Package: General Chemistry with Connect 2-year Access Card
Package: General Chemistry with Connect 2-year Access Card
7th Edition
ISBN: 9781259680458
Author: Raymond Chang Dr.
Publisher: McGraw-Hill Education
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Chapter 4, Problem 4.45QP

(a)

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 compound 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.

(a)

Expert Solution
Check Mark

Explanation of Solution

The oxidation number of Cs in Cs2O

  • According to the rules, the sum of oxidation numbers of all the atoms present in the compound is zero and general oxidation number for oxygen is 2.
  • The oxygen is in 2 oxidation state and Cs is,

    2(X)+1(-2)=02X+(-2)=02X=+2X=+1

  • The oxidation number of Cs in Cs2O is +1.

(b)

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 compound 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.

(b)

Expert Solution
Check Mark

Explanation of Solution

The oxidation number of I in CaI2

  • 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 halides is -1.
  • The I in 1 oxidation state and Ca is,

    X+(2(-1)=0X+(-2)=0X=+2

  • The oxidation number of I in CaI2 is 1

(c)

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 compound 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.

(c)

Expert Solution
Check Mark

Explanation of Solution

The oxidation number of Al in Al2O3

  • The sum of oxidation numbers of all the atoms present in the molecule is zero and the general oxidation number for oxygen is 2.
  • The O in 2 oxidation state and Al is,

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

  • The oxidation number of Al in Al2O3 is +3

(d)

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 compound 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.

(d)

Expert Solution
Check Mark

Explanation of Solution

The oxidation number of As in H3AsO3

  • 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 and hydrogen is +1.
  • The O in 2 and hydrogen is +1 oxidation states so As is,

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

  • The oxidation number of As in H3AsO3 is +3

(e)

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 compound 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.

(e)

Expert Solution
Check Mark

Explanation of Solution

The oxidation number of Ti in TiO2

  • 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 Ti is,

    X+2(-2)=0X+(-4)=0X=+4

  • The oxidation number of Ti in TiO2 is +4

(f)

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 compound 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.

(f)

Expert Solution
Check Mark

Explanation of Solution

The oxidation number of Mo in MoO42-

  • 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 Mo is ,

    1X+4(2-)=-2 X+(-8)=-2X=8-2X=+6

  • The oxidation number of Mo in MoO42- is +6

(g)

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 compound 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.

(g)

Expert Solution
Check Mark

Explanation of Solution

The oxidation number of Pt in PtCl42-

  • 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 halides is -1.
  • The Cl in 1 and oxidation state and Pt is,

    X+4(-1)=-2X+(-4)=-2X=4-2X=+2

  • The oxidation number of Pt in PtCl42- is +2

(h)

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 compound 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.

(h)

Expert Solution
Check Mark

Explanation of Solution

The oxidation number of Pt in PtCl62-

  • 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 halides is -1.
  • The Cl in 1 and oxidation state and Pt is,

    X+6(-1)=-2X+(-6)=-2X=6+(-2)X=+4

  • The oxidation number of Pt in PtCl62- is +4

(i)

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 compound 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.

(i)

Expert Solution
Check Mark

Explanation of Solution

The oxidation number of Sn in SnF2

  • 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 halides is -1.
  • The F in 1 and oxidation state and Pt is,

    X+2(-1)=0X+(-2)=0X=+2

  • The oxidation number of Sn in SnF2 is +2

(j)

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 compound 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.

(j)

Expert Solution
Check Mark

Explanation of Solution

The oxidation number of Cl in ClF3

  • 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 halides is -1 but in the Inter halogen compounds the Fluorine is always having 1 oxidation state.
  • The F in 1 and oxidation state and Cl is,

    X+3(-1)=0X+(-3)=0X=+3

  • The oxidation number of Cl in ClF3 is +3

The oxidation number of Cl in ClF3

(k)

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 compound 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.

(k)

Expert Solution
Check Mark

Explanation of Solution

The oxidation number of I in SbF6-

  • 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 halides is -1 and Fluorine is always having 1 oxidation state.
  • The F in 1 and oxidation state and Sb is,

