Chapter 20, Problem 76A

(a)

Interpretation Introduction

Interpretation: The oxidation number of nitrogen in the species is to be calculated.

Concept Introduction: The charge that an atom seems to have when forming ionic connections with other heteroatoms is known as an atom's oxidation number.

Answer to Problem 76A

The oxidation number of $\text{N}$ is $5$ .

Explanation of Solution

The given species is ${\text{HNO}}_{\text{3}}$ .

The oxidation number of hydrogen is $\text{1}$ .

The oxidation number of oxygen is $\text{2}$ .

Since the species is neutral, the sum of the oxidation number of each element is equal to zero.

Let $x$ be the oxidation number of nitrogen.

The oxidation number of $\text{N}$ in ${\text{HNO}}_{\text{3}}$ is given as follows:

  $\begin{array}{c}{\text{HNO}}_3=1+x+3\left(-2\right)\\ 0=x-5\\ x=5\end{array}$

(b)

Interpretation Introduction

Interpretation: The oxidation number of nitrogen in the species is to be calculated.

Concept Introduction: The charge that an atom seems to have when forming ionic connections with other heteroatoms is known as an atom's oxidation number.

Answer to Problem 76A

The oxidation number of $\text{N}$ is $-3$ .

Explanation of Solution

The given species is ${\text{NH}}_{\text{3}}$ .

The oxidation number of hydrogen is $\text{1}$ .

Since the species is neutral, the sum of the oxidation number of each element is equal to zero.

Let $x$ be the oxidation number of nitrogen.

The oxidation number of $\text{N}$ in ${\text{NH}}_{\text{3}}$ is given as follows:

  $\begin{array}{c}{\text{NH}}_{\text{3}}=x+3\left(1\right)\\ 0=x+3\\ x=-3\end{array}$

(c)

Interpretation Introduction

Interpretation: The oxidation number of nitrogen in the species is to be calculated.

Concept Introduction: The charge that an atom seems to have when forming ionic connections with other heteroatoms is known as an atom's oxidation number.

Answer to Problem 76A

The oxidation number of $\text{N}$ is $3$ .

Explanation of Solution

The given species is ${\text{N}}_2{\text{O}}_{\text{3}}$ .

The oxidation number of oxygen is $\text{2}$ .

Since the species is neutral, the sum of the oxidation number of each element is equal to zero.

Let $x$ be the oxidation number of nitrogen.

The oxidation number of $\text{N}$ in ${\text{N}}_2{\text{O}}_{\text{3}}$ is given as follows:

  $\begin{array}{c}{\text{N}}_2{\text{O}}_{\text{3}}=2x+3\left(-2\right)\\ 0=2x-6\\ 2x=6\\ x=3\end{array}$

(d)

Interpretation Introduction

Interpretation: The oxidation number of nitrogen in the species is to be calculated.

Concept Introduction: The charge that an atom seems to have when forming ionic connections with other heteroatoms is known as an atom's oxidation number.

Answer to Problem 76A

The oxidation number of $\text{N}$ is $3$ .

Explanation of Solution

The given species is ${\text{NO}}_2^{-}$ .

The oxidation number of oxygen is $\text{2}$ .

Since the species has a charge, the sum of the oxidation number of each element is equal to the charge

Let $x$ be the oxidation number of nitrogen.

The oxidation number of $\text{N}$ in ${\text{NO}}_2^{-}$ is given as follows:

  $\begin{array}{c}{\text{NO}}_2^{-}=x+2\left(-2\right)\\ -1=x-4\\ x=-1+4\\ x=3\end{array}$

(e)

Interpretation Introduction

Interpretation: The oxidation number of nitrogen in the species is to be calculated.

Concept Introduction: The charge that an atom seems to have when forming ionic connections with other heteroatoms is known as an atom's oxidation number.

Answer to Problem 76A

The oxidation number of $\text{N}$ is $1$ .

Explanation of Solution

The given species is ${\text{N}}_2\text{O}$ .

The oxidation number of oxygen is $\text{2}$ .

Since the species is neutral, the sum of the oxidation number of each element is equal to zero.

Let $x$ be the oxidation number of nitrogen.

The oxidation number of $\text{N}$ in ${\text{N}}_2\text{O}$ is given as follows:

  $\begin{array}{c}{\text{N}}_2\text{O}=2x+\left(-2\right)\\ 0=2x-2\\ x=1\end{array}$

(f)

Interpretation Introduction

Interpretation: The oxidation number of nitrogen in the species is to be calculated.

Concept Introduction: The charge that an atom seems to have when forming ionic connections with other heteroatoms is known as an atom's oxidation number.

Answer to Problem 76A

The oxidation number of $\text{N}$ is $-3$ .

Explanation of Solution

The given species is ${\text{NH}}_4\text{Cl}$ .

The oxidation number of hydrogen is $\text{1}$ .

The oxidation number of chlorine is $-1$ .

Since the species is neutral, the sum of the oxidation number of each element is equal to zero.

Let $x$ be the oxidation number of nitrogen.

The oxidation number of $\text{N}$ in ${\text{NH}}_4\text{Cl}$ is given as follows:

  $\begin{array}{c}{\text{NH}}_4\text{Cl}=x+4\left(1\right)+\left(-1\right)\\ 0=x+3\\ x=-3\end{array}$

(g)

Interpretation Introduction

Interpretation: The oxidation number of nitrogen in the species is to be calculated.

Concept Introduction: The charge that an atom seems to have when forming ionic connections with other heteroatoms is known as an atom's oxidation number.

Answer to Problem 76A

The oxidation number of $\text{N}$ is $2$ .

Explanation of Solution

The given species is $\text{NO}$ .

The oxidation number of oxygen is $\text{2}$ .

Since the species is neutral, the sum of the oxidation number of each element is equal to zero.

Let $x$ be the oxidation number of nitrogen.

The oxidation number of $\text{N}$ in $\text{NO}$ is given as follows:

  $\begin{array}{c}\text{NO}=x+\left(-2\right)\\ 0=x-2\\ x=2\end{array}$

(h)

Interpretation Introduction

Interpretation: The oxidation number of nitrogen in the species is to be calculated.

Concept Introduction: The charge that an atom seems to have when forming ionic connections with other heteroatoms is known as an atom's oxidation number.

Answer to Problem 76A

The oxidation number of $\text{N}$ is $4$ .

Explanation of Solution

The given species is ${\text{NO}}_2$ .

The oxidation number of oxygen is $\text{2}$ .

Since the species is neutral, the sum of the oxidation number of each element is equal to zero.

Let $x$ be the oxidation number of nitrogen.

The oxidation number of $\text{N}$ in ${\text{NO}}_2$ is given as follows:

  $\begin{array}{c}{\text{NO}}_2=x+2\left(-2\right)\\ 0=x-4\\ x=4\end{array}$