CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT
CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT
14th Edition
ISBN: 9781259327933
Author: Burdge
Publisher: MCG
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Chapter 6.6, Problem 10PPA

Draw three resonance structures for the hydrogen sulfite ion (), one that obeys the octet rule for the central atom and two that expand the octet of the central atom. Calculate the formal charges on all atoms in each structure and determine which, if airy, of the resonance structures has formal charges that are inconsistent with the elements’ electronegativities.

Expert Solution
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Interpretation Introduction

Interpretation:

The resonance structures for hydrogen sulfite ion should be drawn where one resonance structure must obey octet rule and the other two having expanded octet. Also formal charge of all the resonance structure should be drawn and to identify the resonance structure that has an inconsistent with the electronegativities of the elements.

Concept Introduction:

  • Sometimes the chemical bonding of a molecule cannot be represented using a single Lewis structure. In these cases, the chemical bonding are described by delocalization of electrons and is known as resonance.
  • In some molecules, there is possibility of more than one Lewis structure where all the structures are equally acceptable. One of the acceptable Lewis structures of these molecules is called resonance structure. All the possible resonance structures are imaginary whereas the resonance hybrid is real.
  • These structures will differ only in the arrangement of the electrons not in the relative position of the atomic nuclei.
  • The Lewis structure is based on the concept of the octet rule so that the electrons shared in each atom should have 8 electrons in its outer shell.
  • Atoms can be stable even though the number of valence electrons in the atoms in a molecule is more than 8 and is called expanded octet

To identify: the resonance structure of hydrogen sulfite ion

Answer to Problem 10PPA

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 6.6, Problem 10PPA , additional homework tip  1

Explanation of Solution

The resonance structure of HSO3 is drawn below

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 6.6, Problem 10PPA , additional homework tip  2

Delocalization of electrons takes place because of the presence of lone pair of electrons in the terminal atoms and the possibility of double bond. The chemical bonding of a molecule cannot be represented using a single Lewis structure and represented by structures (a), (b) and (c). The chemical bonding are described by delocalization of electrons forming three possible resonance structures. In all the resonance structures the position, over whole charge and chemical framework remains intact. Also in these structures only in the arrangement of the electrons differs not the relative position of the atomic nuclei.

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 6.6, Problem 10PPA , additional homework tip  3

Here sulfur shares a double bond with one of the oxygen atom and shares two single bond with remaining two oxygen atoms. Hence in (a), the central sulfur atom is surrounded by 8 electrons therefore it obeys octet rule.

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 6.6, Problem 10PPA , additional homework tip  4

Here sulfur shares a double bond with one of the oxygen atom and shares two single bond with remaining two oxygen atoms. It also have two lone pairs on it. Hence in (b), the central sulfur atom is surrounded by 10 electrons therefore it obeys does not octet rule and is an expanded octet

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 6.6, Problem 10PPA , additional homework tip  5

Here sulfur shares two double bond with oxygen atoms and shares a single bond with remaining one oxygen atoms. It also have two lone pairs on it. Hence in (c), the central sulfur atom is surrounded by 12 electrons therefore it obeys does not octet rule and is an expanded octet

Expert Solution
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Interpretation Introduction

Interpretation: the formal charge of atoms of the resonance structure of hydrogen sulfite ion should be determined.

Concept Introduction:

  • A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.
  • This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.
  • Formal charge of an atom can be determined by the given formula.

Formalcharge(FC)=(no.ofvalenceelectroninatom)12(no.ofbondingelectrons)(no.ofnon-bondingelectrons)

To determine: the formal charge of atoms of the resonance structure of hydrogen sulfite ion.

Answer to Problem 10PPA

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 6.6, Problem 10PPA , additional homework tip  6 CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 6.6, Problem 10PPA , additional homework tip  7 CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 6.6, Problem 10PPA , additional homework tip  8

Explanation of Solution

For a

The formal charge of atoms in (a) is calculated.

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 6.6, Problem 10PPA , additional homework tip  9

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 6.6, Problem 10PPA , additional homework tip  10 Formalcharge(FC)=(no.ofvalenceelectroninatom)12(no.ofbondingelectrons)(no.ofnon-bondingelectrons)

  • Sulfur atom

Numberofvalenceelectrons=6Numberofbondingelectrons=6Numberofnon-bondingelectrons=2

Substituting,

FC=6(12×6)2=+1

  • Terminal oxygen atoms that has single bond with sulfur atom

Numberofvalenceelectrons=6Numberofbondingelectrons=2Numberofnon-bondingelectrons=6

Substituting,

FC=6(12×2)6=1

  • Oxygen atoms that has single bond with sulfur atom and hydrogen

Numberofvalenceelectrons=6Numberofbondingelectrons=4Numberofnon-bondingelectrons=4

Substituting,

FC=6(12×4)4=0

  • Hydrogen atom

Numberofvalenceelectrons=1Numberofbondingelectrons=2Numberofnon-bondingelectrons=0

Substituting,

FC=1(12×2)=0

For b

The formal charge of atoms in (b) is calculated.

