Chapter 3, Problem 117E

(a)

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.

To determine: The formula of of sulfur difluoride

Explanation of Solution

Sulfur belongs to the Group $\text{VI}$, the oxidation state of sulfur is $+2$.

Fluorine is a non-metal of the $7^{\text{th}}$ group.

Oxidation state $\begin{array}{l}=\text{Group}\text{number}-8\\ =7-8\\ =-1\end{array}$

In ${\text{SF}}_{\text{2}}$ sulfur is cation and fluorine is mono-atomic anion. Here the oxidation state of sulfur is $+2$ because total charge on anion is $-2$. Since compound should be electrically neutral, the formula of the given compound is ${\text{SF}}_{\text{2}}$

Conclusion

The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.

(b)

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.

To determine: The formula of sulfur hexafluoride

Explanation of Solution

Sulfur belongs to the Group $\text{VI}$, the oxidation state of sulfur is $+2,+4,+6$.

Fluorine is a non-metal of the $7^{\text{th}}$ group.

Oxidation state $\begin{array}{l}=\text{Group}\text{number}-8\\ =7-8\\ =-1\end{array}$

In ${\text{SF}}_6$ sulfur is cation and fluorine is mono-atomic anion. Here the oxidation state of sulfur is $+6$ because total charge on anion is $-6$. Since compound should be electrically neutral, the formula of the given compound is ${\text{SF}}_6$

Conclusion

The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.

(c)

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.

To determine: The formula of ${\text{NaH}}_2{\text{PO}}_4$

Explanation of Solution

Sodium belongs to the Group $\text{I}$, hence the oxidation state of sodium is $+1$.

Dihydrogen phosphate is an anion having oxidation state $-1$

So, the formula of the given compound is ${\text{NaH}}_2{\text{PO}}_4$

Conclusion

The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.

(d)

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.

To determine: The formula of lithium nitride

Explanation of Solution

Lithium belongs to the Group $\text{I}$, the oxidation state of lithium is $+1$.

Nitrogen is a non-metal of the ${\text{V}}^{\text{th}}$ group.

Oxidation state $\begin{array}{l}=\text{Group}\text{number}-8\\ =5-8\\ =-3\end{array}$

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

Conclusion

The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.

(e)

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.

To determine: The formula of chromium (III) carbonate.

Explanation of Solution

Chromium belongs to the Group $\text{VI}$, the oxidation state of chromium is $+3$.

Carbonate is an anion having oxidation state $-2$

So, the formula of the given compound is ${\text{Cr}}_{\text{2}}{\left({\text{CO}}_{\text{3}}\right)}_{\text{3}}$

Conclusion

The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.

(f)

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.

To determine: The formula of tin (II) fluoride.

Explanation of Solution

Tin belongs to the Group $\text{14}$, the oxidation state of tin is $+2$.

Fluorine is a non-metal of the $7^{\text{th}}$ group.

Oxidation state $\begin{array}{l}=\text{Group}\text{number}-8\\ =7-8\\ =-1\end{array}$

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

Conclusion

The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.

(g)

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.

To determine: The formula of ammonium acetate.

Explanation of Solution

The oxidation state of ammonium $\left({\text{NH}}_{\text{4}}\right)$ is $+1$.

Acetate $\left({\text{CH}}_{\text{3}}\text{COO}\right)$ is an anion having oxidation state $-1$

So, the formula of the given compound is ${\text{CH}}_{\text{3}}{\text{COONH}}_{\text{4}}$

Conclusion

The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.

(h)

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.

To determine: The formula of ammonium hydrogen sulfate.

Explanation of Solution

The oxidation state of ammonium $\left({\text{NH}}_{\text{4}}\right)$ is $+1$.

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

So, the formula of the given compound is ${\text{NH}}_{\text{4}}{\text{HSO}}_{\text{4}}$

Conclusion

The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.

(i)

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.

To determine: The formula of cobalt (III) nitrate.

Explanation of Solution

The oxidation state of cobalt $\left(\text{Co}\right)$ is $+3$.

Nitrate $\left({\text{NO}}_{\text{3}}\right)$ is an anion having oxidation state $-1$

So, the formula of the given compound is $\text{Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$

Conclusion

The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.

(j)

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.

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

Explanation of Solution

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

Chloride $\left(\text{Cl}\right)$ is an anion having oxidation state $-1$

Two mercury and chlorine ions is joined to form mercury (I) chloride.

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

Conclusion

The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.

(k)

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.

To determine: The formula of potassium chlorate.

Explanation of Solution

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

Chlorate $\left({\text{ClO}}_3\right)$ is an anion having oxidation state $-1$

So, the formula of the given compound is ${\text{KClO}}_{\text{3}}$

Conclusion

The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.

(l)

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.

To determine: The formula of sodium hydride.

Explanation of Solution

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

Hydride $\left(\text{H}\right)$ is an anion having oxidation state $-1$

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

Conclusion

The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.