Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 10, Problem 10.17P

(a)

Interpretation Introduction

Interpretation:

Lewis structure for a resonance form of each ion of BrO3 with the lowest possible formal charges is to be drawn. Also, the oxidation number of the atom is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the molecule are as follows:

Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Calculate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

Formula to calculate the formal charge of the atom is as follows:

  Formalcharge=(numberofvalenceelectrons)((numberofnon-bondingelectrons)+(12)(numberofbondingelectrons))        (1)

The formula to calculate the oxidation number of an atom is as follows:

  Oxidationnumber=(numberofvalenceelectrons)((numberofnon-bondingelectrons)+(numberofbondingelectrons))        (2)

(a)

Expert Solution
Check Mark

Answer to Problem 10.17P

The possible Lewis structures for a resonance form of each ion of BrO3 with the lowest possible formal charges are,

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.17P , additional homework tip  1

The oxidation numbers of Br is +5 and the oxidation number of oxygen is 2.

Explanation of Solution

Lewis structures for a resonance form of BrO3 is,

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.17P , additional homework tip  2

For structure I:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.17P , additional homework tip  3

Substitute 7 for valence electrons, 2 for the number of nonbonded electrons and 6 for the number of bonding electrons in equation (1) to calculate the formal charge on Br atom.

  Formalcharge=(7)((2)+(12)(6))=+2

Substitute 6 for valence electrons, 6 for nonbonded electrons and 2 for the number of bonding electrons in equation (1) to calculate the formal charge on each oxygen atom.

  Formalcharge=(6)((6)+(12)(2))=1

For structure II:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.17P , additional homework tip  4

Substitute 7 for valence electrons, 2 for the number of nonbonded electrons and 8 for the number of bonding electrons in equation (1) to calculate the formal charge on Br atom.

  Formalcharge=(7)((2)+(12)(8))=+1

Substitute 6 for valence electrons, 6 for nonbonded electrons and 2 for the number of bonding electrons in equation (1) to calculate the formal charge on each single bonded oxygen atom.

  Formalcharge=(6)((6)+(12)(2))=1

Substitute 6 for the value of valence electrons, 4 for the number of nonbonded electrons and 4 for the number of bonding electrons in equation (1) to calculate the formal charge on the double bonded oxygen atom.

  Formalcharge=(6)((4)+(12)(4))=0

For structure III:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.17P , additional homework tip  5

Substitute 7 for valence electrons, 2 for the number of nonbonded electrons and 10 for the number of bonding electrons in equation (1) to calculate the formal charge on Br atom.

  Formalcharge=(7)((2)+(12)(10))=0

Substitute 6 for valence electrons, 6 for nonbonded electrons and 2 for number of bonding electrons in equation (1) to calculate the formal charge on the single bonded oxygen atom.

  Formalcharge=(6)((6)+(12)(2))=1

Substitute 6 for the value of valence electrons, 4 for the number of nonbonded electrons and 4 for the number of bonding electrons in equation (1) to calculate the formal charge on each double bonded oxygen atom.

  Formalcharge=(6)((4)+(12)(4))=0.

Therefore, structure II has the more acceptable and reasonable distribution of formal charges.

The oxidation numbers of Br is +5 and the oxidation number of oxygen is 2.

Conclusion

BrO3 has three possible resonating structures. Structure II is more acceptable and has a more distributed formal charge.

(b)

Interpretation Introduction

Interpretation:

Lewis structure for a resonance form of each ion of SO32 with the lowest possible formal charges is to be drawn. Also, the oxidation number of the atom is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the molecule are as follows:

Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Calculate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

Formula to calculate the formal charge of the atom is as follows:

  Formalcharge=(numberofvalenceelectrons)((numberofnon-bondingelectrons)+(12)(numberofbondingelectrons))        (1)

The formula to calculate the oxidation number of an atom is as follows:

  Oxidationnumber=(numberofvalenceelectrons)((numberofnon-bondingelectrons)+(numberofbondingelectrons))        (2)

(b)

Expert Solution
Check Mark

Answer to Problem 10.17P

The possible Lewis structures for a resonance form of each ion of SO32 with the lowest possible formal charges are,

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.17P , additional homework tip  6

The oxidation numbers of. S is +4 and the oxidation number of oxygen is 2.

Explanation of Solution

Lewis structure for a resonance form of SO32 is,

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.17P , additional homework tip  7

For structure I:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.17P , additional homework tip  8

Substitute 6 for valence electrons, 2 for the number of nonbonded electrons and 6 for the number of bonding electrons in equation (1) to calculate the formal charge on S atom.

  Formalcharge=(6)((2)+(12)(6))=+1

Substitute 6 for valence electrons, 6 for nonbonded electrons and 2 for the number of bonding electrons in equation (1) to calculate the formal charge on each oxygen atom.

  Formalcharge=(6)((6)+(12)(2))=1

For structure II:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.17P , additional homework tip  9

Substitute 6 for valence electrons, 2 for the number of nonbonded electrons and 8 for the number of bonding electrons in equation (1) to calculate the formal charge on S atom.

  Formalcharge=(6)((2)+(12)(8))=0

Substitute 6 for valence electrons, 6 for nonbonded electrons and 2 for the number of bonding electrons in equation (1) to calculate the formal charge on each single bonded oxygen atom.

  Formalcharge=(6)((6)+(12)(2))=1

Substitute 6 for the value of valence electrons, 4 for the number of nonbonded electrons and 4 for the number of bonding electrons in equation (1) to calculate the formal charge on the double bonded oxygen atom.

  Formalcharge=(6)((4)+(12)(4))=0.

Therefore, structure II has the more acceptable and reasonable distribution of formal charges.

The oxidation numbers of. S is +4 and the oxidation number of oxygen is 2.

Conclusion

SO32 has two possible resonating structures.

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Chapter 10 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

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