Chapter 6, Problem 84E

(a)

Interpretation Introduction

Interpretation: The oxidation state of given atom in the given molecule has to be calculated.

Concept introduction: The oxidation state is the distinction between the numbers of electrons connected by an atom in a composite as compared with the number of electrons in an atom of the element. The oxidation state is also called oxidation number

Rule 1: The oxidation numeral of an element in its open (uncombined) state is zero.

Rule 2: The oxidation numeral of a monatomic (one-atom) ion is the similar as the indict on the ion.

Answer to Problem 84E

$\begin{array}{l}{\text{SrCr}}_{\text{2}}{\text{O}}_{\text{7}}\text{:Composed}\text{of}{\text{Sr}}^{\text{2+}}\text{and}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}\text{ions}\text{.}\\ \text{Sr}\text{is}\text{+2,}\text{O}\text{is}\text{-2,}\text{Cr}\text{is}\text{+6}\end{array}$

Explanation of Solution

Case 1: To find the oxidation state of $\text{Sr}\text{in}{\text{SrCr}}_{\text{2}}{\text{O}}_{\text{7}}$

Here x is oxidation state of $\text{Sr}$

$\begin{array}{l}\text{x}\text{+2(+6)+7(-2)=0}\\ \text{x+12-14=0}\\ \text{x-2=0}\\ \text{x=+2}\end{array}$

Case 2: To find the oxidation state of $\text{Cr}\text{in}{\text{SrCr}}_{\text{2}}{\text{O}}_{\text{7}}$

Here x is oxidation state of $\text{Cr}$

$\begin{array}{l}\text{+2}\text{+2x+7(-2)=0}\\ \text{2x+12=0}\\ \text{2x=12}\\ \text{x=+6}\end{array}$

Case 3: To find the oxidation state of $\text{O}\text{in}{\text{SrCr}}_{\text{2}}{\text{O}}_{\text{7}}$

Here x is oxidation state of $\text{O}$

$\begin{array}{l}\text{2+2(+6)+7x=0}\\ \text{2+12+7x=0}\\ \text{14+7x=0}\\ \text{7x=-14}\\ \text{x=-2}\end{array}$

The oxidation state of $\text{Cr}\text{in}{\text{SrCr}}_{\text{2}}{\text{O}}_{\text{7}}is+2,$ then the oxidation state of $\text{Cr}\text{in}{\text{SrCr}}_{\text{2}}{\text{O}}_{\text{7}}\text{is}\text{+6}\text{.}$

The oxidation state of $\text{O}\text{in}{\text{SrCr}}_{\text{2}}{\text{O}}_{\text{7}}\text{is}\text{-2}\text{.}$

(b)

Interpretation Introduction

Interpretation: The oxidation state of given atom in the given molecule has to be calculated.

Concept introduction: The oxidation state is the distinction between the numbers of electrons connected by an atom in a composite as compared with the number of electrons in an atom of the element. The oxidation state is also called oxidation number

Rule 1: The oxidation numeral of an element in its open (uncombined) state is zero.

Rule 2: The oxidation numeral of a monatomic (one-atom) ion is the similar as the indict on the ion.

Answer to Problem 84E

$\text{Cu}\text{is }\text{+2,Cl}\text{is}\text{-1}$

Explanation of Solution

Case 1: To find the oxidation state of $\text{Cu}\text{in}{\text{CuCl}}_{\text{2}}$

Here x is oxidation state of $\text{Cu}$

$\begin{array}{l}\text{x+2(-1)=0}\\ \text{x-2=0}\\ \text{x=+2}\end{array}$

Case 2: To find the oxidation state of $\text{Cl}\text{in}{\text{CuCl}}_{\text{2}}$

Here x is oxidation state of $\text{Cl}$

$\begin{array}{l}\text{+2+2x}\text{=0}\\ \text{2x=-2}\\ \text{x=-1}\end{array}$

The oxidation state of $\text{Cu}\text{in}{\text{CuCl}}_{\text{2}}\text{is}\text{+2}$, and then the oxidation state of $\text{Cl}\text{in}{\text{CuCl}}_{\text{2}}$ is -1.

(c)

Interpretation Introduction

Interpretation: The oxidation state of given atom in the given molecule has to be calculated.

Concept introduction: The oxidation state is the distinction between the numbers of electrons connected by an atom in a composite as compared with the number of electrons in an atom of the element. The oxidation state is also called oxidation number

Rule 1: The oxidation numeral of an element in its open (uncombined) state is zero.

