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, 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.

Question

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Answer 1

Question

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, $group number−8$ . 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.

(a)

Expert Solution

Answer to Problem 87E

The formula of sulfur difluoride is $SF2.$

Explanation of Solution

To determine: The formula of of sulfur difluoride

The formula of sulfur difluoride is $SF2$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

In $SF2$ 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 $SF2$

(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, $group number−8$ . 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.

(b)

Expert Solution

Answer to Problem 87E

The formula of sulfur hexafluoride is $SF6$

Explanation of Solution

To determine: The formula of sulfur hexafluoride

The formula of sulfur hexafluoride is $SF6$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

In $SF6$ 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 $SF6.$

(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, $group number−8$ . 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.

(c)

Expert Solution

Answer to Problem 87E

The formula of $NaH2PO4$ is sodium dihydrogen phosphate.

Explanation of Solution

To determine: The formula of $NaH2PO4$

The formula of $NaH2PO4$ is sodium dihydrogen phosphate.

Sodium belongs to the Group $I$ , 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 $NaH2PO4$

(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, $group number−8$ . 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.

(d)

Expert Solution

Answer to Problem 87E

The formula of lithium nitride is $Li3N$

Explanation of Solution

To determine: The formula of lithium nitride

The formula of lithium nitride is $Li3N$

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

Nitrogen is a non-metal of the $Vth$ group.

Oxidation state $= Group number−8= 5−8=−3$

So, the formula of the given compound is $Li3N$

(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, $group number−8$ . 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.

(e)

Expert Solution

Answer to Problem 87E

The formula of chromium (III) carbonate is $Cr2(CO3)3$

Explanation of Solution

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

The formula of chromium (III) carbonate is $Cr2(CO3)3$

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

Carbonate is an anion having oxidation state $−2$

So, the formula of the given compound is $Cr2(CO3)3$

(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, $group number−8$ . 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.

(f)

Expert Solution

Answer to Problem 87E

The formula of tin (II) fluoride is $SnF2$

Explanation of Solution

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

The formula of tin (II) fluoride is $SnF2$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

So, the formula of the given compound is $SnF2$

(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, $group number−8$ . 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.

(g)

Expert Solution

Answer to Problem 87E

The formula of ammonium acetate is $CH3COONH4$

Explanation of Solution

To determine: The formula of ammonium acetate.

The formula of ammonium acetate is $CH3COONH4$

The oxidation state of ammonium $(NH4)$ is $+1$ .

Acetate $(CH3COO)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $CH3COONH4$

(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, $group number−8$ . 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.

(h)

Expert Solution

Answer to Problem 87E

The formula of ammonium hydrogen sulfate is $NH4HSO4$

Explanation of Solution

To determine: The formula of ammonium hydrogen sulfate.

The formula of ammonium hydrogen sulfate is $NH4HSO4$

The oxidation state of ammonium $(NH4)$ is $+1$ .

Hydrogen sulfate $(HSO4)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $NH4HSO4$

(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, $group number−8$ . 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.

(i)

Expert Solution

Answer to Problem 87E

The formula of cobalt (III) nitrate.is $Co(NO3)3$

Explanation of Solution

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

The formula of cobalt (III) nitrate.is $Co(NO3)3$

The oxidation state of cobalt $(Co)$ is $+3$ .

Nitrate $(NO3)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $Co(NO3)3$

(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, $group number−8$ . 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.

(j)

Expert Solution

Answer to Problem 87E

The formula of mercury (I) chloride is $Hg2Cl2$

Explanation of Solution

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

The formula of mercury (I) chloride is $Hg2Cl2$

The oxidation state of mercury $(Hg)$ is $+1$ .

Chloride $(Cl)$ 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 $Hg2Cl2$

(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, $group number−8$ . 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.

(k)

Expert Solution

Answer to Problem 87E

The formula of potassium chlorate is $KClO3$

Explanation of Solution

To determine: The formula of potassium chlorate.

The formula of potassium chlorate is $KClO3$

The oxidation state of potassium $(K)$ is $+1$ .

Chlorate $(ClO3)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $KClO3$

(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, $group number−8$ . 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.

