Chapter 20, Problem 41A

(a)

Interpretation Introduction

Interpretation: The oxidation number of the given chemical species, $C{a}^{2+}$ 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 41A

The oxidation number of $C{a}^{2+}$ is $+2$ .

Explanation of Solution

The given chemical species, $C{a}^{2+}$ is a positively charged ion formed by the loss of two electrons present on them. So the oxidation number of $Ca$ in $C{a}^{2+}$ is $+2$ .

(b)

Interpretation Introduction

Interpretation: The oxidation number of the given chemical species, $A{l}_2{S}_3$ 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 41A

The oxidation number of $Al$ is $+3$ while that of $S$ is $-2$ .

Explanation of Solution

The given chemical species, $A{l}_2{S}_3$ is a neutral binary compound. So the oxidation number can be found by crossing the coefficients in between them. So the oxidation number of $S$ becomes $-2$ and that of Al becomes $+3$ . Sulfur is more electronegative than $Al$ which is why it gets a negative number.

(c)

Interpretation Introduction

Interpretation: The oxidation number of the given chemical species, $N{a}_{2}Cr{O}_4$ 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 elements present in group 1 will get a +1 oxidation state while group 2 gets a +2 oxidation number.

Answer to Problem 41A

The oxidation number of $Cr$ in $N{a}_{2}Cr{O}_4$ is $+6$ .

Explanation of Solution

The given chemical species, $N{a}_{2}Cr{O}_4$ is a neutral chemical species. $Na$ , sodium is a group 1 element so it has a +1 oxidation number. The oxidation number of $Cr$ in $N{a}_{2}Cr{O}_4$ can therefore be calculated as,

  $\begin{array}{c}2+x+\left(-2\times 4\right)=0\\ 2+x+\left(-2\times 4\right)=0\\ 2+x-8=0\\ x=+6\end{array}$

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

(d)

Interpretation Introduction

Interpretation: The oxidation number of the given chemical species, $V_2{O}_5$ 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 41A

The oxidation number of $V$ in $V_2{O}_5$ is $+5$ .

Explanation of Solution

The given chemical species, $V_2{O}_5$ is a neutral chemical compound. The oxidation number of V, vanadium in $V_2{O}_5$ can be calculated as,

  $\begin{array}{c}2x+5\left(-2\right)=0\\ 2x-10=0\\ x=+5\end{array}$

So, the oxidation number of $V$ in $V_2{O}_5$ is $+5$ .

(e)

Interpretation Introduction

Interpretation: The oxidation number of the given chemical species, $Mn{O}_4{}^{-}$ 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.

Answer to Problem 41A

The oxidation number of $Mn$ in $Mn{O}_4{}^{-}$ is +7.

Explanation of Solution

The given chemical species, $Mn{O}_4{}^{-}$ is a charged compound. The oxidation number of $Mn$ in MnO4- can be calculated as,

  $\begin{array}{c}x+4\left(-2\right)=-1\\ x-8=-1\\ x=+7\end{array}$

So, the oxidation number is +7.