Rule 2: The oxidation numeral of a monatomic (one-atom) ion is the similar as the indict on the ion.

Answer to Problem 84E

$\text{O}\text{is }\text{0}$

Explanation of Solution

Case 1: To find the oxidation state of ${\text{O}}_{\text{2}}$

The oxidation state of ${\text{O}}_{\text{2}}$ is Zero due to the Oxygen molecule is homo diatomic molecule.

(d)

Interpretation Introduction

Interpretation: The oxidation state of given atom in the given molecule has to be calculated.

Concept introduction: The oxidation state is the distinction between the numbers of electrons connected by an atom in a composite as compared with the number of electrons in an atom of the element. The oxidation state is also called oxidation number

Rule 1: The oxidation numeral of an element in its open (uncombined) state is zero.

Rule 2: The oxidation numeral of a monatomic (one-atom) ion is the similar as the indict on the ion.

Answer to Problem 84E

$\text{H}\text{is}\text{+1,O}\text{is}\text{-1}$

Explanation of Solution

Case 1: To find the oxidation state of $\text{H}\text{in}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$

Here x is oxidation state of $\text{H}$

$\begin{array}{l}\text{2x+2(-1)=0}\\ \text{2x-2=0}\\ \text{2x=+2}\\ \text{x=+1}\end{array}$

Case 2: To find the oxidation state of $\text{O}\text{in}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$

Here x is oxidation state of $\text{O}$

$\begin{array}{l}\text{+2(+1)+2x}\text{=0}\\ \text{2x=-2}\\ \text{x=-1}\end{array}$

The oxidation state of $\text{O}\text{in}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{is}\text{-1}$, and then the oxidation state of $\text{H}\text{in}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{is}\text{+1}$.

(e)

Interpretation Introduction

Interpretation: The oxidation state of given atom in the given molecule has to be calculated.

Concept introduction: The oxidation state is the distinction between the numbers of electrons connected by an atom in a composite as compared with the number of electrons in an atom of the element. The oxidation state is also called oxidation number

Rule 1: The oxidation numeral of an element in its open (uncombined) state is zero.

Rule 2: The oxidation numeral of a monatomic (one-atom) ion is the similar as the indict on the ion.

Answer to Problem 84E

Answer:

$\text{Mg}\text{is}\text{+2,}\text{O}\text{is}\text{-2,}\text{C}\text{is}\text{+4}$

Explanation of Solution

Explanation:

Case 1: To find the oxidation state of $\text{Mg}\text{in}{\text{MgCO}}_{\text{3}}$

Here x is oxidation state of $\text{Mg}$

$\begin{array}{l}\text{x}\text{+4+3(-2)=0}\\ \text{x+4-6=0}\\ \text{x-2=0}\\ \text{x=+2}\end{array}$

Case 2: To find the oxidation state of $\text{C}\text{in}{\text{MgCO}}_{\text{3}}$

Here x is the oxidation state of $\text{C}$

$\begin{array}{l}\text{+2}\text{+x+3(-2)=0}\\ \text{2+x-6=0}\\ \text{x-4=0}\\ \text{x=+4}\end{array}$

Case 3: To find the oxidation state of $\text{O}\text{in}{\text{MgCO}}_{\text{3}}$

Here x is oxidation state of $\text{O}$

$\begin{array}{l}\text{2+4+3x=0}\\ \text{6+3x=0}\\ \text{3x=-6}\\ \text{x=}\frac{\text{-6}}{\text{3}}\\ \text{x=-2}\end{array}$

The oxidation state of $\text{Mg}\text{in}{\text{MgCO}}_{\text{3}}\text{is}\text{+2}$ then the oxidation state of $\text{C}\text{in}{\text{MgCO}}_{\text{3}}\text{is}\text{+4}$

The oxidation state of $\text{O}\text{in}{\text{MgCO}}_{\text{3}}\text{is}\text{-2}$

(f)

Interpretation Introduction

Interpretation: The oxidation state of given atom in the given molecule has to be calculated.

Concept introduction: The oxidation state is the distinction between the numbers of electrons connected by an atom in a composite as compared with the number of electrons in an atom of the element. The oxidation state is also called oxidation number

Rule 1: The oxidation numeral of an element in its open (uncombined) state is zero.