(l)

Expert Solution

Answer to Problem 87E

The formula of sodium hydride is $NaH$

Explanation of Solution

To determine: The formula of sodium hydride.

The formula of sodium hydride is $NaH$

The oxidation state of sodium $(Na)$ is $+1$ .

Hydride $(H)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $NaH$

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Answer 2

Question

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, $group number−8$ . 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.

(a)

Expert Solution

Answer to Problem 87E

The formula of sulfur difluoride is $SF2.$

Explanation of Solution

To determine: The formula of of sulfur difluoride

The formula of sulfur difluoride is $SF2$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

In $SF2$ 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 $SF2$

(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, $group number−8$ . 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.

(b)

Expert Solution

Answer to Problem 87E

The formula of sulfur hexafluoride is $SF6$

Explanation of Solution

To determine: The formula of sulfur hexafluoride

The formula of sulfur hexafluoride is $SF6$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

In $SF6$ 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 $SF6.$

(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, $group number−8$ . 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.

(c)

Expert Solution

Answer to Problem 87E

The formula of $NaH2PO4$ is sodium dihydrogen phosphate.

Explanation of Solution

To determine: The formula of $NaH2PO4$

The formula of $NaH2PO4$ is sodium dihydrogen phosphate.

Sodium belongs to the Group $I$ , 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 $NaH2PO4$

(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, $group number−8$ . 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.

(d)

Expert Solution

Answer to Problem 87E

The formula of lithium nitride is $Li3N$

Explanation of Solution

To determine: The formula of lithium nitride

The formula of lithium nitride is $Li3N$

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

Nitrogen is a non-metal of the $Vth$ group.

Oxidation state $= Group number−8= 5−8=−3$

So, the formula of the given compound is $Li3N$

(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, $group number−8$ . 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.

(e)

Expert Solution

Answer to Problem 87E

The formula of chromium (III) carbonate is $Cr2(CO3)3$

Explanation of Solution

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

The formula of chromium (III) carbonate is $Cr2(CO3)3$

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

Carbonate is an anion having oxidation state $−2$

So, the formula of the given compound is $Cr2(CO3)3$

(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, $group number−8$ . 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.

(f)

Expert Solution

Answer to Problem 87E

The formula of tin (II) fluoride is $SnF2$

Explanation of Solution

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

The formula of tin (II) fluoride is $SnF2$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

So, the formula of the given compound is $SnF2$

(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, $group number−8$ . 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.

(g)

Expert Solution

Answer to Problem 87E

The formula of ammonium acetate is $CH3COONH4$

Explanation of Solution

To determine: The formula of ammonium acetate.

The formula of ammonium acetate is $CH3COONH4$

The oxidation state of ammonium $(NH4)$ is $+1$ .

Acetate $(CH3COO)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $CH3COONH4$

(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, $group number−8$ . 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.

(h)

Expert Solution

Answer to Problem 87E

The formula of ammonium hydrogen sulfate is $NH4HSO4$

Explanation of Solution

To determine: The formula of ammonium hydrogen sulfate.

The formula of ammonium hydrogen sulfate is $NH4HSO4$

The oxidation state of ammonium $(NH4)$ is $+1$ .

Hydrogen sulfate $(HSO4)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $NH4HSO4$

(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, $group number−8$ . 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.

(i)

Expert Solution

Answer to Problem 87E

The formula of cobalt (III) nitrate.is $Co(NO3)3$

Explanation of Solution

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

The formula of cobalt (III) nitrate.is $Co(NO3)3$

The oxidation state of cobalt $(Co)$ is $+3$ .

Nitrate $(NO3)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $Co(NO3)3$

(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, $group number−8$ . 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.

(j)

Expert Solution

Answer to Problem 87E

The formula of mercury (I) chloride is $Hg2Cl2$

Explanation of Solution

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

The formula of mercury (I) chloride is $Hg2Cl2$

The oxidation state of mercury $(Hg)$ is $+1$ .

Chloride $(Cl)$ 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 $Hg2Cl2$

(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, $group number−8$ . 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.

