:

  $\begin{array}{c}1+x+4\left(-2\right)=0\\ x=7\end{array}$

Therefore, oxidation state of bromine is +7.