EBK FOUNDATIONS OF COLLEGE CHEMISTRY
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781118930144
Author: Willard
Publisher: JOHN WILEY+SONS INC.
bartleby

Videos

Question
Book Icon
Chapter 17, Problem 2PE

(a)

Interpretation Introduction

Interpretation:

Oxidation number of each element in CHF3 has to be determined.

Concept Introduction:

Oxidation number is integer value allotted to every element. It is formal charge occupied by atom if all of its bonds are dissociated heterolytically. Below mentioned are rules to assign oxidation numbers to various elements.

1. Elements present in their free state have zero oxidation number.

2. Oxidation number of hydrogen is generally +1, except for metal hydrides.

3. Oxidation number of oxygen is 2, except for peroxides.

4. Metals have positive oxidation numbers.

5. Negative oxidation numbers are assigned to most electronegative element in covalent compounds.

6. Sum of oxidation numbers of different elements in neutral atom is zero.

7. Sum of oxidation numbers of various elements in polyatomic ion is equal to charge present on ion.

(a)

Expert Solution
Check Mark

Explanation of Solution

Since fluorine is member of halogen group, its oxidation number is 1. Since CHF3 is not hydride compound, oxidation state of H is +1.

Expression for oxidation number in CHF3 is as follows:

  [Oxidation number of C+Oxidation number of H+3(Oxidation number of F)]=0        (1)

Rearrange equation (1) for oxidation number of C.

  Oxidation number of C=[Oxidation number of H3(Oxidation number of F)]        (2)

Substitute 1 for oxidation number of F and +1 for oxidation number of H in equation (2).

  Oxidation number of C=[(+1)3(1)]=1+3=+2

Hence, oxidation number of C is +2, that of H is +1 and that of F is 1.

(b)

Interpretation Introduction

Interpretation:

Oxidation number of each element in P2O5 has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

Since P2O5 is not peroxide, oxidation number of O is 2.

Expression for oxidation number in P2O5 is as follows:

  [2(Oxidation number of P)+5(Oxidation number of O)]=0        (3)

Rearrange equation (3) for oxidation number of P.

  Oxidation number of P=[5(Oxidation number of O)2]        (4)

Substitute 2 for oxidation number of O in equation (4).

  Oxidation number of P=[5(2)2]=+102=+5

Hence, oxidation number of P is +5 and that of O is 2.

(c)

Interpretation Introduction

Interpretation:

Oxidation number of each element in SF6 has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

Since fluorine is member of halogen family, oxidation number of F is 1.

Expression for oxidation number in SF6 is as follows:

  [Oxidation number of S+6(Oxidation number of F)]=0        (5)

Rearrange equation (5) for oxidation number of S.

  Oxidation number of S=6(Oxidation number of F)        (6)

Substitute 1 for oxidation number of F in equation (6).

  Oxidation number of S=6(1)=+6

Hence, oxidation number of F is 1 and that of S is +6.

(d)

Interpretation Introduction

Interpretation:

Oxidation number of each element in SnSO4 has to be determined.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

Since SnSO4 is not peroxide compound, oxidation state of O is 2. Since chlorine is member of halogen group, its oxidation number is 1. Sn is member of carbon family but due to inert pair effect, its oxidation number is +2.

Expression for oxidation number in SnSO4 is as follows:

  [Oxidation number of Sn+Oxidation number of S+4(Oxidation number of O)]=0        (7)

Rearrange equation (7) for oxidation number of S.

  Oxidation number of S=[Oxidation number of Sn4(Oxidation number of O)]        (8)

Substitute 2 for oxidation number of O and +2 for oxidation number of Sn in equation (8).

  Oxidation number of S=[(+2)4(2)]=2+8=+6

Hence, oxidation number of Sn is +2, that of S is +6 and that of O is 2.

(e)

Interpretation Introduction

Interpretation:

Oxidation number of each element in CH3OH has to be determined.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution
Check Mark

Explanation of Solution

Since CH3OH is not peroxide, oxidation number of O is 2. It is not hydride compound so oxidation number of H is +1.

