Chapter 11, Problem 30E

(a)

To determine

The preferred name of the compound ${\text{Al}}_2{\left({\text{CO}}_3\right)}_3$.

Answer to Problem 30E

The preferred name of the compound ${\text{Al}}_2{\left({\text{CO}}_3\right)}_3$ is Aluminium carbonate.

Explanation of Solution

Given info: The compound ${\text{Al}}_2{\left({\text{CO}}_3\right)}_3$ is given.

Explanation:

The provided compound ${\text{Al}}_2{\left({\text{CO}}_3\right)}_3$ has a polyatomic ion ${\text{CO}}_3{}^{2-}$. In this compound, the positive ion is ${\text{Al}}^{3+}$.

The name of ${\text{CO}}_3{}^{2-}$ ion is carbonate ion and ${\text{Al}}^{3+}$ is aluminium ion. While writing name, the name of positive ion is written first followed by the negative ion.

So, the name given to the compound is Aluminium carbonate.

Conclusion:

Therefore, the preferred name of the compound ${\text{Al}}_2{\left({\text{CO}}_3\right)}_3$ is Aluminium carbonate.

(b)

To determine

The preferred name of the compound ${\left({\text{NH}}_4\right)}_2{\text{SO}}_4$.

Answer to Problem 30E

The preferred name of the compound ${\left({\text{NH}}_4\right)}_2{\text{SO}}_4$ is Ammonium sulfate.

Explanation of Solution

Given info: The compound ${\left({\text{NH}}_4\right)}_2{\text{SO}}_4$ is given.

Explanation:

The provided compound ${\left({\text{NH}}_4\right)}_2{\text{SO}}_4$ has two polyatomic ions ${\text{NH}}_4{}^{+}$ and ${\text{SO}}_4{}^{2-}$. In this compound, the positive ion is ${\text{NH}}_4{}^{+}$ and the negative ion is ${\text{SO}}_4{}^{2-}$.

The name of ${\text{NH}}_4{}^{+}$ ion is ammonium ion and ${\text{SO}}_4{}^{2-}$ is sulfate ion. While writing name, the name of positive ion is written first followed by the negative ion.

So, the name given to the compound is Ammonium sulfate.

Conclusion:

Therefore, the preferred name of the compound ${\left({\text{NH}}_4\right)}_2{\text{SO}}_4$ is Ammonium sulfate.

(c)

To determine

The preferred name of the compound ${\text{Li}}_2\text{S}$.

Answer to Problem 30E

The preferred name of the compound ${\text{Li}}_2\text{S}$ is Lithium sulfide.

Explanation of Solution

Given info: The compound ${\text{Li}}_2\text{S}$ is given.

Explanation:

The provided compound ${\text{Li}}_2\text{S}$ is a binary compound of a metal and a non-metal. First, write the name of the metal that is Lithium and then write the name of the non-metal that is sulfur with the suffix ‘ide’.

So, the name of the compound ${\text{Li}}_2\text{S}$ is Lithium sulfide.

The compound ${\text{Li}}_2\text{S}$ has positive ion called cation ${\text{Li}}^{+}$ and a negative ion called anion ${\text{S}}^{2-}$.

Conclusion:

Therefore, the preferred name of the compound ${\text{Li}}_2\text{S}$ is Lithium sulfide.

(d)

To determine

The preferred name of the compound ${\text{SO}}_3$.

Answer to Problem 30E

The preferred name of the compound ${\text{SO}}_3$ is Sulfur trioxide.

Explanation of Solution

Given info: The compound ${\text{SO}}_3$ is given.

Explanation:

The provided compound ${\text{SO}}_3$ is a binary compound of two non-metals. First, write the name of the first non-metal that is sulfur and then write the name of the second non-metal that is oxygen with the suffix ‘ide’.

Also, number of atoms of oxygen is three in the compound. Tri is the Greek prefix used for the number three. Hence, the name of the provided compound ${\text{SO}}_3$ is Sulfur trioxide.

