Chapter 2, Problem 86E

Write the formula for each of the following compounds:

1. a. ammonium hydrogen phosphate
2. b. mercury (I) sulfide
3. c. silicon dioxide
4. d. sodium sulfite
5. e. aluminum hydrogen sulfate
6. f. nitrogen trichloride
7. g. hydiobromic acid
8. h. bromous acid
9. i. perbromic acid
10. j. potassium hydrogen sulfide
11. k. calcium iodide
12. l. cesium perchlorate

Interpretation Introduction

Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as, $\text{group number}-8$. While naming an ionic compound, the cation is named first followed by the naming of the anion. If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 86E

The formula of ammonium hydrogen phosphate is ${\left({\text{NH}}_4\right)}_2{\text{HPO}}_4$

Explanation of Solution

To determine: The formula of ammonium hydrogen phosphate.

The oxidation state of ammonium $\left({\text{NH}}_4\right)$ is $+1$.

Hydrogen phosphate ${\text{HPO}}_4$  is an anion having oxidation state $-1$

So, the formula of the given compound is ${\left({\text{NH}}_4\right)}_2{\text{HPO}}_4$

Conclusion

The formula of ammonium hydrogen phosphate is ${\left({\text{NH}}_4\right)}_2{\text{HPO}}_4$

Interpretation Introduction

Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as, $\text{group number}-8$. While naming an ionic compound, the cation is named first followed by the naming of the anion. If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 86E

The formula of mercury (I) sulfide is ${\text{Hg}}_2\text{S}$

Explanation of Solution

To determine: The formula of mercury (I) sulfide.

The oxidation state of mercury is $+1$.

Sulfide is an anion having oxidation state $-2$.

So, the formula of the given compound is ${\text{Hg}}_2\text{S}$

Conclusion

The formula of mercury(I) sulfide is ${\text{Hg}}_2\text{S}$

Interpretation Introduction

Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as, $\text{group number}-8$. While naming an ionic compound, the cation is named first followed by the naming of the anion. If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 86E

The formula of silicon dioxide is ${\text{SiO}}_2$

Explanation of Solution

To determine: The formula of silicon dioxide

The oxidation state of silicon is $+2$

Dioxide is an anion having oxidation state $-1$

So, the formula of the given compound is ${\text{SiO}}_2$

Conclusion

The formula of silicon dioxide is ${\text{SiO}}_2$

Interpretation Introduction

Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as, $\text{group number}-8$. While naming an ionic compound, the cation is named first followed by the naming of the anion. If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 86E

The formula of sodium sulfite is ${\text{Na}}_2\left({\text{SO}}_3\right)$

Explanation of Solution

To determine: The formula of sodium sulfite

The oxidation state of sodium is $+1$.

Sulfite $\left({\text{SO}}_{\text{3}}\right)$ is an anion having oxidation state $-2$

So, the formula of the given compound is ${\text{Na}}_2\left({\text{SO}}_3\right)$

Conclusion

The formula of sodium sulfite is ${\text{Na}}_2\left({\text{SO}}_3\right)$

Interpretation Introduction

Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as, $\text{group number}-8$. While naming an ionic compound, the cation is named first followed by the naming of the anion. If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 86E

The formula of alumunium hydrogen sulfate is $\text{Al}{\left({\text{HSO}}_{\text{4}}\right)}_3$

Explanation of Solution

To determine: The formula of alumunium hydrogen sulfate.

The oxidation state of alumunium is $+3$.

Hydrogen sulfate $\left({\text{HSO}}_{\text{4}}\right)$ is an anion having oxidation state $-1$

So, the formula of the given compound is $\text{Al}{\left({\text{HSO}}_{\text{4}}\right)}_3$

Conclusion

The formula of alumunium hydrogen sulfate is $\text{Al}{\left({\text{HSO}}_{\text{4}}\right)}_3$

Interpretation Introduction

Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as, $\text{group number}-8$. While naming an ionic compound, the cation is named first followed by the naming of the anion. If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 86E

The formula of nitrogen trichloride is ${\text{NCl}}_3$

Explanation of Solution

To determine: The formula of nitrogen trichloride.

