Chapter 20, Problem 42A

(a)

Interpretation Introduction

Interpretation: The oxidation number of the given chemical species, $O{H}^{-}$ is to be determined.

Concept Introduction: Zero, positive, and negative numbers that are assigned to reactants species to track the electron flow is oxidation number. When electrons are lost (oxidation) from a species positive numbers are given while if electrons are gained(reduction) negative numbers are given. And the reactants that don't involve any change in electron count are assigned with the number zero. Certain rules must also be followed before assigning these numbers.

• The oxidation number of oxygen is -2.
• For a neutral compound, the total sum of the oxidation number is zero.
• For ionic compounds with a charge, the total sum of oxidation number is equal to its charge.

Answer to Problem 42A

The oxidation number of $H$ in $O{H}^{-}$ is $+1$ .

Explanation of Solution

The given chemical species, $O{H}^{-}$ is a negative polyatomic ion. So, the oxidation number of $H$ in $O{H}^{-}$ can be calculated as,

(b)

Interpretation Introduction

Interpretation: The oxidation number of the given chemical species, $P{O}_4{}^{3-}$ is to be determined.

Concept Introduction: Zero, positive, and negative numbers that are assigned to reactants species to track the electron flow is oxidation number. When electrons are lost (oxidation) from a species, positive numbers are given while if electrons are gained(reduction), negative numbers are given. And the reactants that don't involve any change in electron count are assigned with the number zero. Certain rules must also be followed before assigning these numbers.

• The oxidation number of oxygen is -2.
• For a neutral compound, the total sum of the oxidation number is zero.
• For ionic compounds with a charge, the total sum of oxidation number is equal to its charge.

Answer to Problem 42A

The oxidation number of $P$ in $P{O}_4{}^{3-}$ is $+5$ .

Explanation of Solution

The given chemical species, $P{O}_4{}^{3-}$ is a polyatomic charged ion. So, the oxidation number of phosphorus, $P$ in $P{O}_4{}^{3-}$ can be calculated as,

  $\begin{array}{c}x+4\left(-2\right)=-3\\ x-8=-3\\ x=+5\end{array}$

So the oxidation number of $P$ is $+5$ .

(c)

Interpretation Introduction

Interpretation: The oxidation number of the given chemical species, $I{O}_3{}^{-}$ is to be determined.

Concept Introduction: Zero, positive, and negative numbers that are assigned to reactants species to track the electron flow is oxidation number. When electrons are lost (oxidation) from a species, positive numbers are given while if electrons are gained(reduction), negative numbers are given. And the reactants that don't involve any change in electron count are assigned with the number zero. Certain rules must also be followed before assigning these numbers.

• The oxidation number of oxygen is -2.
• For a neutral compound, the total sum of the oxidation number is zero.
• For ionic compounds with a charge, the total sum of oxidation number is equal to its charge.

Answer to Problem 42A

The oxidation number of $I$ in $I{O}_3{}^{-}$ is +5.

Explanation of Solution

The given chemical species, $I{O}_3{}^{-}$ is a polyatomic ion. So the oxidation number of $I$ in $I{O}_3{}^{-}$ can be calculated as,

  $\begin{array}{c}x+3\left(-2\right)=-1\\ x-6=-1\\ x=+5\end{array}$

So the oxidation number of $I$ is $+5$ .

(d)

Interpretation Introduction

Interpretation: The oxidation number of the given chemical species, $H_{2}P{O}_4{}^{-}$ is to be determined.

Concept Introduction: Zero, positive, and negative numbers that are assigned to reactants species to track the electron flow is oxidation number. When electrons are lost (oxidation) from a species, positive numbers are given while if electrons are gained(reduction), negative numbers are given. And the reactants that don't involve any change in electron count are assigned with the number zero. Certain rules must also be followed before assigning these numbers.

• The oxidation number of oxygen is -2.
• For a neutral compound, the total sum of the oxidation number is zero.
• For ionic compounds with a charge, the total sum of oxidation number is equal to its charge.
• The oxidation number of H is +1.

Answer to Problem 42A

The oxidation number of $P$ in $H_{2}P{O}_4{}^{-}$ is +5.

Explanation of Solution

The given chemical species, $H_{2}P{O}_4{}^{-}$ is a polyatomic ion. So, the oxidation number $P$ in $H_{2}P{O}_4{}^{-}$ can be calculated as,

  $\begin{array}{c}x+2+4\left(-2\right)=-1\\ x+2-8=-1\\ x=+5\end{array}$

So, the oxidation number of $P$ is $+5$ .

(e)

Interpretation Introduction

Interpretation: The oxidation number of the given chemical species, $HS{O}_4{}^{-}$ is to be determined.

Concept Introduction: Zero, positive, and negative numbers that are assigned to reactants species to track the electron flow is oxidation number. When electrons are lost (oxidation) from a species, positive numbers are given while if electrons are gained(reduction), negative numbers are given. And the reactants that don't involve any change in electron count are assigned with the number zero. Certain rules must also be followed before assigning these numbers.

• The oxidation number of oxygen is -2.
• For a neutral compound, the total sum of the oxidation number is zero.
• For ionic compounds with a charge, the total sum of oxidation number is equal to its charge.
• The oxidation number of H is +1.

Answer to Problem 42A

The oxidation number of $S$ in $HS{O}_4{}^{-}$ is +6.

Explanation of Solution

The given chemical species, $HS{O}_4{}^{-}$ is a polyatomic ion. So the oxidation number $S$ of $HS{O}_4{}^{-}$ can be calculated as,

  $\begin{array}{c}x+1+4\left(-2\right)=-1\\ x-7=-1\\ x=+6\end{array}$

So, the oxidation number of $S$ is $+6$ .