Chapter 7, Problem 44A

(a)

Interpretation Introduction

Interpretation: The formula for the ions of $\text{KCl}$ needs to be determined.

Concept Introduction: Ions are formed when electrons are lost or gained by an atom to attain noble gas configuration or to complete the octet. Here, metals lose electron/s to form positively charged ions or cations and non-metals gain electron/s to form negatively charged ions or anions. Ionic compounds are formed when these cations and anions are combined in fixed ratios. They are combined in such a way that the total charge of the ionic compound is zero.

Answer to Problem 44A

  ${\text{K}}^{+}$ and ${\mathrm{Cl}}^{-}$

Explanation of Solution

The given ionic compound is KCl. Here, K is metal and Cl is non-metal. Potassium (K) belongs to group 1 thus; K has one valance electron, and it can lose one electron to form ${\text{K}}^{+}$ ion. Similarly, chlorine (Cl) belongs to group 17 and has 7 valance electrons. To attain noble gas configuration, Cl gains one electron to form ${\mathrm{Cl}}^{-}$ ion.

The formula of ions of KCl is ${\text{K}}^{+}$ and ${\mathrm{Cl}}^{-}$ .

(b)

Interpretation Introduction

Interpretation: The formula for the ions of $\text{BaS}$ needs to be determined.

Concept Introduction: Ions are formed when electrons are lost or gained by an atom to attain noble gas configuration or to complete the octet. Here, metals lose electron/s to form positively charged ions or cations and non-metals gain electron/s to form negatively charged ions or anions. Ionic compounds are formed when these cations and anions are combined in fixed ratios. They are combined in such a way that the total charge of the ionic compound is zero.

Answer to Problem 44A

  ${\text{Ba}}^{2+}$ and ${\text{S}}^{2-}$ .

Explanation of Solution

The given ionic compound is $\text{BaS}$ . Here, Ba is metal and S is non-metal. Barium (Ba) belongs to group 2 and has 2 valance electrons. To attain noble gas configuration, Ba can lose two electrons to form ${\text{Ba}}^{2+}$ ion. Similarly, sulphur (S) belongs to group 16 and has 6 valance electrons. It can gain two electrons to attain noble gas configuration or to form ${\text{S}}^{2-}$ ion.

Therefore, the formula of ions of $\text{BaS}$ is ${\text{Ba}}^{2+}$ and ${\text{S}}^{2-}$ .

(c)

Interpretation Introduction

Interpretation: The formula for the ions of ${\text{MgBr}}_2$ needs to be determined.

Concept Introduction: Ions are formed when electrons are lost or gained by an atom to attain noble gas configuration or to complete the octet. Here, metals lose electron/s to form positively charged ions or cations and non-metals gain electron/s to form negatively charged ions or anions. Ionic compounds are formed when these cations and anions are combined in fixed ratios. They are combined in such a way that the total charge of the ionic compound is zero.

Answer to Problem 44A

  ${\text{Mg}}^{2+}$ and ${\text{Br}}^{-}$

Explanation of Solution

The given ionic compound is ${\text{MgBr}}_2$ . Here, Mg is metal and Br is non-metal. Magnesium (Mg) belongs to group 2 and has 2 valance electrons. To attain noble gas configuration, Mg can lose two electrons to form ${\text{Mg}}^{2+}$ ion. Similarly, Bromine (Br) belongs to group 17 and has7 valance electrons. It can gain one electron to attain noble gas configuration or to form ${\text{Br}}^{-}$ ion.

Therefore, the formula of ions of ${\text{MgBr}}_2$ is ${\text{Mg}}^{2+}$ and ${\text{Br}}^{-}$ .

(d)

Interpretation Introduction

Interpretation: The formula for the ions of ${\text{Li}}_{\text{2}}\text{O}$ needs to be determined.

Concept Introduction: Ions are formed when electrons are lost or gained by an atom to attain noble gas configuration or to complete the octet. Here, metals lose electron/s to form positively charged ions or cations and non-metals gain electron/s to form negatively charged ions or anions. Ionic compounds are formed when these cations and anions are combined in fixed ratios. They are combined in such a way that the total charge of the ionic compound is zero.

Answer to Problem 44A

  ${\text{Li}}^{+}$ and ${\text{O}}^{2-}$

Explanation of Solution

The given ionic compound is ${\text{Li}}_{\text{2}}\text{O}$ . Here, Li is metal and O is non-metal. Lithium (Li) belongs to group 1 and has 1 valance electron. To attain noble gas configuration, Li can lose one electron to form ${\text{Li}}^{+}$ ion. Similarly, Oxygen (O) belongs to group 16 and has 6 valance electrons. It can gain two electrons to attain noble gas configuration or to form ${\text{O}}^{2-}$ ion.

Therefore, the formula of ions of ${\text{Li}}_{\text{2}}\text{O}$ is ${\text{Li}}^{+}$ and ${\text{O}}^{2-}$ .