Conclusion:

Therefore, the preferred name of the compound ${\text{SO}}_3$ is Sulfur trioxide.

(e)

To determine

The preferred name of the compound ${\text{Ba}}_3{\text{N}}_2$.

Answer to Problem 30E

The preferred name of the compound ${\text{Ba}}_3{\text{N}}_2$ is Barium nitride.

Explanation of Solution

Given info: The compound ${\text{Ba}}_3{\text{N}}_2$ is given.

Explanation:

The provided compound ${\text{Ba}}_3{\text{N}}_2$ is a binary compound of a metal and a non-metal. First, write the name of the metal that is Barium and then write the name of the non-metal that is nitrogen with the suffix ‘ide’.

So, the name of the compound ${\text{Ba}}_3{\text{N}}_2$ is Barium nitride.

The compound ${\text{Ba}}_3{\text{N}}_2$ has positive ion called cation ${\text{Ba}}^{2+}$ and a negative ion called anion ${\text{N}}^{3-}$.

Conclusion:

Therefore, the preferred name of the compound ${\text{Ba}}_3{\text{N}}_2$ is Barium nitride.

(f)

To determine

The preferred name of the compound $\text{Ba}{\left({\text{NO}}_3\right)}_2$.

Answer to Problem 30E

The preferred name of the compound $\text{Ba}{\left({\text{NO}}_3\right)}_2$ is Barium nitrate.

Explanation of Solution

Given info: The compound $\text{Ba}{\left({\text{NO}}_3\right)}_2$ is given.

Explanation:

The provided compound $\text{Ba}{\left({\text{NO}}_3\right)}_2$ has a polyatomic ion ${\text{NO}}_3{}^{-}$. In this compound, the positive ion is ${\text{Ba}}^{2+}$.

The name of ${\text{NO}}_3{}^{-}$ ion is nitrate ion and ${\text{Ba}}^{2+}$ is barium ion. While writing name, the name of positive ion is written first followed by the negative ion.

So, the name given to the compound is Barium nitrate.

Conclusion:

Therefore, the preferred name of the compound $\text{Ba}{\left({\text{NO}}_3\right)}_2$ is Barium nitrate.

(g)

To determine

The preferred name of the compound ${\text{SiF}}_4$.

Answer to Problem 30E

The preferred name of the compound ${\text{SiF}}_4$ is Silicon tetrafluoride.

Explanation of Solution

Given info: The compound ${\text{SiF}}_4$ is given.

Explanation:

The provided compound ${\text{SiF}}_4$ is a binary compound of two non-metals. First, write the name of the first non-metal that is silicon and then write the name of the second non-metal that is fluorine with the suffix ‘ide’.

Also, number of atoms of fluorine is four in the compound. Tetra is the Greek prefix used for the number four. Hence, the name of the provided compound ${\text{SiF}}_4$ is Silicon tetrafluoride.

Conclusion:

Therefore, the preferred name of the compound ${\text{SiF}}_4$ is Silicon tetrafluoride.

(h)

To determine

The preferred name of the compound ${\text{S}}_2{\text{Cl}}_2$.

Answer to Problem 30E

The preferred name of the compound ${\text{S}}_2{\text{Cl}}_2$ is Disulfur Dichloride.

Explanation of Solution

Given info: The compound ${\text{S}}_2{\text{Cl}}_2$ is given.

Explanation:

The provided compound ${\text{S}}_2{\text{Cl}}_2$ is a binary compound of two non-metals. First, write the name of the first non-metal that is sulfur and then write the name of the second non-metal that is chlorine with the suffix ‘ide’.

Also, the number of atoms of sulfur is two. Number of atoms of chlorine is two in the compound. Di is the Greek prefix used for the number two. Hence, the name of the provided compound ${\text{S}}_2{\text{Cl}}_2$ is Disulfur Dichloride.

Conclusion:

Therefore, the preferred name of the compound ${\text{S}}_2{\text{Cl}}_2$ is Disulfur Dichloride.