(k)

Expert Solution

Answer to Problem 87E

The formula of potassium chlorate is $KClO3$

Explanation of Solution

To determine: The formula of potassium chlorate.

The formula of potassium chlorate is $KClO3$

The oxidation state of potassium $(K)$ is $+1$ .

Chlorate $(ClO3)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $KClO3$

(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, $group number−8$ . 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.

(l)

Expert Solution

Answer to Problem 87E

The formula of sodium hydride is $NaH$

Explanation of Solution

To determine: The formula of sodium hydride.

The formula of sodium hydride is $NaH$

The oxidation state of sodium $(Na)$ is $+1$ .

Hydride $(H)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $NaH$

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Answer 3

Question

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, $group number−8$ . 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.

(a)

Expert Solution

Answer to Problem 87E

The formula of sulfur difluoride is $SF2.$

Explanation of Solution

To determine: The formula of of sulfur difluoride

The formula of sulfur difluoride is $SF2$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

In $SF2$ 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 $SF2$

(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, $group number−8$ . 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.

(b)

Expert Solution

Answer to Problem 87E

The formula of sulfur hexafluoride is $SF6$

Explanation of Solution

To determine: The formula of sulfur hexafluoride

The formula of sulfur hexafluoride is $SF6$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

In $SF6$ 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 $SF6.$

(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, $group number−8$ . 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.

(c)

Expert Solution

Answer to Problem 87E

The formula of $NaH2PO4$ is sodium dihydrogen phosphate.

Explanation of Solution

To determine: The formula of $NaH2PO4$

The formula of $NaH2PO4$ is sodium dihydrogen phosphate.

Sodium belongs to the Group $I$ , 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 $NaH2PO4$

(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, $group number−8$ . 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.

(d)

Expert Solution

Answer to Problem 87E

The formula of lithium nitride is $Li3N$

Explanation of Solution

To determine: The formula of lithium nitride

The formula of lithium nitride is $Li3N$

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

Nitrogen is a non-metal of the $Vth$ group.

Oxidation state $= Group number−8= 5−8=−3$

So, the formula of the given compound is $Li3N$

(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, $group number−8$ . 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.

(e)

Expert Solution

Answer to Problem 87E

The formula of chromium (III) carbonate is $Cr2(CO3)3$

Explanation of Solution

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

The formula of chromium (III) carbonate is $Cr2(CO3)3$

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

Carbonate is an anion having oxidation state $−2$

So, the formula of the given compound is $Cr2(CO3)3$

(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, $group number−8$ . 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.

(f)

Expert Solution

Answer to Problem 87E

The formula of tin (II) fluoride is $SnF2$

Explanation of Solution

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

The formula of tin (II) fluoride is $SnF2$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

So, the formula of the given compound is $SnF2$

(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, $group number−8$ . 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.

(g)

Expert Solution

Answer to Problem 87E

The formula of ammonium acetate is $CH3COONH4$

Explanation of Solution

To determine: The formula of ammonium acetate.

The formula of ammonium acetate is $CH3COONH4$

The oxidation state of ammonium $(NH4)$ is $+1$ .

Acetate $(CH3COO)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $CH3COONH4$

(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, $group number−8$ . 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.

(h)

Expert Solution

Answer to Problem 87E

The formula of ammonium hydrogen sulfate is $NH4HSO4$

Explanation of Solution

To determine: The formula of ammonium hydrogen sulfate.

The formula of ammonium hydrogen sulfate is $NH4HSO4$

The oxidation state of ammonium $(NH4)$ is $+1$ .

Hydrogen sulfate $(HSO4)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $NH4HSO4$

(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, $group number−8$ . 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.

(i)

Expert Solution

Answer to Problem 87E

The formula of cobalt (III) nitrate.is $Co(NO3)3$

Explanation of Solution

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

The formula of cobalt (III) nitrate.is $Co(NO3)3$

The oxidation state of cobalt $(Co)$ is $+3$ .

Nitrate $(NO3)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $Co(NO3)3$

(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, $group number−8$ . 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.