Rule 2: The oxidation numeral of a monatomic (one-atom) ion is the similar as the indict on the ion.

Answer to Problem 84E

$\text{Ag}\text{is}\text{0}$

Explanation of Solution

Case 1: To find the oxidation state of $\text{Ag}$

The oxidation states of $\text{Ag}$ is zero due to the elements contain 0 oxidation.

(g)

Interpretation Introduction

Interpretation: The oxidation state of given atom in the given molecule has to be calculated.

Concept introduction: The oxidation state is the distinction between the numbers of electrons connected by an atom in a composite as compared with the number of electrons in an atom of the element. The oxidation state is also called oxidation number

Rule 1: The oxidation numeral of an element in its open (uncombined) state is zero.

Rule 2: The oxidation numeral of a monatomic (one-atom) ion is the similar as the indict on the ion.

Answer to Problem 84E

$\text{Pb}\text{is}\text{+2,}\text{O}\text{is}\text{-2,}\text{S}\text{is}\text{+4}$

Explanation of Solution

Case 1: To find the oxidation state of $\text{Pb}\text{in}{\text{PbSO}}_{\text{3}}$

Here x is oxidation state of $\text{Pb}$

$\begin{array}{l}\text{x}\text{+4+3(-2)=0}\\ \text{x+4-6=0}\\ \text{x-2=0}\\ \text{x=+2}\end{array}$

Case 2: To find the oxidation state of $\text{S}\text{in}{\text{PbSO}}_{\text{3}}$

Here x is the oxidation state of $\text{S}$

$\begin{array}{l}\text{+2}\text{+x+3(-2)=0}\\ \text{2+x-6=0}\\ \text{x-4=0}\\ \text{x=+4}\end{array}$

Case 3: To find the oxidation state of $\text{O}\text{in}{\text{PbSO}}_{\text{3}}$

Here x is oxidation state of $\text{O}$

$\begin{array}{l}\text{2+4+3x=0}\\ \text{6+3x=0}\\ \text{3x=-6}\\ \text{x=}\frac{\text{-6}}{\text{3}}\\ \text{x=-2}\end{array}$

The oxidation state of $\text{Pb}\text{in}\text{PbSO4}\text{is}\text{+2}$ then the oxidation state of $\text{S}\text{in}{\text{PbSO}}_{\text{3}}\text{is}\text{+4}$

The oxidation state of $\text{O}\text{in}{\text{PbSO}}_{\text{3}}\text{is}\text{-2}$

(h)

Interpretation Introduction

Interpretation: The oxidation state of given atom in the given molecule has to be calculated.

Concept introduction: The oxidation state is the distinction between the numbers of electrons connected by an atom in a composite as compared with the number of electrons in an atom of the element. The oxidation state is also called oxidation number

Rule 1: The oxidation numeral of an element in its open (uncombined) state is zero.

Rule 2: The oxidation numeral of a monatomic (one-atom) ion is the similar as the indict on the ion.

Answer to Problem 84E

$\text{O}\text{is}\text{-2,}\text{Pb}\text{is}\text{+4}$

Explanation of Solution

Case 1: To find the oxidation state of $\text{Pb}\text{in}{\text{PbO}}_{\text{2}}$

Here x is oxidation state of $\text{Pb}$

$\begin{array}{l}\text{x}\text{+2(-2)=0}\\ \text{x-4=0}\\ \text{x=+4}\end{array}$

Case 2: To find the oxidation state of $\text{O}\text{in}{\text{PbO}}_{\text{2}}$

Here x is the oxidation state of $\text{O}$

$\begin{array}{l}\text{+4}\text{+2x=0}\\ \text{2x=-4}\\ \text{x=}\frac{\text{-4}}{\text{2}}\\ \text{x=-2}\end{array}$

The oxidation state of $\text{Pb}\text{in}{\text{PbO}}_{\text{2}}\text{is}\text{+4}$ and the oxidation state of $\text{O}\text{in}{\text{PbO}}_{\text{2}}\text{is}\text{-2}$

(i)

Interpretation Introduction

Interpretation: The oxidation state of given atom in the given molecule has to be calculated.

Concept introduction: The oxidation state is the distinction between the numbers of electrons connected by an atom in a composite as compared with the number of electrons in an atom of the element. The oxidation state is also called oxidation number

Rule 1: The oxidation numeral of an element in its open (uncombined) state is zero.

