Chapter 6, Problem 26E

Interpretation Introduction

(a)

Interpretation:

The chemical formula of binary ionic compound, $\text{copper}\left(\text{II}\right)\text{sulfide}$ is to be predicted.

Concept introduction:

Ionic compounds are made up of ions that are bonded by electrostatic forces of attraction. In ionic compounds, cations are defined as the positively charged ions, whereas anions are the negatively charged ions.

Answer to Problem 26E

The chemical formula of binary ionic compound, $\text{copper}\left(\text{II}\right)\text{sulfide}$ is $\text{CuS}$.

Explanation of Solution

The IUPAC system of naming is used to provide a proper naming to binary ionic compound. According to this system of naming, the cation of the binary ionic compound is named as the same as atom and the anions of the compound ends with suffix -ide. In case of transition elements, the charge over the cation is written in roman numbers inside the parentheses.

The given binary ionic compound is $\text{copper}\left(\text{II}\right)\text{sulfide}$. This compound contains cobalt ion as cation with $+2$ charge and sulfide ion as anion with $-2$ charge.

The balanced chemical reaction between copper ion and sulfide ion is given below.

${\text{Cu}}^{2+}+{\text{S}}^{2-}\to \text{CuS}$

Therefore, the chemical formula of $\text{copper}\left(\text{II}\right)\text{sulfide}$ is $\text{CuS}$.

Conclusion

The chemical formula of binary ionic compound, $\text{copper}\left(\text{II}\right)\text{sulfide}$ is $\text{CuS}$.

Interpretation Introduction

(b)

Interpretation:

The chemical formula of binary ionic compound, $\text{iron}\left(\text{III}\right)\text{phosphide}$ is to be predicted.

Concept introduction:

Ionic compounds are made up of ions that are bonded by electrostatic forces of attraction. In ionic compounds, cations are defined as the positively charged ions, whereas anions are the negatively charged ions.

Answer to Problem 26E

The chemical formula of binary ionic compound, $\text{iron}\left(\text{III}\right)\text{phosphide}$ is $\text{FeP}$.

Explanation of Solution

The IUPAC system of naming is used to provide a proper naming to binary ionic compound. According to this system of naming, the cation of the binary ionic compound is named as the same as atom and the anions of the compound ends with suffix -ide. In case of transition elements, the charge over the cation is written in roman numbers inside the parentheses.

The given binary ionic compound is $\text{iron}\left(\text{III}\right)\text{phosphide}$. This compound contains iron ion as cation with $+3$ charge and phosphide ion as anion with $-3$ charge.

The balanced chemical reaction between iron ion and phosphide ion is given below.

${\text{Fe}}^{3+}+{\text{P}}^{3-}\to \text{FeP}$

Therefore, the chemical formula of $\text{iron}\left(\text{III}\right)\text{phosphide}$ is ${\text{Fe}}_{\text{3}}{\text{P}}_{\text{2}}$.

Conclusion

The chemical formula of binary ionic compound, $\text{iron}\left(\text{III}\right)\text{phosphide}$ is $\text{FeP}$.

Interpretation Introduction

(c)

Interpretation:

The chemical formula of binary ionic compound, $\text{mercury}\left(\text{I}\right)\text{iodide}$ is to be predicted.

Concept introduction:

Ionic compounds are made up of ions that are bonded by electrostatic forces of attraction. In ionic compounds, cations are defined as the positively charged ions, whereas anions are the negatively charged ions.

Answer to Problem 26E

The chemical formula of binary ionic compound, $\text{mercury}\left(\text{I}\right)\text{iodide}$ is $\text{HgI}$.

Explanation of Solution

The given name is $\text{mercury}\left(\text{I}\right)\text{iodide}$. The given compound is an example of ionic compound. The given compound consists of copper and sulfide ions.

The IUPAC system of naming is used to provide a proper naming to binary ionic compound. According to this system of naming, the cation of the binary ionic compound is named as the same as atom and the anions of the compound ends with suffix -ide. In case of transition elements, the charge over the cation is written in roman numbers inside the parentheses.

The given binary ionic compound is $\text{mercury}\left(\text{I}\right)\text{iodide}$. This compound contains mercury ion as cation with $+1$ charge and iodide ion as anion with $-1$ charge.

The balanced chemical reaction between mercury and iodine ion is given below.

${\text{Hg}}^{+}+{\text{I}}^{-}\to \text{HgI}$

Therefore, the chemical formula of $\text{mercury}\left(\text{I}\right)\text{iodide}$ is $\text{HgI}$.

Conclusion

The chemical formula of binary ionic compound, $\text{mercury}\left(\text{I}\right)\text{iodide}$ is $\text{HgI}$.

Interpretation Introduction

(d)

Interpretation:

The chemical formula of binary ionic compound, $\text{lead}\left(\text{II}\right)\text{oxide}$ is to be predicted.

Concept introduction:

Ionic compounds are made up of ions that are bonded by electrostatic forces of attraction. In ionic compounds, cations are defined as the positively charged ions, whereas anions are the negatively charged ions.

Answer to Problem 26E

The chemical formula of binary ionic compound, $\text{lead}\left(\text{II}\right)\text{oxide}$ is $\text{PbO}$.

Explanation of Solution

The IUPAC system of naming is used to provide a proper naming to binary ionic compound. According to this system of naming, the cation of the binary ionic compound is named as the same as atom and the anions of the compound ends with suffix -ide. In case of transition elements, the charge over the cation is written in roman numbers inside the parentheses.

The given binary ionic compound is $\text{lead}\left(\text{II}\right)\text{oxide}$. This compound contains lead ion as cation with $+2$ charge and oxygen ion as anion with $-2$ charge.

The balanced chemical reaction between lead and oxide ion is given below.

${\text{Pb}}^{2+}+{\text{O}}^{2-}\to \text{PbO}$

Therefore, the chemical formula of $\text{lead}\left(\text{II}\right)\text{oxide}$ is $\text{PbO}$.

Conclusion

The chemical formula of binary ionic compound, $\text{lead}\left(\text{II}\right)\text{oxide}$ is $\text{PbO}$.