(j)

Expert Solution

Answer to Problem 87E

The formula of mercury (I) chloride is $Hg2Cl2$

Explanation of Solution

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

The formula of mercury (I) chloride is $Hg2Cl2$

The oxidation state of mercury $(Hg)$ is $+1$ .

Chloride $(Cl)$ 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 $Hg2Cl2$

(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, $group number−8$ . 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.

(k)

Expert Solution

Answer to Problem 87E

The formula of potassium chlorate is $KClO3$

Explanation of Solution

To determine: The formula of potassium chlorate.

The formula of potassium chlorate is $KClO3$

The oxidation state of potassium $(K)$ is $+1$ .

Chlorate $(ClO3)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $KClO3$

(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, $group number−8$ . 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.

(l)

Expert Solution

Answer to Problem 87E

The formula of sodium hydride is $NaH$

Explanation of Solution

To determine: The formula of sodium hydride.

The formula of sodium hydride is $NaH$

The oxidation state of sodium $(Na)$ is $+1$ .

Hydride $(H)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $NaH$

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Answer 4

Question

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, $group number−8$ . 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.

(a)

Expert Solution

Answer to Problem 87E

The formula of sulfur difluoride is $SF2.$

Explanation of Solution

To determine: The formula of of sulfur difluoride

The formula of sulfur difluoride is $SF2$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

In $SF2$ 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 $SF2$

(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, $group number−8$ . 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.

(b)

Expert Solution

Answer to Problem 87E

The formula of sulfur hexafluoride is $SF6$

Explanation of Solution

To determine: The formula of sulfur hexafluoride

The formula of sulfur hexafluoride is $SF6$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

In $SF6$ 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 $SF6.$

(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, $group number−8$ . 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.

(c)

Expert Solution

Answer to Problem 87E

The formula of $NaH2PO4$ is sodium dihydrogen phosphate.

Explanation of Solution

To determine: The formula of $NaH2PO4$

The formula of $NaH2PO4$ is sodium dihydrogen phosphate.

Sodium belongs to the Group $I$ , 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 $NaH2PO4$

(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, $group number−8$ . 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.

(d)

Expert Solution

Answer to Problem 87E

The formula of lithium nitride is $Li3N$

Explanation of Solution

To determine: The formula of lithium nitride

The formula of lithium nitride is $Li3N$

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

Nitrogen is a non-metal of the $Vth$ group.

Oxidation state $= Group number−8= 5−8=−3$

So, the formula of the given compound is $Li3N$

(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, $group number−8$ . 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.

(e)

Expert Solution

Answer to Problem 87E

The formula of chromium (III) carbonate is $Cr2(CO3)3$

Explanation of Solution

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

The formula of chromium (III) carbonate is $Cr2(CO3)3$

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

Carbonate is an anion having oxidation state $−2$

So, the formula of the given compound is $Cr2(CO3)3$

(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, $group number−8$ . 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.

(f)

Expert Solution

Answer to Problem 87E

The formula of tin (II) fluoride is $SnF2$

Explanation of Solution

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

The formula of tin (II) fluoride is $SnF2$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

So, the formula of the given compound is $SnF2$

(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, $group number−8$ . 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.

(g)

Expert Solution

Answer to Problem 87E

The formula of ammonium acetate is $CH3COONH4$

Explanation of Solution

To determine: The formula of ammonium acetate.

The formula of ammonium acetate is $CH3COONH4$

The oxidation state of ammonium $(NH4)$ is $+1$ .

Acetate $(CH3COO)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $CH3COONH4$

(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, $group number−8$ . 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.

(h)

Expert Solution

Answer to Problem 87E

The formula of ammonium hydrogen sulfate is $NH4HSO4$

Explanation of Solution

To determine: The formula of ammonium hydrogen sulfate.

The formula of ammonium hydrogen sulfate is $NH4HSO4$

The oxidation state of ammonium $(NH4)$ is $+1$ .

Hydrogen sulfate $(HSO4)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $NH4HSO4$

(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, $group number−8$ . 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.