Rule 2: The oxidation numeral of a monatomic (one-atom) ion is the similar as the indict on the ion.

Answer to Problem 84E

$\text{Na}\text{is}\text{+1,}\text{O}\text{is}\text{-2,}\text{C}\text{is}\text{+3}$

Explanation of Solution

Case 1: To find the oxidation state of $\text{Na}\text{in}{\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$

Here x is oxidation state of $\text{Na}$

$\begin{array}{l}\text{2x}\text{+2(+3)+4(-2)=0}\\ \text{2x+6-8=0}\\ \text{2x-2=0}\\ \text{x=}\frac{\text{+2}}{\text{2}}\\ \text{x=+1}\end{array}$

Case 2: To find the oxidation state of $\text{C}\text{in}{\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$

Here x is the oxidation state of $\text{C}$

$\begin{array}{l}\text{2(+1)}\text{+2x+4(-2)=0}\\ \text{2+2x-8=0}\\ \text{2x-6=0}\\ \text{2x=+6}\\ \text{x=}\frac{\text{6}}{\text{2}}\\ \text{x=+3}\end{array}$

Case 3: To find the oxidation state of $\text{O}\text{in}{\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$

Here x is oxidation state of $\text{O}$

$\begin{array}{l}\text{2(+1)+2(+3)+4x=0}\\ \text{2+6+4x=0}\\ \text{8+4x=0}\\ \text{4x=-8}\\ \text{x=}\frac{\text{-8}}{\text{4}}\\ \text{x=-2}\end{array}$

The oxidation state of $\text{Na}\text{in}{\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{is}\text{+1}$ then the oxidation state of $\text{C}\text{in}{\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{is}\text{+3}$

The oxidation state of $\text{O}\text{in}{\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{is}\text{-2}$

(j)

Interpretation Introduction

Interpretation: The oxidation state of given atom in the given molecule has to be calculated.

Concept introduction: The oxidation state is the distinction between the numbers of electrons connected by an atom in a composite as compared with the number of electrons in an atom of the element. The oxidation state is also called oxidation number

Rule 1: The oxidation numeral of an element in its open (uncombined) state is zero.

Rule 2: The oxidation numeral of a monatomic (one-atom) ion is the similar as the indict on the ion.

Answer to Problem 84E

$\text{O}\text{is}\text{-2,}\text{C}\text{is}\text{+4}$

Explanation of Solution

Case 1: To find the oxidation state of $\text{C}\text{in}{\text{CO}}_{\text{2}}$

Here x is oxidation state of $\text{C}$

$\begin{array}{l}\text{x}\text{+2(-2)=0}\\ \text{x-4=0}\\ \text{x=+4}\end{array}$

Case 2: To find the oxidation state of $\text{O}\text{in}{\text{CO}}_{\text{2}}$

Here x is the oxidation state of $\text{O}$

$\begin{array}{l}\text{+4}\text{+2x=0}\\ \text{2x=-4}\\ \text{x=}\frac{\text{-4}}{\text{2}}\\ \text{x=-2}\end{array}$

The oxidation state of $\text{C}\text{in}{\text{CO}}_{\text{2}}\text{is}\text{+4}$ and the oxidation state of $\text{O}\text{in}{\text{CO}}_{\text{2}}\text{is}\text{-2}$

(k)

Interpretation Introduction

Interpretation: The oxidation state of given atom in the given molecule has to be calculated.

Concept introduction: The oxidation state is the distinction between the numbers of electrons connected by an atom in a composite as compared with the number of electrons in an atom of the element. The oxidation state is also called oxidation number

Rule 1: The oxidation numeral of an element in its open (uncombined) state is zero.

Rule 2: The oxidation numeral of a monatomic (one-atom) ion is the similar as the indict on the ion.

