Chapter 4, Problem 4.32E

Name the following binary ionic compounds, using a Roman numeral to indicate the charge on the metal ion:

a. $\text{PbO}$ and ${\text{PbO}}_2$

b. $\text{CuCl}$ and ${\text{CuCl}}_2$

c. ${\text{Au}}_{\text{2}}\text{S}$ and ${\text{Au}}_{\text{2}}{\text{S}}_3$

d. $\text{CoO}$ and ${\text{Co}}_2{\text{O}}_3$

Interpretation Introduction

(a)

Interpretation:

The name of the given binary ionic compounds is to be stated by using a Roman numeral.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. The binary ionic compounds are composed of only one cation and one anion.

The naming of non-metal ions is done by the addition of suffix –ide at the end of the element’s name. The naming of binary ionic compound is represented by the name of metal followed by the name of non-metal.

Answer to Problem 4.32E

The name of the binary ionic compound $\text{PbO}$ is $\text{lead}\left(\text{II}\right)\text{oxide}$ and ${\text{PbO}}_2$ is $\text{lead}\left(\text{IV}\right)\text{oxide}$.

Explanation of Solution

Roman numeral system represents the ionic charge of the metal ion.

The given binary ionic compounds are $\text{PbO}$ and ${\text{PbO}}_2$. Both the compounds contain lead as a metal and oxygen as a non-metal. The lead can exists as ${\text{Pb}}^{2+}$ and ${\text{Pb}}^{4+}$ ions and oxygen as ${\text{O}}^{2-}$ ion.

The naming of the binary ionic compound is given by,

$\text{name}=\text{metal}\left(\text{charge}\text{in}\text{roman}\right)+\text{nonmetal}\text{stem}+\text{-}ide$

The suffix –ide is added after “ox” of oxygen to obtain oxide.

Hence, the name of the binary ionic compound $\text{PbO}$ is $\text{lead}\left(\text{II}\right)\text{oxide}$ and ${\text{PbO}}_2$ is $\text{lead}\left(\text{IV}\right)\text{oxide}$.

Conclusion

The name of the binary ionic compound $\text{PbO}$ is $\text{lead}\left(\text{II}\right)\text{oxide}$ and ${\text{PbO}}_2$ is $\text{lead}\left(\text{IV}\right)\text{oxide}$.

Interpretation Introduction

(b)

Interpretation:

The name of the given binary ionic compound is to be stated by using a Roman numeral.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. The binary ionic compounds are composed of only one cation and one anion.

The naming of non-metal ions is done by the addition of suffix –ide at the end of the element’s name. The naming of binary ionic compound is represented by the name of metal followed by the name of non-metal.

Answer to Problem 4.32E

The name of the binary ionic compound $\text{CuCl}$ is $\text{copper}\left(\text{I}\right)\text{chloride}$ and ${\text{CuCl}}_2$ is $\text{copper}\left(\text{II}\right)\text{chloride}$.

Explanation of Solution

Roman numeral system represents the ionic charge of the metal ion.

The binary ionic compounds are $\text{CuCl}$ and ${\text{CuCl}}_2$. Both the compounds contain copper as a metal and chlorine as a non-metal. The copper can exists as ${\text{Cu}}^{+}$ and ${\text{Cu}}^{+2}$ ions and chlorine as ${\text{Cl}}^{-}$ ion.

The naming of the binary ionic compound is given by,

$\text{name}=\text{metal}\left(\text{charge}\text{in}\text{roman}\right)+\text{nonmetal}\text{stem}+\text{-}ide$

The suffix –ide is added after “chlor” of chlorine to obtain chloride.

Hence, the name of the binary ionic compound $\text{CuCl}$ is $\text{copper}\left(\text{I}\right)\text{chloride}$ and ${\text{CuCl}}_2$ is $\text{copper}\left(\text{II}\right)\text{chloride}$.

Conclusion

The name of the binary ionic compound $\text{CuCl}$ is $\text{copper}\left(\text{I}\right)\text{chloride}$ and ${\text{CuCl}}_2$ is $\text{copper}\left(\text{II}\right)\text{chloride}$.