The oxidation state of nitrogen is $-3$.

The oxidation state of chlorine is $-1$

So, the formula of the given compound is ${\text{NCl}}_3$

Conclusion

The formula of nitrogen trichloride is ${\text{NCl}}_3$

Interpretation Introduction

Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as, $\text{group number}-8$. While naming an ionic compound, the cation is named first followed by the naming of the anion. If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 86E

The formula of hydrobromic acid is $\text{HBr}$

Explanation of Solution

To determine: The formula of of hydrobromic acid

The oxidation state of hydrogen $\left(\text{H}\right)$ is $+1$.

Bromine is an anion having oxidation state $-1$

So, the formula of the given compound is $\text{HBr}$

Conclusion

The formula of hydrobromic acid is $\text{HBr}$

Interpretation Introduction

Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as, $\text{group number}-8$. While naming an ionic compound, the cation is named first followed by the naming of the anion. If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 86E

The formula of bromous acid is ${\text{HBrO}}_2$

Explanation of Solution

To determine: The formula of bromous acid

The oxidation state of hydrogen $\left(\text{H}\right)$ is $+1$.

Bromite $\left({\text{BrO}}_2\right)$ is an anion having oxidation state $-1$

So, the formula of the given compound is ${\text{HBrO}}_2$

Conclusion

The formula of bromous acid is ${\text{HBrO}}_2$

Interpretation Introduction

Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as, $\text{group number}-8$. While naming an ionic compound, the cation is named first followed by the naming of the anion. If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 86E

The formula of perbromic acid is ${\text{HBrO}}_4$

Explanation of Solution

To determine: The formula of perbromic acid

The oxidation state of hydrogen $\left(\text{H}\right)$ is $+1$.

Perbromate $\left({\text{BrO}}_4\right)$ is an anion having oxidation state $-1$

So, the formula of the given compound is ${\text{HBrO}}_4$

Conclusion

The formula of perbromic acid is ${\text{HBrO}}_4$

Interpretation Introduction

Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as, $\text{group number}-8$. While naming an ionic compound, the cation is named first followed by the naming of the anion. If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 86E

The formula of potassium hydrogen sulfide is $\text{KHS}$

Explanation of Solution

To determine: The formula of potassium hydrogen sulfide

The oxidation state of potassium $\left(\text{K}\right)$ is $+1$.

Hydrogen sulfide is an anion having oxidation state $-1$

So, the formula of the given compound is $\text{KHS}$

Conclusion

The formula of potassium hydrogen sulfide is $\text{KHS}$

Interpretation Introduction

Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as, $\text{group number}-8$. While naming an ionic compound, the cation is named first followed by the naming of the anion. If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 86E

The formula of calcium iodide is ${\text{CaI}}_2$

Explanation of Solution

To determine: The formula of of calcium iodide

The oxidation state of calcium $\left(\text{Ca}\right)$ is $+2$.

Iodine is an anion having oxidation state $-1$

So, the formula of the given compound is ${\text{CaI}}_2$

Conclusion

The formula of calcium iodide is ${\text{CaI}}_2$

Interpretation Introduction

Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as, $\text{group number}-8$. While naming an ionic compound, the cation is named first followed by the naming of the anion. If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 86E

The formula of cesium perchlorate is ${\text{CsClO}}_4$

Explanation of Solution

To determine: The formula of of cesium perchlorate.

The oxidation state of cesium $\left(\text{Cs}\right)$ is $+1$.

Perchlorate $\left({\text{ClO}}_4\right)$ is an anion having oxidation state $-1$

So, the formula of the given compound is ${\text{CsClO}}_4$

Conclusion

The formula of cesium perchlorate is ${\text{CsClO}}_4$