Answer to Problem 84E

${\text{NH}}_{\text{4}}^{\text{+}}\text{is}\text{+1,}{\text{SO}}_{\text{4}}^{\text{2-}}\text{is}\text{-2,H}\text{is}\text{+1,}\text{N}\text{is}\text{-3,O}\text{is}\text{-2,S}\text{is}\text{+6}$

Explanation of Solution

Ammonium ion has a $+1$ charge ${\text{(NH}}_{\text{4}}^{\text{+}}\text{)}$, and sulfate ion has a $-2$ charge $\left({\text{SO}}_{\text{4}}^{\text{2-}}\right)\text{.}$

Therefore, the oxidation state of cerium must be

Case 1: To find the oxidation state of $\text{H }\text{in}{\text{(NH}}_{\text{4}}^{\text{+}}\text{)}$

Here x is the oxidation state of $\text{H}$

$\begin{array}{l}\text{-3+4x=+1}\\ \text{4x=+1+3}\\ \text{4x=4}\\ \text{x=}\frac{\text{+4}}{\text{4}}\\ \text{x=+1}\end{array}$

Case 2: To find the oxidation state of $\text{N }\text{in}{\text{(NH}}_{\text{4}}^{\text{+}}\text{)}$

Here x is the oxidation state of $\text{N}$

$\begin{array}{l}\text{x+4(+1)=1}\\ \text{x+4=1}\\ \text{x=+1-4}\\ \text{x=-3}\end{array}$

Case 3: To find the oxidation state of $\text{S}\text{in}\left({\text{SO}}_{\text{4}}^{\text{2-}}\right)$

Here x is the oxidation state of $\text{S}$

$\begin{array}{l}\text{x+4(-2)=-2}\\ \text{x-8=-2}\\ \text{x=-2+8}\\ \text{x=+6}\end{array}$

Case 4: To find the oxidation state of $\text{O}\text{in}\left({\text{SO}}_{\text{4}}^{\text{2-}}\right)$

Here x is the oxidation state of $\text{O}$

$\begin{array}{l}\text{6+4x=-2}\\ \text{6-4x=-2}\\ \text{4x=-2-6}\\ \text{4x=-8}\\ \text{x=}\frac{\text{-8}}{\text{4}}\\ \text{x=-2}\end{array}$

Case 5: To find the oxidation state of $\text{Ce}\text{in}{{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}\text{Ce}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$

Here x is the oxidation state of $\text{Ce}$

$\begin{array}{l}\text{two}\text{ammonium}\text{ion}\text{is}\text{ +2,three sulfate ion}\text{is -6, }\text{so}\\ \text{+2+x-6=0}\\ \text{x=-2+6}\\ \text{x=+4}\end{array}$

The oxidation state of $\text{H }\text{in}{\text{(NH}}_{\text{4}}^{\text{+}}\text{)}\text{is }+1$, the oxidation state of $\text{N }\text{in}{\text{(NH}}_{\text{4}}^{\text{+}}\text{)}\text{is}\text{-3}$ then the oxidation state of $\text{S}\text{in}\left({\text{SO}}_{\text{4}}^{\text{2-}}\right)\text{is}\text{+6}$, the oxidation state of $\text{O}\text{in}\left({\text{SO}}_{\text{4}}^{\text{2-}}\right)\text{}\text{}\text{is}\text{-2}$ and the oxidation state of $\text{Ce}\text{in}{{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}\text{Ce}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\text{is}\text{+4}$

(l)

Interpretation Introduction

Interpretation: The oxidation state of given atom in the given molecule has to be calculated.

Concept introduction: The oxidation state is the distinction between the numbers of electrons connected by an atom in a composite as compared with the number of electrons in an atom of the element. The oxidation state is also called oxidation number

Rule 1: The oxidation numeral of an element in its open (uncombined) state is zero.

Rule 2: The oxidation numeral of a monatomic (one-atom) ion is the similar as the indict on the ion.

Answer to Problem 84E

$\text{O}\text{is}\text{-2}\text{,}\text{Cr}\text{is}\text{+3}$

Explanation of Solution

Case 1: To find the oxidation state of $\text{Cr}\text{in}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}$

Here x is oxidation state of $\text{Cr}$

$\begin{array}{l}\text{2x}\text{+3(-2)=0}\\ \text{2x-6=0}\\ \text{2x=+6}\\ \text{x=}\frac{\text{+6}}{\text{2}}\\ \text{x=+3}\end{array}$

Case 2: To find the oxidation state of $\text{O}\text{in}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}$

Here x is the oxidation state of $\text{O}$

$\begin{array}{l}\text{+3x+2(+3)=0}\\ \text{3x+6=0}\\ \text{3x=+6}\\ \text{x=}\frac{\text{6}}{\text{3}}\\ \text{x=-2}\end{array}$

The oxidation state of $\text{Cr}\text{in}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{is}\text{+3}$ then the oxidation state of $\text{O}\text{in}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{is}\text{-2}$