Interpretation Introduction

(c)

Interpretation:

The name of the given binary ionic compound is to be stated by using a Roman numeral.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. The binary ionic compounds are composed of only one cation and one anion.

The naming of non-metal ions is done by the addition of suffix –ide at the end of the element’s name. The naming of binary ionic compound is represented by the name of metal followed by the name of non-metal.

Answer to Problem 4.32E

The name of the binary ionic compound ${\text{Au}}_{\text{2}}\text{S}$ is $\text{gold}\left(\text{I}\right)\text{sulfide}$ and ${\text{Au}}_{\text{2}}{\text{S}}_3$ is $\text{gold}\left(\text{III}\right)\text{sulfide}$.

Explanation of Solution

Roman numeral system represents the ionic charge of the metal ion.

The given binary ionic compounds are ${\text{Au}}_{\text{2}}\text{S}$ and ${\text{Au}}_{\text{2}}{\text{S}}_3$. Both the compounds contain gold as a metal and sulfur as a non-metal. The gold exists as ${\text{Au}}^{+}$ and ${\text{Au}}^{3+}$ ions and sulfide as ${\text{S}}^{2-}$ ion in ionic form.

The naming of the binary ionic compound is given by,

$\text{name}=\text{metal}\left(\text{charge}\text{in}\text{roman}\right)+\text{nonmetal}\text{stem}+\text{-}ide$

The suffix –ide is added after “sulf” of sulfur to obtain sulfide.

Hence, the name of the binary ionic compound ${\text{Au}}_{\text{2}}\text{S}$ is $\text{gold}\left(\text{I}\right)\text{sulfide}$ and ${\text{Au}}_{\text{2}}{\text{S}}_3$ is $\text{gold}\left(\text{III}\right)\text{sulfide}$.

Conclusion

The name of the binary ionic compound ${\text{Au}}_{\text{2}}\text{S}$ is $\text{gold}\left(\text{I}\right)\text{sulfide}$ and ${\text{Au}}_{\text{2}}{\text{S}}_3$ is $\text{gold}\left(\text{III}\right)\text{sulfide}$.

Interpretation Introduction

(d)

Interpretation:

The name of the given binary ionic compound is to be stated by using a Roman numeral.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. The binary ionic compounds are composed of only one cation and one anion.

The naming of non-metal ions is done by the addition of suffix –ide at the end of the element’s name. The naming of binary ionic compound is represented by the name of metal followed by the name of non-metal.

Answer to Problem 4.32E

The name of the binary ionic compound $\text{CoO}$ is $\text{cobalt}\left(\text{II}\right)\text{oxide}$ and ${\text{Co}}_2{\text{O}}_3$ is $\text{cobalt}\left(\text{III}\right)\text{oxide}$.

Explanation of Solution

Roman numeral system represents the ionic charge of the metal ion.

The given binary ionic compounds are $\text{CoO}$ and ${\text{Co}}_2{\text{O}}_3$. Both the compounds cobalt as a metal and oxygen as a non-metal. The cobalt exists as ${\text{Co}}^{2+}$ and ${\text{Co}}^{3+}$ ions and oxygen as ${\text{O}}^{2-}$ ions in ionic form.

The naming of the binary ionic compound is given by,

$\text{name}=\text{metal}\left(\text{charge}\text{in}\text{roman}\right)+\text{nonmetal}\text{stem}+\text{-}ide$

The suffix –ide is added after “ox” of oxygen to obtain oxide.

Hence, the name of the binary ionic compound $\text{CoO}$ is $\text{cobalt}\left(\text{II}\right)\text{oxide}$ and ${\text{Co}}_2{\text{O}}_3$ is $\text{cobalt}\left(\text{III}\right)\text{oxide}$.

Conclusion

The name of the binary ionic compound $\text{CoO}$ is $\text{cobalt}\left(\text{II}\right)\text{oxide}$ and ${\text{Co}}_2{\text{O}}_3$ is $\text{cobalt}\left(\text{III}\right)\text{oxide}$.