Chapter 2, Problem 84E

(a)

Interpretation Introduction

Interpretation:

The naming of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 84E

The naming of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is acetic acid.

Explanation of Solution

To determine: The  naming of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$

The naming of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is acetic acid.

Rules for binary compound:

• In naming of binary compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.
• Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state.
• When oxygen is not present in anion, then acids is named by using “hydro-” in prefix.
• If the name of anions ends in “ate-” then acid name ends in –ic or –ric
• If the name of anion ends in “-ite”, then the name of acid ends in –ous.

In the acidic compound ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ the cation is hydrogen $\left({\text{H}}^{\text{+}}\right)$ and anion is acetate $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}\right)$

Hence, the name of given compound is acetic acid.

Conclusion

The naming of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is acetic acid.

(b)

Interpretation Introduction

Interpretation:

The naming of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 84E

The naming of ${\text{NH}}_4{\text{NO}}_{\text{2}}$ is ammonium nitrite.

Explanation of Solution

To determine: The  naming of ${\text{NH}}_4{\text{NO}}_{\text{2}}$

The naming of ${\text{NH}}_4{\text{NO}}_{\text{2}}$ is ammonium nitrite.

Rules for naming binary compound:

• In naming of binary compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.
• Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state.
• When oxygen is not present in anion, then acids is named by using “hydro-” in prefix.
• If the name of anions ends in “ate-” then acid name ends in –ic or –ric
• If the name of anion ends in “-ite”, then the name of acid ends in –ous.

In the acidic compound ${\text{NH}}_4{\text{NO}}_{\text{2}}$ the cation is ammonium $\left({\text{NH}}_4{}^{\text{+}}\right)$ and anion is nitrite $\left({\text{NO}}_{\text{2}}{}^{-}\right)$

Hence, the name of given compound is ammonium nitrite.

Conclusion

The naming of ${\text{NH}}_4{\text{NO}}_{\text{2}}$ is ammonium nitrite.

(c)

Interpretation Introduction

Interpretation:

The naming of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 84E

The naming of ${\text{Co}}_2{\text{S}}_3$ is cobalt (III) sulfide.

Explanation of Solution

To determine: The  naming of ${\text{Co}}_2{\text{S}}_3$

The naming of ${\text{Co}}_2{\text{S}}_3$ is cobalt (III) sulfide.

Rules for naming binary compound:

• In naming of binary compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.
• In the binary II compound roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state.
• When oxygen is not present in anion, then acids is named by using “hydro-” in prefix.
• If the name of anions ends in “ate-” then acid name ends in –ic or –ric
• If the name of anion ends in “-ite”, then the name of acid ends in –ous.

In the binary II compound ${\text{Co}}_2{\text{S}}_3$ the cation is cobalt $\left({\text{Co}}^{\text{3+}}\right)$ having oxidation state $+3$ and anion is sulfur $\left({\text{S}}^{2-}\right)$

Hence, the name of given compound is cobalt (III) sulfide

Conclusion

The naming of ${\text{Co}}_2{\text{S}}_3$ is cobalt (III) sulfide.

(d)

Interpretation Introduction

Interpretation:

The naming of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 84E

The naming of $\text{ICl}$ is iodine monochloride.

Explanation of Solution

To determine: The  naming of $\text{ICl}$

The naming of $\text{ICl}$ is iodine monochloride.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide.

The compound $\text{ICl}$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Hence, the name of $\text{ICl}$ compound is iodine monochloride.

Conclusion

The naming of $\text{ICl}$ is iodine monochloride.

(e)

Interpretation Introduction

Interpretation:

The naming of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 84E

The naming of ${\text{Pb}}_3{\left({\text{PO}}_4\right)}_3$ is lead (II) phosphate.

Explanation of Solution

To determine: The  naming of ${\text{Pb}}_3{\left({\text{PO}}_4\right)}_3$

The naming of ${\text{Pb}}_3{\left({\text{PO}}_4\right)}_3$ is lead (II) phosphate.

Rules for naming binary compound:

• In naming of binary compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.
• In the binary II compound roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state.
• When oxygen is not present in anion, then acids is named by using “hydro-” in prefix.
• If the name of anions ends in “ate-” then acid name ends in –ic or –ric
• If the name of anion ends in “-ite”, then the name of acid ends in –ous.

In the binary II compound ${\text{Pb}}_3{\left({\text{PO}}_4\right)}_3$ the cation is lead $\left({\text{Pb}}^{\text{3+}}\right)$ having oxidation state $+2$ and anion is phosphate $\left({\text{PO}}_4{}^{3-}\right)$

Hence, the name of given compound is lead (II) phosphate.

Conclusion

The naming of ${\text{Pb}}_3{\left({\text{PO}}_4\right)}_3$ is lead (II) phosphate.

(f)

Interpretation Introduction

Interpretation:

The naming of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 84E

The naming of ${\text{KClO}}_{\text{3}}$ is potassium chlorate.

Explanation of Solution

To determine: The  naming of ${\text{KClO}}_{\text{3}}$

The naming of ${\text{KClO}}_{\text{3}}$ is potassium chlorate.

Rules for naming binary compound:

• In naming of binary compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Polyatomic anions are named with the ending -ate.

In the binary compound ${\text{KClO}}_{\text{3}}$ the cation is potassium $\left({\text{K}}^{\text{+}}\right)$ anion is chlorate $\left({\text{ClO}}_3{}^{-}\right)$

Hence, the name of given compound is potassium chlorate.

