Chemistry: Atoms First
Chemistry: Atoms First
2nd Edition
ISBN: 9780073511184
Author: Julia Burdge, Jason Overby Professor
Publisher: McGraw-Hill Education
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Chapter 9, Problem 9.48QP

Give the oxidation numbers for the underlined atoms in the following molecules and ions: (a) Mg3N2, (b) CsO2, (c) CaC2, (d) CO32−, (e) C2O42−, (f) ZnO22−, (g) NaBH4, (h) WO42−.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

  • The oxidation number of atom in the elemental form is zero.
  • The sum of oxidation numbers of all the atoms present in the molecule is zero.
  • The general oxidation number for oxygen is 2 and hydrogen is +1.
  • The combination of any atoms in group 7 the oxidation number is varied.
  • The Inter halogen compounds the Fluorine is always having 1 oxidation state.
  • In the oxohalide compounds oxidation number of halide is in positive.
  • The general oxidation state of halides is -1.
  • The earth metals always having +1 oxidation number only.
  • The oxidation state of hydrogen in metal hydride is 1.

Explanation of Solution

The oxidation number of N2 in Mg3N2

  • According to the rules, the alkaline earth metals always having +2 oxidation number.
  • The sum of oxidation numbers of all the atoms present in the molecule is zero

3(2+)+2X=0(6+)+2X=02X=-6X=-3

  • The oxidation number of N2 in Mg3N2 is 3

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

  • The oxidation number of atom in the elemental form is zero.
  • The sum of oxidation numbers of all the atoms present in the molecule is zero.
  • The general oxidation number for oxygen is 2 and hydrogen is +1.
  • The combination of any atoms in group 7 the oxidation number is varied.
  • The Inter halogen compounds the Fluorine is always having 1 oxidation state.
  • In the oxohalide compounds oxidation number of halide is in positive.
  • The general oxidation state of halides is -1.
  • The earth metals always having +1 oxidation number only.
  • The oxidation state of hydrogen in metal hydride is 1.

Explanation of Solution

The oxidation number of O2 in CsO2is

  • According to the rules, the sum of oxidation numbers of all the atoms present in the molecule is zero and the earth metals always having +1 oxidation number only
  • The Cs in +1 oxidation state and O is,
  • (1+)+2(X)=02X=-1X=-12

  • The oxidation number of O in CsO2 is 12

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

  • The oxidation number of atom in the elemental form is zero.
  • The sum of oxidation numbers of all the atoms present in the molecule is zero.
  • The general oxidation number for oxygen is 2 and hydrogen is +1.
  • The combination of any atoms in group 7 the oxidation number is varied.
  • The Inter halogen compounds the Fluorine is always having 1 oxidation state.
  • In the oxohalide compounds oxidation number of halide is in positive.
  • The general oxidation state of halides is -1.
  • The earth metals always having +1 oxidation number only.
  • The oxidation state of hydrogen in metal hydride is 1.

Explanation of Solution

The oxidation number of C in CaC2

  • According to the rules, the alkaline earth metals always having +2 oxidation number.
  • The sum of oxidation numbers of all the atoms present in the molecule is zero
  • The Ca in +2 oxidation state and C is,
  • 1(2+)+2(X)=0(2+)+2(X)=02X=-2X=-1C=-1

  • The oxidation number of C in CaC2 is 1

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

  • The oxidation number of atom in the elemental form is zero.
  • The sum of oxidation numbers of all the atoms present in the molecule is zero.
  • The general oxidation number for oxygen is 2 and hydrogen is +1.
  • The combination of any atoms in group 7 the oxidation number is varied.
  • The Inter halogen compounds the Fluorine is always having 1 oxidation state.
  • In the oxohalide compounds oxidation number of halide is in positive.
  • The general oxidation state of halides is -1.
  • The earth metals always having +1 oxidation number only.
  • The oxidation state of hydrogen in metal hydride is 1.