Expression for oxidation number in CH3OH is as follows:

  [Oxidation number of C+4(Oxidation number of H)+Oxidation number of O]=0        (9)

Rearrange equation (9) for oxidation number of C.

  Oxidation number of C=[4(Oxidation number of H)Oxidation number of O]        (10)

Substitute 2 for oxidation number of O and +1 for oxidation number of H in equation (10).

  Oxidation number of C=[4(+1)(2)]=4+2=2

Hence, oxidation state of O is 2, that of H is +1 and that of C is 2.

(f)

Interpretation Introduction

Interpretation:

Oxidation number of each element in H3PO4 has to be determined.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution
Check Mark

Explanation of Solution

Since H3PO4 is neither hydride nor peroxide compound, oxidation number of H is +1 and that of O is 2.

Expression for oxidation number in H3PO4 is as follows:

  [3(Oxidation number of H)+Oxidation number of P+4(Oxidation number of O)]=0        (11)

Rearrange equation (11) for oxidation number of P.

  Oxidation number of P=[3(Oxidation number of H)4(Oxidation number of O)]        (12)

Substitute +1 for oxidation number of H and 2 for oxidation number of O in equation (12).

  Oxidation number of P=[3(+1)4(2)]=3+8=+5

Hence, oxidation number of H is +1, that of P is +5 and that of O is 2.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
The acid-base chemistry of both EDTA and EBT are important to ensuring that the reactions proceed as desired, thus the pH is controlled using a buffer. What percent of the EBT indicator will be in the desired HIn2- state at pH = 10.5. pKa1 = 6.2 and pKa2 = 11.6 of EBT
What does the phrase 'fit for purpose' mean in relation to analytical chemistry? Please provide examples too.
For each of the substituted benzene molecules below, determine the inductive and resonance effects the substituent will have on the benzene ring, as well as the overall electron-density of the ring compared to unsubstituted benzene. Molecule Inductive Effects Resonance Effects Overall Electron-Density × NO2 ○ donating O donating O withdrawing O withdrawing O electron-rich electron-deficient no inductive effects O no resonance effects O similar to benzene E [ CI O donating withdrawing O no inductive effects Explanation Check ○ donating withdrawing no resonance effects electron-rich electron-deficient O similar to benzene © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Acces

Chapter 17 Solutions

EBK FOUNDATIONS OF COLLEGE CHEMISTRY

Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
  • Text book image
    Introductory Chemistry: A Foundation
    Chemistry
    ISBN:9781337399425
    Author:Steven S. Zumdahl, Donald J. DeCoste
    Publisher:Cengage Learning
    Text book image
    Introductory Chemistry: A Foundation
    Chemistry
    ISBN:9781285199030
    Author:Steven S. Zumdahl, Donald J. DeCoste
    Publisher:Cengage Learning
    Text book image
    World of Chemistry, 3rd edition
    Chemistry
    ISBN:9781133109655
    Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
    Publisher:Brooks / Cole / Cengage Learning
  • Text book image
    Chemistry: Matter and Change
    Chemistry
    ISBN:9780078746376
    Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
    Publisher:Glencoe/McGraw-Hill School Pub Co
    Text book image
    Chemistry
    Chemistry
    ISBN:9781305957404
    Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
    Publisher:Cengage Learning
    Text book image
    Chemistry & Chemical Reactivity
    Chemistry
    ISBN:9781337399074
    Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
    Publisher:Cengage Learning
Text book image
Introductory Chemistry: A Foundation
Chemistry
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Introductory Chemistry: A Foundation
Chemistry
ISBN:9781285199030
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
World of Chemistry, 3rd edition
Chemistry
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:Brooks / Cole / Cengage Learning
Text book image
Chemistry: Matter and Change
Chemistry
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Glencoe/McGraw-Hill School Pub Co
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
How to Calculate Oxidation Numbers Introduction; Author: Tyler DeWitt;https://www.youtube.com/watch?v=-a2ckxhfDjQ;License: Standard YouTube License, CC-BY