Conclusion

The naming of ${\text{KClO}}_{\text{3}}$ is potassium chlorate.

(g)

Interpretation Introduction

Interpretation:

The naming of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 84E

The naming of ${\text{H}}_2{\text{SO}}_4$ is tetra sulfuric acid.

Explanation of Solution

To determine: The  naming of ${\text{H}}_2{\text{SO}}_4$

The naming of ${\text{H}}_2{\text{SO}}_4$ is tetra sulfuric acid.

Rules for naming binary compound:

• In naming of binary compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.
• In the binary II compound roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state.
• When oxygen is not present in anion, then acids is named by using “hydro-” in prefix.
• If the name of anions ends in “ate-” then acid name ends in –ic or –ric
• If the name of anion ends in “-ite”, then the name of acid ends in –ous.

In the binary acidic compound ${\text{H}}_2{\text{SO}}_4$ the cation is hydrogen $\left({\text{H}}^{\text{+}}\right)$ and anion is sulphate $\left({\text{SO}}_4{}^{2-}\right)$

Hence, the name of given compound is sulfuric acid.

Conclusion

The naming of ${\text{H}}_2{\text{SO}}_4$ is tetra sulfuric acid.

(h)

Interpretation Introduction

Interpretation:

The naming of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 84E

The naming of ${\text{Sr}}_3{\text{N}}_2$ is strontium nitride.

Explanation of Solution

To determine: The  naming of ${\text{Sr}}_3{\text{N}}_2$

The naming of ${\text{Sr}}_3{\text{N}}_2$ is strontium nitride.

Rules for naming binary compound:

• In naming of binary compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monoatomic anions are named with the ending -ide.

In the binary compound ${\text{Sr}}_3{\text{N}}_2$ the cation is strontium $\left({\text{Sr}}^{\text{2+}}\right)$ anion is nitrogen $\left({\text{N}}^{3-}\right)$

Hence, the name of given compound is strontium nitride.

Conclusion

The naming of ${\text{Sr}}_3{\text{N}}_2$ is strontium nitride.

(i)

Interpretation Introduction

Interpretation:

The naming of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 84E

The naming of ${\text{Al}}_2{\left({\text{SO}}_3\right)}_3$ is alumunium sulfite.

Explanation of Solution

To determine: The  naming of ${\text{Al}}_2{\left({\text{SO}}_3\right)}_3$

The naming of ${\text{Al}}_2{\left({\text{SO}}_3\right)}_3$ is alumunium sulfite.

Rules for naming binary compound:

• In naming of binary compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Polyatomic anions are named with the ending -ite.

In the binary compound ${\text{Al}}_2{\left({\text{SO}}_3\right)}_3$ the cation is alumunium $\left({\text{Al}}^{3+}\right)$ anion is sulfite $\left({\text{SO}}_3{}^{2-}\right)$

Hence, the name of given compound is alumunium sulfite.

Conclusion

The naming of ${\text{Al}}_2{\left({\text{SO}}_3\right)}_3$ is alumunium sulfite.

(j)

Interpretation Introduction

Interpretation:

The naming of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 84E

The naming of ${\text{SnO}}_2$ is tin (IV) oxide.

Explanation of Solution

To determine: The  naming of ${\text{SnO}}_2$

The naming of ${\text{SnO}}_2$ is tin (IV) oxide.

Rules for naming binary compound:

• In naming of binary compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.
• In the binary II compound roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state.

In the binary II compound ${\text{SnO}}_2$ the cation is tin $\left({\text{Sn}}^{\text{+}}\right)$ having oxidation state $+4$ and anion is oxygen $\left({\text{O}}^{2-}\right)$

Hence, the name of given compound is tin (IV) oxide.

Conclusion

The naming of ${\text{SnO}}_2$ is tin (IV) oxide.

(k)

Interpretation Introduction

Interpretation:

The naming of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 84E

The naming of ${\text{Na}}_2{\text{CrO}}_4$ is sodium chromate.

Explanation of Solution

To determine: The  naming of ${\text{Na}}_2{\text{CrO}}_4$

The naming of ${\text{Na}}_2{\text{CrO}}_4$ is sodium chromate.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygens ends in -ite, and the anion with the most ends in -ate.

Here sodium ion $\left({\text{Na}}^{+}\right)$ is cation and chromate ion $\left({\text{CrO}}_4{}^{\text{2-}}\right)$ is anion.

Hence the name of ${\text{Na}}_2{\text{CrO}}_4$ is sodium chromate.

Conclusion

The naming of ${\text{Na}}_2{\text{CrO}}_4$ is sodium chromate.

 (l)

Interpretation Introduction

Interpretation:

The naming of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 84E

The naming of $\text{HClO}$ is hypochlorous acid.

Explanation of Solution

To determine: The  naming of $\text{HClO}$

The naming of $\text{HClO}$ is hypochlorous acid.

Rules for naming binary compound:

• In naming of binary compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Polyatomic anions are named with the ending -ate.
• If the name of anions ends in “ate-” then acid name ends in –ic or –ric
• If the name of anion ends in “-ite”, then the name of acid ends in –ous.
• If more than two oxyanion is present in compound then “hypo-” (less than) and “per-” (more than) is used in prefix.

Hence, the name of given compound is hypochlorous acid.

Conclusion

The naming of $\text{HClO}$ is hypochlorous acid.