    X+6(-1)=-1X+(-6)=-1X=6-1X=+5

  • The oxidation number of I in SbF6- is +5

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Chapter 4 Solutions

Package: General Chemistry with Connect 2-year Access Card

Ch. 4.5 - Prob. 1PECh. 4.5 - Prob. 2PECh. 4.5 - Prob. 3PECh. 4.5 - Prob. 1RCCh. 4.6 - Prob. 1PECh. 4.6 - Prob. 1RCCh. 4.6 - Prob. 2PECh. 4.6 - Prob. 3PECh. 4 - Prob. 4.1QPCh. 4 - Prob. 4.2QPCh. 4 - Prob. 4.3QPCh. 4 - 4.4 What is the difference between the following...Ch. 4 - 4.5 Water is an extremely weak electrolyte and...Ch. 4 - Prob. 4.6QPCh. 4 - Prob. 4.7QPCh. 4 - 4.8 Which of the following diagrams best...Ch. 4 - Prob. 4.9QPCh. 4 - Prob. 4.10QPCh. 4 - Prob. 4.11QPCh. 4 - Prob. 4.12QPCh. 4 - Prob. 4.13QPCh. 4 - Prob. 4.14QPCh. 4 - Prob. 4.15QPCh. 4 - Prob. 4.16QPCh. 4 - Prob. 4.17QPCh. 4 - Prob. 4.18QPCh. 4 - Prob. 4.19QPCh. 4 - Prob. 4.20QPCh. 4 - 4.21 Write ionic and net ionic equations for the...Ch. 4 - Prob. 4.22QPCh. 4 - Prob. 4.23QPCh. 4 - Prob. 4.24QPCh. 4 - Prob. 4.25QPCh. 4 - Prob. 4.26QPCh. 4 - Prob. 4.27QPCh. 4 - Prob. 4.28QPCh. 4 - Prob. 4.29QPCh. 4 - Prob. 4.30QPCh. 4 - Prob. 4.31QPCh. 4 - Prob. 4.32QPCh. 4 - Prob. 4.33QPCh. 4 - Prob. 4.34QPCh. 4 - Prob. 4.35QPCh. 4 - Prob. 4.36QPCh. 4 - Prob. 4.37QPCh. 4 - Prob. 4.38QPCh. 4 - 4.39 For the complete redox reactions given here,...Ch. 4 - Prob. 4.40QPCh. 4 - Prob. 4.41QPCh. 4 - Prob. 4.42QPCh. 4 - Prob. 4.43QPCh. 4 - Prob. 4.44QPCh. 4 - Prob. 4.45QPCh. 4 - Prob. 4.46QPCh. 4 - Prob. 4.47QPCh. 4 - Prob. 4.48QPCh. 4 - Prob. 4.49QPCh. 4 - Prob. 4.50QPCh. 4 - Prob. 4.51QPCh. 4 - Prob. 4.52QPCh. 4 - Prob. 4.53QPCh. 4 - Prob. 4.54QPCh. 4 - Prob. 4.55QPCh. 4 - Prob. 4.56QPCh. 4 - Prob. 4.57QPCh. 4 - Prob. 4.58QPCh. 4 - Prob. 4.59QPCh. 4 - Prob. 4.60QPCh. 4 - Prob. 4.61QPCh. 4 - Prob. 4.62QPCh. 4 - Prob. 4.63QPCh. 4 - Prob. 4.64QPCh. 4 - Prob. 4.65QPCh. 4 - Prob. 4.66QPCh. 4 - Prob. 4.67QPCh. 4 - Prob. 4.68QPCh. 4 - Prob. 4.69QPCh. 4 - 4.70 Distilled water must be used in the...Ch. 4 - 4.71 If 30.0 mL of 0.150 M CaCl2 is added to 15.0...Ch. 4 - Prob. 4.72QPCh. 4 - Prob. 4.73QPCh. 4 - Prob. 4.74QPCh. 4 - Prob. 4.75QPCh. 4 - Prob. 4.76QPCh. 4 - Prob. 4.77QPCh. 4 - Prob. 4.78QPCh. 4 - Prob. 4.79QPCh. 4 - Prob. 4.80QPCh. 4 - Prob. 4.81QPCh. 4 - Prob. 4.82QPCh. 4 - Prob. 4.83QPCh. 4 - Prob. 4.84QPCh. 4 - Prob. 4.85QPCh. 4 - Prob. 4.86QPCh. 4 - Prob. 4.87QPCh. 4 - Prob. 4.88QPCh. 4 - Prob. 4.89QPCh. 4 - Prob. 4.90QPCh. 4 - Prob. 4.91QPCh. 4 - Prob. 4.92QPCh. 4 - Prob. 4.93QPCh. 4 - 4.74 The molecular formula of malonic acid is...Ch. 4 - Prob. 4.95QPCh. 4 - Prob. 4.96QPCh. 4 - Prob. 4.97QPCh. 4 - Prob. 4.98QPCh. 4 - Prob. 4.99QPCh. 4 - Prob. 4.100QPCh. 4 - Prob. 4.101QPCh. 4 - Prob. 4.102QPCh. 4 - 4.103 These are common household compounds: table...Ch. 4 - Prob. 4.104QPCh. 4 - Prob. 4.105QPCh. 4 - Prob. 4.106QPCh. 4 - 4.107 A number of metals are involved in redox...Ch. 4 - Prob. 4.108QPCh. 4 - Prob. 4.109QPCh. 4 - Prob. 4.110QPCh. 4 - Prob. 4.111QPCh. 4 - Prob. 4.112QPCh. 4 - Prob. 4.114SPCh. 4 - Prob. 4.115SPCh. 4 - Prob. 4.116SPCh. 4 - Prob. 4.117SPCh. 4 - Prob. 4.118SP
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How to Calculate Oxidation Numbers Introduction; Author: Tyler DeWitt;https://www.youtube.com/watch?v=-a2ckxhfDjQ;License: Standard YouTube License, CC-BY