(i)

Expert Solution

Answer to Problem 87E

The formula of cobalt (III) nitrate.is $Co(NO3)3$

Explanation of Solution

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

The formula of cobalt (III) nitrate.is $Co(NO3)3$

The oxidation state of cobalt $(Co)$ is $+3$ .

Nitrate $(NO3)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $Co(NO3)3$

(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, $group number−8$ . 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.

(j)

Expert Solution

Answer to Problem 87E

The formula of mercury (I) chloride is $Hg2Cl2$

Explanation of Solution

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

The formula of mercury (I) chloride is $Hg2Cl2$

The oxidation state of mercury $(Hg)$ is $+1$ .

Chloride $(Cl)$ 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 $Hg2Cl2$

(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, $group number−8$ . 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.

(k)

Expert Solution

Answer to Problem 87E

The formula of potassium chlorate is $KClO3$

Explanation of Solution

To determine: The formula of potassium chlorate.

The formula of potassium chlorate is $KClO3$

The oxidation state of potassium $(K)$ is $+1$ .

Chlorate $(ClO3)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $KClO3$

(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, $group number−8$ . 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.

(l)

Expert Solution

Answer to Problem 87E

The formula of sodium hydride is $NaH$

Explanation of Solution

To determine: The formula of sodium hydride.

The formula of sodium hydride is $NaH$

The oxidation state of sodium $(Na)$ is $+1$ .

Hydride $(H)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $NaH$

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Answer 5

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, $group number−8$ . 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.

(a)

Expert Solution

Answer to Problem 87E

The formula of sulfur difluoride is $SF2.$

Explanation of Solution

To determine: The formula of of sulfur difluoride

The formula of sulfur difluoride is $SF2$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

In $SF2$ 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 $SF2$

(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, $group number−8$ . 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.

(b)

Expert Solution

Answer to Problem 87E

The formula of sulfur hexafluoride is $SF6$

Explanation of Solution

To determine: The formula of sulfur hexafluoride

The formula of sulfur hexafluoride is $SF6$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

In $SF6$ 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 $SF6.$

(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, $group number−8$ . 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.

(c)

Expert Solution

Answer to Problem 87E

The formula of $NaH2PO4$ is sodium dihydrogen phosphate.

Explanation of Solution

To determine: The formula of $NaH2PO4$

The formula of $NaH2PO4$ is sodium dihydrogen phosphate.

Sodium belongs to the Group $I$ , 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 $NaH2PO4$

(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, $group number−8$ . 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.

(d)

Expert Solution

Answer to Problem 87E

The formula of lithium nitride is $Li3N$

Explanation of Solution

To determine: The formula of lithium nitride

The formula of lithium nitride is $Li3N$

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

Nitrogen is a non-metal of the $Vth$ group.

Oxidation state $= Group number−8= 5−8=−3$

So, the formula of the given compound is $Li3N$

(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, $group number−8$ . 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.

(e)

Expert Solution

Answer to Problem 87E

The formula of chromium (III) carbonate is $Cr2(CO3)3$

Explanation of Solution

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

The formula of chromium (III) carbonate is $Cr2(CO3)3$

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

Carbonate is an anion having oxidation state $−2$

So, the formula of the given compound is $Cr2(CO3)3$

(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, $group number−8$ . 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.

(f)

Expert Solution

Answer to Problem 87E

The formula of tin (II) fluoride is $SnF2$

Explanation of Solution

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

The formula of tin (II) fluoride is $SnF2$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

So, the formula of the given compound is $SnF2$

(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, $group number−8$ . 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.

(g)

Expert Solution

Answer to Problem 87E

The formula of ammonium acetate is $CH3COONH4$

Explanation of Solution

To determine: The formula of ammonium acetate.

The formula of ammonium acetate is $CH3COONH4$

The oxidation state of ammonium $(NH4)$ is $+1$ .

Acetate $(CH3COO)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $CH3COONH4$

(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, $group number−8$ . 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.

(h)

Expert Solution

Answer to Problem 87E

The formula of ammonium hydrogen sulfate is $NH4HSO4$

Explanation of Solution

To determine: The formula of ammonium hydrogen sulfate.

