Chapter 12, Problem 18E

(a)

To determine

Formula for given ionic compound rubidium iodide.

Answer to Problem 18E

The formula for ionic compound rubidium iodide is $\text{RbI}$.

Explanation of Solution

Given Info: Refer to Table 11.6 in the textbook.

Explanation:

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

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

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 metal is used as a subscript for the anion. The charge on the anion is used as a subscript for the cation.

If ${\text{Rb}}^{1+}$ combines with ${\text{I}}^{-}$, the formula for the ionic compound is $\text{RbI}$ because $\left(1\right)$ is used as a subscript for cation and $\left(1\right)$ is used as a subscript for the anion.

Conclusion:

Therefore, the formula for ionic compound rubidium iodide is $\text{RbI}$.

(b)

To determine

Formula for given ionic compound cesium phosphide.

Answer to Problem 18E

The formula for ionic compound cesium phosphide is ${\text{Cs}}_3\text{P}$.

Explanation of Solution

Given Info: Refer to Table 11.6 in the textbook.

Explanation:

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

An element from the group $\text{5A}$ has an ionic charge $-3$. Phosphide has a charge of $-3$.

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 metal is used as a subscript for the anion. The charge on the anion is used as a subscript for the cation.

If ${\text{Cs}}^{1+}$ combines with ${\text{P}}^{3-}$, the formula for the ionic compound is ${\text{Cs}}_3\text{P}$ because $\left(3\right)$ is used as a subscript for cation and $\left(1\right)$ is used as a subscript for the anion.

Conclusion:

Therefore, the formula for ionic compound cesium phosphide is ${\text{Cs}}_3\text{P}$.

(c)

To determine

Formula for given ionic compound lithium sulfate.

Answer to Problem 18E

The formula for ionic compound lithium sulfate is ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$.

Explanation of Solution

Given Info: Refer to Table 11.6 in the textbook.

Explanation:

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

Anion sulfate has a charge of $-2$.

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 metal is used as a subscript for the anion. The charge on the anion is used as a subscript for the cation.

If ${\text{Li}}^{1+}$ combines with ${\text{SO}}_4{}^{2-}$, the formula for the ionic compound is ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ because $\left(2\right)$ is used as a subscript for cation and $\left(1\right)$ is used as a subscript for the anion ${\text{SO}}_4{}^{2-}$.

Conclusion:

Therefore, the formula for ionic compound lithium sulfate is ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$.

(d)

To determine

Formula for given ionic compound silver carbonate.

Answer to Problem 18E

The formula for ionic compound silver carbonate is ${\text{Ag}}_2{\text{CO}}_3$.

Explanation of Solution

Given Info: Refer to Table 11.6 in the textbook.

Explanation:

Silver has a charge of $+1$.

Anion carbonate has a charge of $-2$.

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 metal is used as a subscript for the anion. The charge on the anion is used as a subscript for the cation.

If ${\text{Ag}}^{1+}$ combines with ${\text{CO}}_3{}^{2-}$, the formula for the ionic compound is ${\text{Ag}}_2{\text{CO}}_3$ because $\left(2\right)$ is used as a subscript for cation and $\left(1\right)$ is used as a subscript for the anion ${\text{CO}}_3{}^{2-}$.

Conclusion:

Therefore, the formula for ionic compound silver carbonate is ${\text{Ag}}_2{\text{CO}}_3$.

(e)

To determine

Formula for given ionic compound zinc phosphate.

Answer to Problem 18E

The formula for ionic compound zinc phosphate is ${\text{Zn}}_3{\left({\text{PO}}_4\right)}_2$.

Explanation of Solution

Given Info: Refer to Table 11.6 in the textbook.

Explanation:

Zinc has a charge of $+2$.

Anion phosphate has a charge of $-3$.

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 metal is used as a subscript for the anion. The charge on the anion is used as a subscript for the cation.

If ${\text{Zn}}^{2+}$ combines with ${\text{PO}}_4{}^{3-}$, the formula for the ionic compound is ${\text{Zn}}_3{\left({\text{PO}}_4\right)}_2$ because $\left(3\right)$ is used as a subscript for cation and $\left(2\right)$ is used as a subscript for the anion ${\text{PO}}_4{}^{3-}$.

Conclusion:

Therefore, the formula for ionic compound zinc phosphate is ${\text{Zn}}_3{\left({\text{PO}}_4\right)}_2$.

(f)

To determine

Formula for given ionic compound aluminium nitrate.

Answer to Problem 18E

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

Explanation of Solution

Given Info: Refer to Table 11.6 in the textbook.

Explanation:

Aluminium has a charge of $+3$.

Anion nitrate has a charge of $-1$.

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 metal is used as a subscript for the anion. The charge on the anion is used as a subscript for the cation.

If ${\text{Al}}^{3+}$ combines with ${\text{NO}}_3{}^{1-}$, the formula for the ionic compound is $\text{Al}{\left({\text{NO}}_3\right)}_3$ because $\left(1\right)$ is used as a subscript for cation and $\left(3\right)$ is used as a subscript for the anion ${\text{NO}}_3{}^{1-}$.

Conclusion:

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