Chapter 2, Problem 89E

(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 89E

The formula of sodium oxide is ${\text{Na}}_2\text{O}$

Explanation of Solution

To determine: The formula of sodium oxide.

The formula of sodium oxide is ${\text{Na}}_2\text{O}$

The oxidation state of sodium is $+1$.

Oxygen is an anion having oxidation state $-2$

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

(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 89E

The formula of sodium peroxide is ${\text{Na}}_2{\text{O}}_2$

Explanation of Solution

To determine: The formula of sodium peroxide.

The formula of sodium peroxide is ${\text{Na}}_2{\text{O}}_2$

The oxidation state of sodium is $+1$.

Oxygen is an anion having oxidation state $-1$. It is exception case which occurs only in peroxide condition.

Sodium peroxide is formed when two sodium and oxygen atoms is combined together.

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

(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 89E

The formula of potassium cyanide is $\text{KCN}$

Explanation of Solution

To determine: The formula of potassium cyanide

The formula of potassium cyanide is $\text{KCN}$

The oxidation state of potassium is $+1$

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

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

(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 89E

The formula of copper (II) nitrate is $\text{Cu}{\left({\text{NO}}_3\right)}_2$

Explanation of Solution

To determine: The formula of copper (II) nitrate

The formula of copper (II) nitrate is $\text{Cu}{\left({\text{NO}}_3\right)}_2$

The oxidation state of copper is $+2$.

Nitrate ${\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is an anion having oxidation state $-1$

So, the formula of the given compound is $\text{Cu}{\left({\text{NO}}_3\right)}_2$

(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 89E

The formula of selenium tetrabromide is ${\text{SeBr}}_4$

Explanation of Solution

To determine: The formula of selenium tetrabromide.

The formula of selenium tetrabromide is ${\text{SeBr}}_4$

The oxidation state of selenium is $+4$.

Bromine is an anion having oxidation state $-1$

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

 (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 89E

The formula of iodous acid is ${\text{HIO}}_2$

Explanation of Solution

To determine: The formula of iodous acid

The formula of iodous acid is ${\text{HIO}}_2$

The oxidation state of hydrogen cation is $+1$.

Iodite $\left({\text{IO}}_{\text{2}}\right)$ is anion having oxidation state $-1$

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

(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 89E

The formula of lead (IV) sulfide is ${\text{PbS}}_2$

Explanation of Solution

To determine: The formula of of lead (IV) sulfide.

The formula of lead (IV) sulfide is ${\text{PbS}}_2$

The oxidation state of lead $\left(\text{Pb}\right)$ is $+4$.

Sulfide is an anion having oxidation state $-2$

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

(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 89E

The formula of copper (I) chloride is $\text{CuCl}$

Explanation of Solution

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

The formula of copper (I) chloride is $\text{CuCl}$

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

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

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

 (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 89E

The formula of gallium arsenide is $\text{GaAs}$

Explanation of Solution

To determine: The formula of gallium arsenide

The formula of gallium arsenide is $\text{GaAs}$

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

Arsenic $\left(\text{As}\right)$ is an anion having oxidation state $+3$

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

 (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 89E

The formula of cadmium selenide is $\text{CdSe}$

Explanation of Solution

To determine: The formula of cadmium selenide.

The formula of cadmium selenide is $\text{CdSe}$

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

Selenium $\left(\text{Se}\right)$ is an anion having oxidation state $-2$

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

 (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 89E

The formula of zinc sulfide is $\text{ZnS}$

Explanation of Solution

To determine: The formula of of zinc sulfide.

The formula of zinc sulfide is $\text{ZnS}$

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

Sulfide is an anion having oxidation state $-2$

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

(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 89E

The formula of nitrous acid is ${\text{HNO}}_2$

Explanation of Solution

To determine: The formula of of nitrous acid.

The formula of nitrous acid is ${\text{HNO}}_2$

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

Nitrous $\left({\text{NO}}_{\text{2}}\right)$ is an anion having oxidation state $-1$

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

(m)

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 89E

The formula of diphosphorous pentaoxide is ${\text{P}}_2{\text{O}}_5$

Explanation of Solution

To determine: The formula of of diphosphorous pentaoxide.

The formula of diphosphorous pentaoxide is ${\text{P}}_2{\text{O}}_5$

The oxidation state of phosphorous $\left(\text{P}\right)$ is $+5$.

Oxygen is an anion having oxidation state $-2$

So, the formula of the given compound is ${\text{P}}_2{\text{O}}_5$