Chapter 2, Problem 87E

(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.

Answer to Problem 87E

The formula of sulfur difluoride is ${\text{SF}}_{\text{2}}.$

Explanation of Solution

To determine: The formula of of sulfur difluoride

The formula of sulfur difluoride is ${\text{SF}}_{\text{2}}$

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 ${\text{SF}}_{\text{2}}$

(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.

Answer to Problem 87E

The formula of sulfur hexafluoride is ${\text{SF}}_6$

Explanation of Solution

To determine: The formula of sulfur hexafluoride

The formula of sulfur hexafluoride is ${\text{SF}}_6$

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 ${\text{SF}}_6.$

(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.

Answer to Problem 87E

The formula of ${\text{NaH}}_2{\text{PO}}_4$ is sodium dihydrogen phosphate.

Explanation of Solution

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

The formula of ${\text{NaH}}_2{\text{PO}}_4$ is sodium dihydrogen phosphate.

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 ${\text{NaH}}_2{\text{PO}}_4$

(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.

Answer to Problem 87E

The formula of lithium nitride is ${\text{Li}}_3\text{N}$

Explanation of Solution

To determine: The formula of lithium nitride

The formula of lithium nitride is ${\text{Li}}_3\text{N}$

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 ${\text{Li}}_3\text{N}$

(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.

Answer to Problem 87E

The formula of chromium (III) carbonate is ${\text{Cr}}_{\text{2}}{\left({\text{CO}}_{\text{3}}\right)}_{\text{3}}$

Explanation of Solution

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

The formula of chromium (III) carbonate is ${\text{Cr}}_{\text{2}}{\left({\text{CO}}_{\text{3}}\right)}_{\text{3}}$

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 ${\text{Cr}}_{\text{2}}{\left({\text{CO}}_{\text{3}}\right)}_{\text{3}}$

(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.

Answer to Problem 87E

The formula of tin (II) fluoride is ${\text{SnF}}_2$

Explanation of Solution

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

The formula of tin (II) fluoride is ${\text{SnF}}_2$

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 ${\text{SnF}}_2$

(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.

Answer to Problem 87E

The formula of ammonium acetate is ${\text{CH}}_{\text{3}}{\text{COONH}}_{\text{4}}$

Explanation of Solution

To determine: The formula of ammonium acetate.

The formula of ammonium acetate is ${\text{CH}}_{\text{3}}{\text{COONH}}_{\text{4}}$

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 ${\text{CH}}_{\text{3}}{\text{COONH}}_{\text{4}}$

(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.

Answer to Problem 87E

The formula of ammonium hydrogen sulfate is ${\text{NH}}_{\text{4}}{\text{HSO}}_{\text{4}}$

Explanation of Solution

To determine: The formula of ammonium hydrogen sulfate.

The formula of ammonium hydrogen sulfate is ${\text{NH}}_{\text{4}}{\text{HSO}}_{\text{4}}$

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 ${\text{NH}}_{\text{4}}{\text{HSO}}_{\text{4}}$

(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.

Answer to Problem 87E

The formula of cobalt (III) nitrate.is $\text{Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$

Explanation of Solution

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

The formula of cobalt (III) nitrate.is $\text{Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$

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 $\text{Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$

(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.

Answer to Problem 87E

The formula of mercury (I) chloride is ${\text{Hg}}_2{\text{Cl}}_2$

Explanation of Solution

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

The formula of mercury (I) chloride is ${\text{Hg}}_2{\text{Cl}}_2$

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 ${\text{Hg}}_2{\text{Cl}}_2$

(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.

Answer to Problem 87E

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

Explanation of Solution

To determine: The formula of potassium chlorate.

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

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 ${\text{KClO}}_{\text{3}}$

(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.

Answer to Problem 87E

The formula of sodium hydride is $\text{NaH}$

Explanation of Solution

To determine: The formula of sodium hydride.

The formula of sodium hydride is $\text{NaH}$

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 $\text{NaH}$