Chapter 2, Problem 79E

(a)

Interpretation Introduction

Interpretation:

The name 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 79E

The naming of $\text{CuI}$ is copper (I) iodide.

Explanation of Solution

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

The name of $\text{CuI}$ is copper(I) iodide.

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. 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. Copper exhibits $+1\text{and}+2$ oxidation state.

Here copper $\left({\text{Cu}}^{2+}\right)$ ion is cation and iodine $\left({\text{I}}^{-}\right)$ ion is mono-atomic anion. The oxidation state of copper is $+1$ because total charge on anion is $-1$. Since compound should be electrically neutral, the name of $\text{CuI}$ compound is copper (I) iodide.

Conclusion

The name of $\text{CuI}$ is copper(I) iodide.

(b)

Interpretation Introduction

Interpretation:

The name 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 79E

The naming of ${\text{CuI}}_2$ is copper (II) iodide.

Explanation of Solution

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

The name of ${\text{CuI}}_2$ is copper(II) iodide.

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. 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. Copper exhibits $+1\text{and}+2$ oxidation state.

Here copper $\left({\text{Cu}}^{2+}\right)$ ion is cation and iodine $\left({\text{I}}^{-}\right)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $-2$. Since compound should be electrically neutral, the name of ${\text{CuI}}_2$ compound is copper (II) iodide.

Conclusion

The name of ${\text{CuI}}_2$ is copper(II) iodide.

(c)

Interpretation Introduction

Interpretation:

The name 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 79E

The naming of ${\text{CuI}}_2$ is cobalt (II) iodide.

Explanation of Solution

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

The name of ${\text{CuI}}_2$ is cobalt (II) iodide.

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. 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. Cobalt exhibits $+2\text{and}+3$ oxidation state.

Here cobalt $\left({\text{Cu}}^{2+}\right)$ ion is cation and iodine $\left({\text{I}}^{\text{-}}\right)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $-2$. Since compound should be electrically neutral, the name of ${\text{CoI}}_2$ compound is cobalt (II) iodide.

Conclusion

The name of ${\text{CuI}}_2$ is cobalt (II) iodide.

(d)

Interpretation Introduction

Interpretation:

The name 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 79E

The naming of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is sodium carbonate

Explanation of Solution

To determine: The  naming of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$

The name of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is sodium carbonate

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 ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $\left({\text{Na}}^{+}\right)$ is cation and carbonate ion $\left({\text{CO}}_3{}^{\text{2-}}\right)$ is anion. The total charge on cation is $+2$ and total charge on anion is $-2$. Since compound should be electrically neutral, the name of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ compound is sodium carbonate

Conclusion

The name of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is sodium carbonate

(e)

Interpretation Introduction

Interpretation:

The name 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 79E

The naming of ${\text{NaHCO}}_{\text{3}}$ is sodium bicarbonate.

Explanation of Solution

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

The name of ${\text{NaHCO}}_{\text{3}}$ is sodium bicarbonate

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 ${\text{NaHCO}}_{\text{3}}$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $\left({\text{Na}}^{+}\right)$ is cation and bicarbonate ion $\left({\text{HCO}}_3{}^{-}\right)$ is anion. The total charge on cation is $+1$ and total charge on anion is $-1$. Since compound should be electrically neutral, the name of ${\text{NaHCO}}_{\text{3}}$ compound is sodium bicarbonate

Conclusion

The name of ${\text{NaHCO}}_{\text{3}}$ is sodium bicarbonate

(f)

Interpretation Introduction

Interpretation:

The name 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 79E

The naming of ${\text{S}}_4{\text{N}}_{\text{4}}$ is tetra sulfur tetra nitride.

Explanation of Solution

To determine: The  naming of ${\text{S}}_4{\text{N}}_{\text{4}}$

The name of ${\text{S}}_4{\text{N}}_{\text{4}}$ is tetra sulfur tetra nitride.

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 prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $\left({\text{S}}^{2-}\right)$ is cation having four atoms and nitrogen ion $\left({\text{N}}^{\text{3-}}\right)$ is poly-atomic anion having four atoms. Hence the name of ${\text{S}}_4{\text{N}}_{\text{4}}$ compound is tetra sulfur tetra nitride.

Conclusion

The name of ${\text{S}}_4{\text{N}}_{\text{4}}$ is tetra sulfur tetra nitride.

(g)

Interpretation Introduction

Interpretation:

The name 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 79E

The naming of ${\text{SF}}_6$ is tetra sulfur tetra nitride.

Explanation of Solution

To determine: The  naming of ${\text{SF}}_6$

The name of ${\text{SF}}_6$ is tetra sulfur tetra nitride.

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 prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $\left({\text{S}}^{2-}\right)$ is cation having single atom and fluorine ion $\left({\text{F}}^{\text{-}}\right)$ is mono-atomic anion having six atoms. Hence the name of ${\text{SF}}_6$ compound is sulfur hexafluoride.

Conclusion

The name of ${\text{SF}}_6$ is tetra sulfur tetra nitride.

(h)

Interpretation Introduction

Interpretation:

The name 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 79E

The naming of $\text{NaOCl}$ is sodium Hypochlorite.

Explanation of Solution

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

The name of $\text{NaOCl}$ is sodium Hypochlorite.

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 oxygen’s ends in -ite, and the anion with the most ends in -ate.

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 $\text{NaOCl}$ compound is sodium Hypochlorite.

Conclusion

The name of $\text{NaOCl}$ is sodium Hypochlorite.

(i)

Interpretation Introduction

Interpretation:

The name 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 79E

The naming of ${\text{BaCrO}}_4$ is barium chromate.

Explanation of Solution

To determine: The  naming of ${\text{BaCrO}}_4$

The name of ${\text{BaCrO}}_4$ is barium 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 oxygen ends in -ite, and the anion with the most ends in -ate.

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

Hence the name of ${\text{BaCrO}}_4$ is barium chromate.

Conclusion

The name of ${\text{BaCrO}}_4$ is barium chromate.

(j)

Interpretation Introduction

Interpretation:

The name 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 79E

The naming of ${\text{NH}}_4{\text{NO}}_{\text{3}}$ is ammonium nitrate.

Explanation of Solution

To determine: The  naming of ${\text{NH}}_4{\text{NO}}_{\text{3}}$

The name of ${\text{NH}}_4{\text{NO}}_{\text{3}}$ is ammonium nitrate.

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 oxygen ends in -ite, and the anion with the most ends in -ate.

Here ammonium $\left({\text{NH}}_4{}^{+}\right)$ is cation and nitratee ion $\left({\text{NO}}_3{}^{\text{-}}\right)$ is anion.

Hence the name of ${\text{NH}}_4{\text{NO}}_{\text{3}}$ is ammonium nitrate.

Conclusion

The name of ${\text{NH}}_4{\text{NO}}_{\text{3}}$ is ammonium nitrate.