The formula of ammonium hydrogen sulfate is $NH4HSO4$

The oxidation state of ammonium $(NH4)$ is $+1$ .

Hydrogen sulfate $(HSO4)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $NH4HSO4$

(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, $group number−8$ . 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.

(i)

Expert Solution

Answer to Problem 87E

The formula of cobalt (III) nitrate.is $Co(NO3)3$

Explanation of Solution

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

The formula of cobalt (III) nitrate.is $Co(NO3)3$

The oxidation state of cobalt $(Co)$ is $+3$ .

Nitrate $(NO3)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $Co(NO3)3$

(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, $group number−8$ . 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.

(j)

Expert Solution

Answer to Problem 87E

The formula of mercury (I) chloride is $Hg2Cl2$

Explanation of Solution

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

The formula of mercury (I) chloride is $Hg2Cl2$

The oxidation state of mercury $(Hg)$ is $+1$ .

Chloride $(Cl)$ 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 $Hg2Cl2$

(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, $group number−8$ . 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.

(k)

Expert Solution

Answer to Problem 87E

The formula of potassium chlorate is $KClO3$

Explanation of Solution

To determine: The formula of potassium chlorate.

The formula of potassium chlorate is $KClO3$

The oxidation state of potassium $(K)$ is $+1$ .

Chlorate $(ClO3)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $KClO3$

(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, $group number−8$ . 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.

(l)

Expert Solution

Answer to Problem 87E

The formula of sodium hydride is $NaH$

Explanation of Solution

To determine: The formula of sodium hydride.

The formula of sodium hydride is $NaH$

The oxidation state of sodium $(Na)$ is $+1$ .

Hydride $(H)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $NaH$

Answer 6

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, $group number−8$ . 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.

(a)

Expert Solution

Answer to Problem 87E

The formula of sulfur difluoride is $SF2.$

Explanation of Solution

To determine: The formula of of sulfur difluoride

The formula of sulfur difluoride is $SF2$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

In $SF2$ 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 $SF2$

Answer 7

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, $group number−8$ . 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 8

(a)

Expert Solution

Answer to Problem 87E

The formula of sulfur difluoride is $SF2.$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

To determine: The formula of of sulfur difluoride

The formula of sulfur difluoride is $SF2$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

In $SF2$ 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 $SF2$

Answer 13

(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, $group number−8$ . 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.

(b)

Expert Solution

Answer to Problem 87E

The formula of sulfur hexafluoride is $SF6$

Explanation of Solution

To determine: The formula of sulfur hexafluoride

The formula of sulfur hexafluoride is $SF6$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

In $SF6$ 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 $SF6.$

Answer 14

(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, $group number−8$ . 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 15

(b)

Expert Solution

Answer to Problem 87E

The formula of sulfur hexafluoride is $SF6$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

To determine: The formula of sulfur hexafluoride

The formula of sulfur hexafluoride is $SF6$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

In $SF6$ 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 $SF6.$

Answer 20

(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, $group number−8$ . 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.

(c)

Expert Solution

Answer to Problem 87E

The formula of $NaH2PO4$ is sodium dihydrogen phosphate.

Explanation of Solution

To determine: The formula of $NaH2PO4$

The formula of $NaH2PO4$ is sodium dihydrogen phosphate.

Sodium belongs to the Group $I$ , 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 $NaH2PO4$

Answer 21

(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, $group number−8$ . 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 22

(c)

Expert Solution

Answer to Problem 87E

The formula of $NaH2PO4$ is sodium dihydrogen phosphate.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

To determine: The formula of $NaH2PO4$

The formula of $NaH2PO4$ is sodium dihydrogen phosphate.

Sodium belongs to the Group $I$ , 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 $NaH2PO4$

Answer 27

(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, $group number−8$ . 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.

(d)

Expert Solution

Answer to Problem 87E

The formula of lithium nitride is $Li3N$

Explanation of Solution

To determine: The formula of lithium nitride

The formula of lithium nitride is $Li3N$

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

Nitrogen is a non-metal of the $Vth$ group.

