General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 4, Problem 4.46QP

(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 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.

(a)

Expert Solution
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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)

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.

(b)

Expert Solution
Check Mark

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)

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.

(c)

Expert Solution
Check Mark

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)

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.

(d)

Expert Solution
Check Mark

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)

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.

(e)

Expert Solution
Check Mark

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)

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.

(f)

Expert Solution
Check Mark

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)

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.

(g)

Expert Solution
Check Mark

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)

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.

(h)

Expert Solution
Check Mark

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 4 Solutions

General Chemistry

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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