Chapter 7, Problem 81A

(a)

Interpretation Introduction

Interpretation:

The formula of ionic compound calcium iodide needs to be determined.

Concept introduction:

Ionic bonds are formed when there is complete transfer of electrons from one atom to another. These atoms either lose or gain electrons to become negatively or positively charged ions. The forces of attraction between these ions cause the ionic bond formation. Ionic compounds are formed only when there is a balance between their charges.

Answer to Problem 81A

The chemical formula for ionic compound calcium iodide is ${\text{ CaI}}_2$

Explanation of Solution

The element Calcium belongs to Group 2 and has atomic number of 20. Its electronic configuration is ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{4s}}^2$. Hence it can donate 2 of its outermost electrons to give ${\text{Ca}}^{\text{2+}}$.

The element Iodine belongs to Halogen group or group 17 having atomic number as 9. The electronic configuration is ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^5$. It has 7 electrons in outermost shell and requires one electron and become ${\text{I}}^{-}$.

To combine ${\text{Ca}}^{\text{2+}}$ and ${\text{ I}}^{-}$, it will give rise to following compound

Hence, the formula of the compound calcium iodide is ${\text{ CaI}}_2$

(b)

Interpretation Introduction

Interpretation:

The formula of ionic compound silver bromide needs to be determined.

Concept introduction:

Ionic bonds are formed when there is complete transfer of electrons from one atom to another. These atoms either lose or gain electrons to become negatively or positively charged ions. The forces of attraction between these ions cause the ionic bond formation. Ionic compounds are formed only when there is a balance between their charges.

Answer to Problem 81A

The chemical formula for ionic compound Silver Bromide is $\text{AgBr}$

Explanation of Solution

The element Silver belongs to Group 11 and has atomic number of 47. Its electronic configuration is ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{4s}}^2{\text{3d}}^{10}{\text{4p}}^6{\text{5s}}^2{\text{4d}}^9$. Hence it can donate 1 of its outermost electrons to give ${\text{Ag}}^{+}$.

The element Bromide belongs to Halogen group or group 17 having atomic number as 35. The electronic configuration is ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{4s}}^2{\text{3d}}^{10}{\text{4p}}^5$. It has 7 electrons in outermost shell and requires one electron and become ${\text{Br}}^{-}$.

To combine ${\text{Ag}}^{+}$ and ${\text{Br}}^{-}$, it will give rise to following compound

Hence, the formula of the compound Silver Bromide is $\text{AgBr}$

(c)

Interpretation Introduction

Interpretation:

The formula of ionic compound copper chloride needs to be determined.

Concept introduction:

Ionic bonds are formed when there is complete transfer of electrons from one atom to another. These atoms either lose or gain electrons to become negatively or positively charged ions. The forces of attraction between these ions causes the ionic bond formation. Ionic compounds are formed only when there is a balance between their charges.

Answer to Problem 81A

The chemical formula for ionic compound Copper chloride is ${\text{CuCl}}_2{}^{}$

Explanation of Solution

The element Copper belongs to Group 11 and has atomic number of 29. Its electronic configuration is ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{4s}}^2{\text{3d}}^9$. Hence it can donate 2 of its outermost electrons to give ${\text{Cu}}^{\text{2+}}$.

The element Chlorine belongs to Halogen group or group 17 having atomic number as 17. The electronic configuration is ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^2{\text{3p}}^5$. It has 7 electrons in outermost shell and requires one electron and become ${\text{Cl}}^{-}$.

To combine ${\text{Cu}}^{\text{2+}}$ and ${\text{Cl}}^{-}$, it will give rise to following compound

Hence, the formula of the compound Copper chloride is ${\text{CuCl}}_2{}^{}$

(d)

Interpretation Introduction

Interpretation:

The formula of ionic compound potassium periodate needs to be determined.

Concept introduction:

Ionic bonds are formed when there is complete transfer of electrons from one atom to another. These atoms either lose or gain electrons to become negatively or positively charged ions. The forces of attraction between these ions cause the ionic bond formation. Ionic compounds are formed only when there is a balance between their charges.

Answer to Problem 81A

The chemical formula for ionic compound potassium periodate is ${\text{KIO}}_4$

Explanation of Solution

The element Potassium belongs to Group 1 and has atomic number of 19. Its electronic configuration is ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{4s}}^1$. Hence it can donate 1of its outermost electrons to give ${\text{K}}^{\text{+}}$.

The element Iodine belongs to Halogen group or group 17 having atomic number as 9. The electronic configuration is ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^5$. It has 7 electrons in outermost shell and requires one electron and become ${\text{I}}^{-}$.

Periodate is oxianion of Iodine and is denoted as ${\text{IO}}_4{}^{-}$. Its structure is as below having a charge of -1.

To combine ${\text{K}}^{\text{+}}$ and ${\text{IO}}_4{}^{-}$, it will give rise to following compound

Hence, the formula of the compound potassium periodate is ${\text{KIO}}_4$

(e)

Interpretation Introduction

Interpretation:

The formula of ionic compound Silver acetate needs to be determined.

Concept introduction:

Ionic bonds are formed when there is complete transfer of electrons from one atom to another. These atoms either lose or gain electrons to become negatively or positively charged ions. The forces of attraction between these ions cause the ionic bond formation. Ionic compounds are formed only when there is a balance between their charges.

Answer to Problem 81A

The chemical formula for ionic compound Silver acetate is ${\text{CH}}_{\text{3}}\text{COOAg}$

Explanation of Solution

The element Silver belongs to Group 11 and has atomic number of 47. Its electronic configuration is ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{4s}}^2{\text{3d}}^{10}{\text{4p}}^6{\text{5s}}^2{\text{4d}}^9$. Hence it can donate 1 of its outermost electrons to give ${\text{Ag}}^{+}$.

The acetate ion is conjugate base of acetic acid. It is depicted as ${\text{CH}}_{\text{3}}\text{COO-}$. It has one negative charge.

To combine ${\text{Ag}}^{+}$ and ${\text{CH}}_{\text{3}}\text{COO-}$, it will give rise to following compound

Hence, the formula of the compound Silver acetate is ${\text{CH}}_{\text{3}}\text{COOAg}$