Oxidation state $= Group number−8= 5−8=−3$

So, the formula of the given compound is $Li3N$

Answer 28

(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, $group number−8$ . 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 29

(d)

Expert Solution

Answer to Problem 87E

The formula of lithium nitride is $Li3N$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

To determine: The formula of lithium nitride

The formula of lithium nitride is $Li3N$

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

Nitrogen is a non-metal of the $Vth$ group.

Oxidation state $= Group number−8= 5−8=−3$

So, the formula of the given compound is $Li3N$

Answer 34

(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, $group number−8$ . 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.

(e)

Expert Solution

Answer to Problem 87E

The formula of chromium (III) carbonate is $Cr2(CO3)3$

Explanation of Solution

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

The formula of chromium (III) carbonate is $Cr2(CO3)3$

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

Carbonate is an anion having oxidation state $−2$

So, the formula of the given compound is $Cr2(CO3)3$

Answer 35

(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, $group number−8$ . 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 36

(e)

Expert Solution

Answer to Problem 87E

The formula of chromium (III) carbonate is $Cr2(CO3)3$

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

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

The formula of chromium (III) carbonate is $Cr2(CO3)3$

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

Carbonate is an anion having oxidation state $−2$

So, the formula of the given compound is $Cr2(CO3)3$

Answer 41

(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, $group number−8$ . 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.

(f)

Expert Solution

Answer to Problem 87E

The formula of tin (II) fluoride is $SnF2$

Explanation of Solution

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

The formula of tin (II) fluoride is $SnF2$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

So, the formula of the given compound is $SnF2$

Answer 42

(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, $group number−8$ . 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 43

(f)

Expert Solution

Answer to Problem 87E

The formula of tin (II) fluoride is $SnF2$

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

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

The formula of tin (II) fluoride is $SnF2$

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

Fluorine is a non-metal of the $7th$ group.

Oxidation state $= Group number−8= 7−8=−1$

So, the formula of the given compound is $SnF2$

Answer 48

(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, $group number−8$ . 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.

(g)

Expert Solution

Answer to Problem 87E

The formula of ammonium acetate is $CH3COONH4$

Explanation of Solution

To determine: The formula of ammonium acetate.

The formula of ammonium acetate is $CH3COONH4$

The oxidation state of ammonium $(NH4)$ is $+1$ .

Acetate $(CH3COO)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $CH3COONH4$

Answer 49

(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, $group number−8$ . 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 50

(g)

Expert Solution

Answer to Problem 87E

The formula of ammonium acetate is $CH3COONH4$

Answer 51

(g)

Expert Solution

Answer 52

(g)

Expert Solution

Answer 53

Expert Solution

Answer 54

Explanation of Solution

To determine: The formula of ammonium acetate.

The formula of ammonium acetate is $CH3COONH4$

The oxidation state of ammonium $(NH4)$ is $+1$ .

Acetate $(CH3COO)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $CH3COONH4$

Answer 55

(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, $group number−8$ . 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.

(h)

Expert Solution

Answer to Problem 87E

The formula of ammonium hydrogen sulfate is $NH4HSO4$

Explanation of Solution

To determine: The formula of ammonium hydrogen sulfate.

The formula of ammonium hydrogen sulfate is $NH4HSO4$

The oxidation state of ammonium $(NH4)$ is $+1$ .

Hydrogen sulfate $(HSO4)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $NH4HSO4$

Answer 56

(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, $group number−8$ . 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 57

(h)

Expert Solution

Answer to Problem 87E

The formula of ammonium hydrogen sulfate is $NH4HSO4$

Answer 58

(h)

Expert Solution

Answer 59

(h)

Expert Solution

Answer 60

Expert Solution

Answer 61

Explanation of Solution

To determine: The formula of ammonium hydrogen sulfate.

The formula of ammonium hydrogen sulfate is $NH4HSO4$

The oxidation state of ammonium $(NH4)$ is $+1$ .

Hydrogen sulfate $(HSO4)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $NH4HSO4$

Answer 62

(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, $group number−8$ . 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.

(i)

Expert Solution

Answer to Problem 87E

The formula of cobalt (III) nitrate.is $Co(NO3)3$

Explanation of Solution

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

The formula of cobalt (III) nitrate.is $Co(NO3)3$

The oxidation state of cobalt $(Co)$ is $+3$ .

