Chapter 7, Problem 70A

(a)

Interpretation Introduction

Interpretation: The formula of ionic compound formed from the combination of ${\text{Fe}}^{3+}$ and ${\text{O}}^{2-}$ needs to be determined.

Concept Introduction: Ionic compounds are formed by the chemical combination of a metal and a non-metal. Here, metals tend to be electron-rich; thus they lose electron/s to form a cation. Similarly, non-metals tend to be electron deficient; thus, they gain electron/s to form an anion. These cations and anions can combine in a fixed ratio to form ionic compounds.

Explanation of Solution

The given ions are ${\text{Fe}}^{3+}$ and ${\text{O}}^{2-}$ . Here, Fe is a cation (positive ion) with a +3 charge and O is an anion (negative charge) with a -2 charge.

Since ionic compounds are electrically neutral (charge zero), thus, Fe and O ions combine in such a way that the total charge of the compound is 0.

Here, Fe and O ions combine in 2:3 ratio to form ${\text{Fe}}_2{\text{O}}_3$ .

(b)

Interpretation Introduction

Interpretation: The formula of ionic compound formed from the combination of ${\text{Pb}}^{4+}$ and ${\text{O}}^{2-}$ needs to be determined.

Concept Introduction: Ionic compounds are formed by the chemical combination of a metal and a non-metal. Here, metals tend to be electron-rich thus they lose electron/s to form a cation. Similarly, non-metals tend to be electron deficient thus, they gain electron/s to form an anion. These cations and anions can combine in a fixed ratio to form ionic compounds.

Explanation of Solution

The given ions are ${\text{Pb}}^{4+}$ and ${\text{O}}^{2-}$ . Here, Pb is a cation (positive ion) with a +4 charge and O is an anion (negative charge) with a -2 charge.

Since ionic compounds are electrically neutral (charge zero) thus, Pb and O ions combine in such a way that the total charge of the compound is 0.

Here, Pb and O ions combine in a 2:4 ratio to form ${\text{Pb}}_2{\text{O}}_4$ .

(c)

Interpretation Introduction

Interpretation: The formula of ionic compound formed from a combination of ${\text{Li}}^{+}$ and ${\text{O}}^{2-}$ needs to be determined.

Concept Introduction: Ionic compounds are formed by the chemical combination of a metal and a non-metal. Here, metals tend to be electron-rich thus they lose electron/s to form a cation. Similarly, non-metals tend to be electron deficient thus, they gain electron/s to form an anion. These cations and anions can combine in a fixed ratio to form ionic compounds.

Explanation of Solution

The given ions are ${\text{Li}}^{+}$ and ${\text{O}}^{2-}$ . Here, Li is a cation (positive ion) with a +1 charge and O is an anion (negative charge) with a -2 charge.

Since ionic compounds are electrically neutral (charge zero) thus, Li and O ions combine in such a way that the total charge of the compound is 0.

Here, Li and O ions combine in a 2:1 ratio to form ${\text{Li}}_2\text{O}$ .

(d)

Interpretation Introduction

Interpretation: The formula of ionic compound formed from the combination of ${\text{Mg}}^{2+}$ and ${\text{O}}^{2-}$ needs to be determined.

Concept Introduction: Ionic compounds are formed by the chemical combination of a metal and a non-metal. Here, metals tend to be electron-rich; thus, they lose electron/s to form a cation. Similarly, non-metals tend to be electron deficient thus, they gain electron/s to form an anion. These cations and anions can combine in a fixed ratio to form ionic compounds.

Explanation of Solution

The given ions are ${\text{Mg}}^{2+}$ and ${\text{O}}^{2-}$ . Here, Mg is a cation (positive ion) with a +2 charge and O is an anion (negative charge) with a -2 charge.

Since ionic compounds are electrically neutral (charge zero), thus, Mg and O ions combine in such a way that the total charge of the compound is 0.

Here, Mg and O ions combine in a 1:1 ratio to form $\text{MgO}$ .