Chapter 7.2, Problem 15LC

(a)

Interpretation Introduction

Interpretation:The formula for barium chloride needs to be determined.

Concept Introduction:Ionic compound is formed by the chemical combination of a metal and a non-metal. Here, metal tends to be electron-rich, and non-metal tends to be electron deficient. Thus, metals donate electron/s to form a positively charged ion or cation and non-metals accept electron/s to form a negatively charged ion or anion.

Answer to Problem 15LC

  ${\text{BaCl}}_{\text{2}}$

Explanation of Solution

The given ionic compound is barium chloride. Here, barium belongs to group 2 thus, it can give 2 electrons to form ${\text{Ba}}^{2+}$ ion. Chlorine belongs to group 17 with 7 valance electrons. It can accept 1 electron to complete the octet and form ${\text{Cl}}^{-}$ ion. Since the total charge of the ionic compound is zero thus, 1 ${\text{Ba}}^{2+}$ ion combines with 2 ${\text{Cl}}^{-}$ ions. The formula of the ionic compound will be ${\text{BaCl}}_{\text{2}}$ .

(b)

Interpretation Introduction

Interpretation: The formula for magnesium oxide needs to be determined.

Concept Introduction:Ionic compound is formed by the chemical combination of a metal and a non-metal. Here, metal tends to be electron-rich, and non-metal tends to be electron deficient. Thus, metals donate electron/s to form a positively charged ion or cation and non-metals accept electron/s to form a negatively charged ion or anion.

Answer to Problem 15LC

  $\text{MgO}$

Explanation of Solution

The given ionic compound is magnesium oxide. Here, magnesium belongs to group 2 thus, it can give 2 electrons to form ${\text{Mg}}^{2+}$ ion. Oxygen belongs to group 16 with 6 valance electrons. It can accept 2 electrons to complete the octet and form ${\text{O}}^{2-}$ ion. Since the total charge of the ionic compound is zero thus, 1 ${\text{Mg}}^{2+}$ ion combines with 1 ${\text{O}}^{2-}$ ion. The formula of the ionic compound will be $\text{MgO}$ .

(c)

Interpretation Introduction

Interpretation: The formula for lithium oxide needs to be determined.

Concept Introduction:Ionic compound is formed by the chemical combination of a metal and a non-metal. Here, metal tends to be electron-rich, and non-metal tends to be electron deficient. Thus, metals donate electron/s to form a positively charged ion or cation and non-metals accept electron/s to form a negatively charged ion or anion.

Answer to Problem 15LC

  ${\text{Li}}_{\text{2}}\text{O}$

Explanation of Solution

The given ionic compound is lithium oxide. Here, lithium belongs to group 1 thus, it can give 1 electron to form ${\text{Li}}^{+}$ ion. Oxygen belongs to group 16 with 6 valance electrons. It can accept 2 electrons to complete the octet and form ${\text{O}}^{2-}$ ion. Since the total charge of the ionic compound is zero thus, 2 ${\text{Li}}^{+}$ ions combine with 1 ${\text{O}}^{2-}$ ion. The formula of the ionic compound will be ${\text{Li}}_{\text{2}}\text{O}$ .

(d)

Interpretation Introduction

Interpretation: The formula for calcium fluoride needs to be determined.

Concept Introduction:Ionic compound is formed by the chemical combination of a metal and a non-metal. Here, metal tends to be electron-rich, and non-metal tends to be electron deficient. Thus, metals donate electron/s to form a positively charged ion or cation and non-metals accept electron/s to form a negatively charged ion or anion.

Answer to Problem 15LC

  ${\text{CaF}}_{\text{2}}$

Explanation of Solution

The given ionic compound is calcium fluoride. Here, calcium belongs to group 2 thus, it can give 2 electrons to form ${\text{Ca}}^{2+}$ ion. Fluoride belongs to group 17 with 7 valance electrons. It can accept 1 electron to complete the octet and form ${\text{F}}^{-}$ ion. Since the total charge of the ionic compound is zero; thus, 1 ${\text{Ca}}^{2+}$ ion combines with 2 ${\text{F}}^{-}$ ions. The formula of the ionic compound will be ${\text{CaF}}_{\text{2}}$ .