Nitrate $(NO3)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $Co(NO3)3$

Answer 63

(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, $group number−8$ . 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 64

(i)

Expert Solution

Answer to Problem 87E

The formula of cobalt (III) nitrate.is $Co(NO3)3$

Answer 65

(i)

Expert Solution

Answer 66

(i)

Expert Solution

Answer 67

Expert Solution

Answer 68

Explanation of Solution

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

The formula of cobalt (III) nitrate.is $Co(NO3)3$

The oxidation state of cobalt $(Co)$ is $+3$ .

Nitrate $(NO3)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $Co(NO3)3$

Answer 69

(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, $group number−8$ . 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.

(j)

Expert Solution

Answer to Problem 87E

The formula of mercury (I) chloride is $Hg2Cl2$

Explanation of Solution

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

The formula of mercury (I) chloride is $Hg2Cl2$

The oxidation state of mercury $(Hg)$ is $+1$ .

Chloride $(Cl)$ 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 $Hg2Cl2$

Answer 70

(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, $group number−8$ . 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 71

(j)

Expert Solution

Answer to Problem 87E

The formula of mercury (I) chloride is $Hg2Cl2$

Answer 72

(j)

Expert Solution

Answer 73

(j)

Expert Solution

Answer 74

Expert Solution

Answer 75

Explanation of Solution

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

The formula of mercury (I) chloride is $Hg2Cl2$

The oxidation state of mercury $(Hg)$ is $+1$ .

Chloride $(Cl)$ 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 $Hg2Cl2$

Answer 76

(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, $group number−8$ . 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.

(k)

Expert Solution

Answer to Problem 87E

The formula of potassium chlorate is $KClO3$

Explanation of Solution

To determine: The formula of potassium chlorate.

The formula of potassium chlorate is $KClO3$

The oxidation state of potassium $(K)$ is $+1$ .

Chlorate $(ClO3)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $KClO3$

Answer 77

(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, $group number−8$ . 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 78

(k)

Expert Solution

Answer to Problem 87E

The formula of potassium chlorate is $KClO3$

Answer 79

(k)

Expert Solution

Answer 80

(k)

Expert Solution

Answer 81

Expert Solution

Answer 82

Explanation of Solution

To determine: The formula of potassium chlorate.

The formula of potassium chlorate is $KClO3$

The oxidation state of potassium $(K)$ is $+1$ .

Chlorate $(ClO3)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $KClO3$

Answer 83

(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, $group number−8$ . 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.

(l)

Expert Solution

Answer to Problem 87E

The formula of sodium hydride is $NaH$

Explanation of Solution

To determine: The formula of sodium hydride.

The formula of sodium hydride is $NaH$

The oxidation state of sodium $(Na)$ is $+1$ .

Hydride $(H)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $NaH$

Answer 84

(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, $group number−8$ . 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 85

(l)

Expert Solution

Answer to Problem 87E

The formula of sodium hydride is $NaH$

Answer 86

(l)

Expert Solution

Answer 87

(l)

Expert Solution

Answer 88

Expert Solution

Answer 89

Explanation of Solution

To determine: The formula of sodium hydride.

The formula of sodium hydride is $NaH$

The oxidation state of sodium $(Na)$ is $+1$ .

Hydride $(H)$ is an anion having oxidation state $−1$

So, the formula of the given compound is $NaH$

Answer 90

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Answer to Problem 87E

Explanation of Solution

Answer to Problem 87E

Explanation of Solution

Answer to Problem 87E

Explanation of Solution

Answer to Problem 87E

Explanation of Solution

Answer to Problem 87E

Explanation of Solution

Answer to Problem 87E

Explanation of Solution

Answer to Problem 87E

Explanation of Solution

Answer to Problem 87E

Explanation of Solution

Answer to Problem 87E

Explanation of Solution

Answer to Problem 87E

Explanation of Solution

Answer to Problem 87E

Explanation of Solution

Answer to Problem 87E

Explanation of Solution

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