Explanation of Solution

The oxidation number of C in CO32-

  • According to the rules, the sum of oxidation numbers of all the atoms present in the molecule is -2 and the general oxidation state of oxygen is 2.
  • The O in 2 oxidation state and C is,
  • 1(X)+3(-2)=-2(X)+(-6)=-2X=+4

  • The oxidation number of C in CO32- is +4

(e)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

  • The oxidation number of atom in the elemental form is zero.
  • The sum of oxidation numbers of all the atoms present in the molecule is zero.
  • The general oxidation number for oxygen is 2 and hydrogen is +1.
  • The combination of any atoms in group 7 the oxidation number is varied.
  • The Inter halogen compounds the Fluorine is always having 1 oxidation state.
  • In the oxohalide compounds oxidation number of halide is in positive.
  • The general oxidation state of halides is -1.
  • The earth metals always having +1 oxidation number only.
  • The oxidation state of hydrogen in metal hydride is 1.

Explanation of Solution

The oxidation number of C in C2O42-

  • According to the rules, the sum of oxidation numbers of all the atoms present in the molecule is -2 and the general oxidation state of oxygen is 2.
  • The O in 2 oxidation state and C is,
  • 4(-2)+2(X)=-2(-8)+2(X)=02X=+6C=+3

  • The oxidation number of C in C2O42- is +3

(f)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

  • The oxidation number of atom in the elemental form is zero.
  • The sum of oxidation numbers of all the atoms present in the molecule is zero.
  • The general oxidation number for oxygen is 2 and hydrogen is +1.
  • The combination of any atoms in group 7 the oxidation number is varied.
  • The Inter halogen compounds the Fluorine is always having 1 oxidation state.
  • In the oxohalide compounds oxidation number of halide is in positive.
  • The general oxidation state of halides is -1.
  • The earth metals always having +1 oxidation number only.
  • The oxidation state of hydrogen in metal hydride is 1.

Explanation of Solution

The oxidation number of O in ZnO22-

  • According to the rules, the sum of oxidation numbers of all the atoms present in the molecule is zero and the general oxidation state of Oxygen is 2.
  • The stable  oxidation state and Zn in +2 is and O is,
  • 1(+2)+2(X)=-2(+2)+2X=-22X=4X = -2

  • The oxidation number of O in ZnO22- is 2

(g)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

  • The oxidation number of atom in the elemental form is zero.
  • The sum of oxidation numbers of all the atoms present in the molecule is zero.
  • The general oxidation number for oxygen is 2 and hydrogen is +1.
  • The combination of any atoms in group 7 the oxidation number is varied.
  • The Inter halogen compounds the Fluorine is always having 1 oxidation state.
  • In the oxohalide compounds oxidation number of halide is in positive.
  • The general oxidation state of halides is -1.
  • The earth metals always having +1 oxidation number only.
  • The oxidation state of hydrogen in metal hydride is 1.

Explanation of Solution

The oxidation number of B in NaBH4

  • According to the rules, the sum of oxidation numbers of all the atoms present in the molecule is zero and general oxidation number for oxygen is 2 and hydrogen is +1 but in the hydride compound general oxidation number of hydrogen is 1
  • The H in 1 and Na is in +1 oxidation states and B is,
  • 1(+1)+1(X)+4(-1)=0(+1)+X+(-4)=0X=+3

  • The oxidation number of B in NaBH4 is +3

(h)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

  • The oxidation number of atom in the elemental form is zero.
  • The sum of oxidation numbers of all the atoms present in the molecule is zero.
  • The general oxidation number for oxygen is 2 and hydrogen is +1.
  • The combination of any atoms in group 7 the oxidation number is varied.
  • The Inter halogen compounds the Fluorine is always having 1 oxidation state.
  • In the oxohalide compounds oxidation number of halide is in positive.
  • The general oxidation state of halides is -1.
  • The earth metals always having +1 oxidation number only.
  • The oxidation state of hydrogen in metal hydride is 1.

Explanation of Solution

The oxidation number of W in WO42-

  • According to the rules, the sum of oxidation numbers of all the atoms present in the molecule is zero and the general oxidation state of Oxygen is 2.
  • The O in 2 oxidation state and W is,
  • X+4(-2)=-2X+(-8)=-2X=8+(-2)W=+6

  • The oxidation number of W in WO42- is +6.

