Chapter 12, Problem 17E

(a)

To determine

The formula for given ionic compound cesium iodide.

Answer to Problem 17E

The formula for ionic compound cesium iodide is $\text{CsI}$.

Explanation of Solution

For a neutral ionic compound, the number of negative and positive charge of ions should be equal. To write the formula of the compound the charge neutrality is maintained. Signs of cation and anion charges are not considered while writing the formula for the compound. The charge on the cation is used as a subscript for the anion and the charge on the anion is used as a subscript for the cation.

An element from the group $\text{1A}$ has an ionic charge $+1$. Therefore, cesium has a charge of $+1$.

An element from the group $\text{7A}$ has an ionic charge $-1$. Therefore, iodine has a charge of $-1$.

The charge on anion $\left({\text{I}}^{-\text{1}}\right)$ is used as a subscript for the cesium and the charge on cation $\left({\text{Cs}}^{+1}\right)$ is used as a subscript for the iodine thus the formula for the ionic compound is $\text{CsI}$.

Conclusion:

Therefore, the formula for ionic compound cesium iodide is $\text{CsI}$.

(b)

To determine

The formula for given ionic compound barium fluoride.

Answer to Problem 17E

The formula for ionic compound barium fluoride is ${\text{BaF}}_{\text{2}}$.

Explanation of Solution

For a neutral ionic compound, the number of negative and positive charge of ions should be equal. To write the formula of the compound the charge neutrality is maintained. Signs of cation and anion charges are not considered while writing the formula for the compound. The charge on the cation is used as a subscript for the anion and the charge on the anion is used as a subscript for the cation.

An element from the group $\text{2A}$ has an ionic charge $+2$. Barium has a charge of $+2$.

An element from the group $\text{7A}$ has an ionic charge $-1$. Therefore, iodine has a charge of $-1$.

The charge on anion $\left({\text{F}}^{-1}\right)$ is used as a subscript for the barium and the charge on cation $\left({\text{Ba}}^{2+}\right)$ is used as a subscript for the fluorine thus the formula for the ionic compound is ${\text{BaF}}_{\text{2}}$.

Conclusion:

Therefore, the formula for ionic compound barium fluoride is ${\text{BaF}}_{\text{2}}$

(c)

To determine

The formula for given ionic compound aluminium nitrate.

Answer to Problem 17E

The formula for ionic compound aluminium nitrate is $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$.

Explanation of Solution

Given Info: Refer to Table 11.6 in the textbook.

Explanation:

For a neutral ionic compound, the number of negative and positive charge of ions should be equal. To write the formula of the compound the charge neutrality is maintained. Signs of cation and anion charges are not considered while writing the formula for the compound. The charge on the cation is used as a subscript for the anion and the charge on the anion is used as a subscript for the cation.

Aluminium belongs to the group $\text{3A}$ and has a charge of $+3$.

The nitrate ion is a polyatomic anion and has a charge of $-1$.

The charge on anion $\left({\text{NO}}_3{}^{-1}\right)$ is used as a subscript for aluminium and the charge on cation $\left({\text{Al}}^{3+}\right)$ is used as a subscript for the nitrate thus the formula for the ionic compound is $\text{Al}{\left({\text{NO}}_3\right)}_3$.

Conclusion:

Therefore, the formula for ionic compound aluminium nitrate is $\text{Al}{\left({\text{NO}}_3\right)}_3$.

(d)

To determine

The formula for given ionic compound lithium sulfide.

Answer to Problem 17E

The formula for ionic compound lithium sulfide is ${\text{Li}}_{\text{2}}\text{S}$.

Explanation of Solution

For a neutral ionic compound, the number of negative and positive charge of ions should be equal. To write the formula of the compound the charge neutrality is maintained. Signs of cation and anion charges are not considered while writing the formula for the compound. The charge on the cation is used as a subscript for the anion and the charge on the anion is used as a subscript for the cation.

An element from the group $\text{1A}$ has an ionic charge $+1$. Lithium has a charge of $+1$.

An element from the group $\text{6A}$ has an ionic charge $-2$. Therefore, sulfide ion has a charge of $-2$.

The charge on anion $\left({\text{S}}^{-2}\right)$ is used as a subscript for the lithium and the charge on cation $\left({\text{Li}}^{1+}\right)$ is used as a subscript for the sulfide thus the formula for the ionic compound is ${\text{Li}}_{\text{2}}\text{S}$.

Conclusion:

Therefore, the formula for ionic compound lithium sulfide is ${\text{Li}}_{\text{2}}\text{S}$.

(e)

To determine

The formula for given ionic compound beryllium oxide.

Answer to Problem 17E

The formula for ionic compound beryllium oxide is $\text{BeO}$.

Explanation of Solution

For a neutral ionic compound, the number of negative and positive charge of ions should be equal. To write the formula of the compound the charge neutrality is maintained. Signs of cation and anion charges are not considered while writing the formula for the compound. The charge on the cation is used as a subscript for the anion and the charge on the anion is used as a subscript for the cation.

An element from the group $\text{2A}$ has an ionic charge $+2$. Beryllium has a charge of $+2$.

An element from the group $\text{6A}$ has an ionic charge $-2$. Therefore, oxide ion has a charge of $-2$.

The charge on anion $\left({\text{O}}^{-2}\right)$ is used as a subscript for the beryllium and the charge on cation $\left({\text{Be}}^{2+}\right)$ is used as a subscript for the oxide thus the formula for the ionic compound is ${\text{Be}}_2{\text{O}}_2$. However, the smallest possible ratio of ions is taken, therefore, the formula is $\text{BeO}$.

Conclusion:

Therefore, the formula for ionic compound beryllium oxide is $\text{BeO}$.

(f)

To determine

The formula for given ionic compound ammonium sulfate.

Answer to Problem 17E

The formula for ionic compound ammonium sulfate is ${\left({\text{NH}}_4\right)}_2\left({\text{SO}}_4\right)$.

Explanation of Solution

Given Info: Refer to Table 11.6 in the textbook.

Explanation:

For a neutral ionic compound, the number of negative and positive charge of ions should be equal. To write the formula of the compound the charge neutrality is maintained. Signs of cation and anion charges are not considered while writing the formula for the compound. The charge on the cation is used as a subscript for the anion and the charge on the anion is used as a subscript for the cation.

Ammonium and sulfate ions are polyatomic ions. Ammonium ion has a charge of $+1$.

Sulfate ion has a charge of $-2$.

The charge on anion $\left({\text{SO}}_4{}^{-2}\right)$ is used as a subscript for the ammonium and the charge on cation $\left({\text{NH}}_4{}^{+1}\right)$ is used as a subscript for the sulfate thus the formula for the ionic compound is ${\left({\text{NH}}_4\right)}_2\left({\text{SO}}_4\right)$. Each sulfate ion is associated with two ammonium ion to maintain neutrality

Conclusion:

Therefore, the formula for ionic compound ammonium sulfate is ${\left({\text{NH}}_4\right)}_2\left({\text{SO}}_4\right)$.