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Chapter 9 Solutions

Chemistry: Atoms First

Ch. 9.2 - Using Tables 9.2 and 9.3, identify a compound that...Ch. 9.2 - Prob. 9.3WECh. 9.2 - Prob. 3PPACh. 9.2 - Prob. 3PPBCh. 9.2 - Which diagram best represents the result when...Ch. 9.2 - Prob. 9.2.1SRCh. 9.2 - Prob. 9.2.2SRCh. 9.2 - Prob. 9.2.3SRCh. 9.2 - Prob. 9.2.4SRCh. 9.2 - Prob. 9.2.5SRCh. 9.3 - Prob. 9.4WECh. 9.3 - Prob. 4PPACh. 9.3 - Prob. 4PPBCh. 9.3 - Prob. 4PPCCh. 9.3 - Prob. 9.3.1SRCh. 9.3 - Prob. 9.3.2SRCh. 9.3 - Prob. 9.3.3SRCh. 9.3 - Prob. 9.3.4SRCh. 9.4 - Prob. 9.5WECh. 9.4 - Prob. 5PPACh. 9.4 - Prob. 5PPBCh. 9.4 - Write the balanced equation for the reaction...Ch. 9.4 - Prob. 9.6WECh. 9.4 - Using the activity series, predict which of the...Ch. 9.4 - Prob. 6PPBCh. 9.4 - Prob. 6PPCCh. 9.4 - Prob. 9.7WECh. 9.4 - Predict which of the following reactions will...Ch. 9.4 - Prob. 7PPBCh. 9.4 - Prob. 7PPCCh. 9.4 - Prob. 9.4.1SRCh. 9.4 - Prob. 9.4.2SRCh. 9.4 - Prob. 9.4.3SRCh. 9.4 - Prob. 9.4.4SRCh. 9.5 - Prob. 9.8WECh. 9.5 - Prob. 8PPACh. 9.5 - Prob. 8PPBCh. 9.5 - Prob. 8PPCCh. 9.5 - Prob. 9.9WECh. 9.5 - Prob. 9PPACh. 9.5 - Prob. 9PPBCh. 9.5 - Prob. 9PPCCh. 9.5 - Starting with a 2.0-M stock solution of...Ch. 9.5 - Starting with a 6.552-M stock solution of HNO3,...Ch. 9.5 - Five standard solutions of HBr are prepared by...Ch. 9.5 - Prob. 10PPCCh. 9.5 - Prob. 9.11WECh. 9.5 - Prob. 11PPACh. 9.5 - Prob. 11PPBCh. 9.5 - Prob. 11PPCCh. 9.5 - Prob. 9.12WECh. 9.5 - Calculate the hydronium ion concentration in a...Ch. 9.5 - Prob. 12PPBCh. 9.5 - Prob. 12PPCCh. 9.5 - Prob. 9.13WECh. 9.5 - Prob. 13PPACh. 9.5 - Prob. 13PPBCh. 9.5 - Prob. 13PPCCh. 9.5 - Prob. 9.5.1SRCh. 9.5 - Prob. 9.5.2SRCh. 9.5 - Prob. 9.5.3SRCh. 9.5 - Prob. 9.5.4SRCh. 9.5 - Prob. 9.5.5SRCh. 9.5 - Prob. 9.5.6SRCh. 9.6 - Prob. 9.14WECh. 9.6 - Prob. 14PPACh. 9.6 - Prob. 14PPBCh. 9.6 - Which diagram best represents the solution...Ch. 9.6 - Prob. 9.15WECh. 9.6 - Prob. 15PPACh. 9.6 - What volume (in mL) of a 0.2550 M NaOH solution...Ch. 9.6 - Prob. 15PPCCh. 9.6 - Prob. 9.16WECh. 9.6 - Prob. 16PPACh. 9.6 - Prob. 16PPBCh. 9.6 - Prob. 16PPCCh. 9.6 - Prob. 9.17WECh. 9.6 - Prob. 17PPACh. 9.6 - What is the molar mass of a diprotic acid if 30.5...Ch. 9.6 - Prob. 17PPCCh. 9.6 - Prob. 9.6.1SRCh. 9.6 - Prob. 9.6.2SRCh. 9.6 - Prob. 9.6.3SRCh. 9.6 - Prob. 9.6.4SRCh. 9 - Define solute, solvent, and solution by describing...Ch. 9 - What is the difference between a nonelectrolyte...Ch. 9 - Prob. 9.3QPCh. 9 - Prob. 9.4QPCh. 9 - Prob. 9.5QPCh. 9 - Prob. 9.6QPCh. 9 - You are given a water-soluble compound X. 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