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 6.6, Problem 10PPA , additional homework tip  11

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 6.6, Problem 10PPA , additional homework tip  12 Formalcharge(FC)=(no.ofvalenceelectroninatom)12(no.ofbondingelectrons)(no.ofnon-bondingelectrons)

  • Sulfur atom

Numberofvalenceelectrons=6Numberofbondingelectrons=8Numberofnon-bondingelectrons=2

Substituting,

FC=6(12×8)2=0

  • Terminal oxygen atom that has single bond with sulfur atom

Numberofvalenceelectrons=6Numberofbondingelectrons=2Numberofnon-bondingelectrons=6

Substituting,

FC=6(12×2)6=1

  • Terminal oxygen atom that has double bond with sulfur atom

Numberofvalenceelectrons=6Numberofbondingelectrons=4Numberofnon-bondingelectrons=4

Substituting,

FC=6(12×4)4=0

  • Oxygen atoms that has single bond with sulfur atom and hydrogen

Numberofvalenceelectrons=6Numberofbondingelectrons=4Numberofnon-bondingelectrons=4

Substituting,

FC=6(12×4)4=0

  • Hydrogen atom

Numberofvalenceelectrons=1Numberofbondingelectrons=2Numberofnon-bondingelectrons=0

Substituting,

FC=1(12×2)=0

For c

The formal charge of atoms in (c) is calculated.

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 6.6, Problem 10PPA , additional homework tip  13

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 6.6, Problem 10PPA , additional homework tip  14 Formalcharge(FC)=(no.ofvalenceelectroninatom)12(no.ofbondingelectrons)(no.ofnon-bondingelectrons)

  • Sulfur atom

Numberofvalenceelectrons=6Numberofbondingelectrons=10Numberofnon-bondingelectrons=2

Substituting,

FC=6(12×10)2=1

  • Terminal oxygen atoms that has double bond with sulfur atom

Numberofvalenceelectrons=6Numberofbondingelectrons=Numberofnon-bondingelectrons=6

Substituting,

FC=6(12×4)4=0

Oxygen atoms that has single bond with sulfur atom and hydrogen

Numberofvalenceelectrons=6Numberofbondingelectrons=4Numberofnon-bondingelectrons=4

Substituting,

FC=6(12×4)4=0

Hydrogen atom

Numberofvalenceelectrons=1Numberofbondingelectrons=2Numberofnon-bondingelectrons=0

Substituting,

FC=1(12×2)=0

  • The electronegativity of oxygen is higher than sulfur so usually more electronegative receives negative formal charge. But in the case of resonance structure (c), the less electronegative sulfur atom has -1 formal charge where oxygen gets zero formal charge. Therefore resonance structure (c) has formal charges that are inconsistent with the electronegativities of the elements.

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Chapter 6 Solutions

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT

Ch. 6.2 - Prob. 3PPBCh. 6.2 - Prob. 3PPCCh. 6.2 - Prob. 6.2.1SRCh. 6.2 - Prob. 6.2.2SRCh. 6.2 - Prob. 6.2.3SRCh. 6.2 - Prob. 6.2.4SRCh. 6.3 - Draw the Lewis structure for carbon disulfide...Ch. 6.3 - Prob. 4PPACh. 6.3 - Prob. 4PPBCh. 6.3 - Prob. 4PPCCh. 6.3 - Prob. 6.3.1SRCh. 6.3 - Prob. 6.3.2SRCh. 6.4 - The widespread use of fertilizers has resulted in...Ch. 6.4 - Prob. 5PPACh. 6.4 - Prob. 5PPBCh. 6.4 - Prob. 5PPCCh. 6.4 - Formaldehyde (CH2O), which can be used 10 preserve...Ch. 6.4 - Prob. 6PPACh. 6.4 - Prob. 6PPBCh. 6.4 - Prob. 6PPCCh. 6.4 - Prob. 6.4.1SRCh. 6.4 - Prob. 6.4.2SRCh. 6.5 - Prob. 6.7WECh. 6.5 - Prob. 7PPACh. 6.5 - Prob. 7PPBCh. 6.5 - Prob. 7PPCCh. 6.5 - Prob. 6.5.1SRCh. 6.5 - Prob. 6.5.2SRCh. 6.6 - Prob. 6.8WECh. 6.6 - Prob. 8PPACh. 6.6 - Prob. 8PPBCh. 6.6 - Prob. 8PPCCh. 6.6 - Prob. 6.9WECh. 6.6 - Prob. 9PPACh. 6.6 - Prob. 9PPBCh. 6.6 - Elements in the same group exhibit similar...Ch. 6.6 - Prob. 6.10WECh. 6.6 - Draw three resonance structures for the hydrogen...Ch. 6.6 - Draw two resonance structures for each speciesone...Ch. 6.6 - Prob. 10PPCCh. 6.6 - Prob. 6.6.1SRCh. 6.6 - Prob. 6.6.2SRCh. 6.6 - Prob. 6.6.3SRCh. 6.6 - Prob. 6.6.4SRCh. 6 - Prob. 6.1QPCh. 6 - Prob. 6.2QPCh. 6 - Prob. 6.3QPCh. 6 - Prob. 6.4QPCh. 6 - Prob. 6.5QPCh. 6 - Prob. 6.6QPCh. 6 - Prob. 6.7QPCh. 6 - Prob. 6.8QPCh. 6 - Prob. 6.9QPCh. 6 - Define electronegativity and explain the...Ch. 6 - Prob. 6.11QPCh. 6 - Prob. 6.12QPCh. 6 - Prob. 6.13QPCh. 6 - Prob. 6.14QPCh. 6 - Prob. 6.15QPCh. 6 - Prob. 6.16QPCh. 6 - Arrange the following bonds in order of increasing...Ch. 6 - Prob. 6.18QPCh. 6 - Prob. 6.19QPCh. 6 - Prob. 6.20QPCh. 6 - Prob. 6.21QPCh. 6 - Prob. 6.22QPCh. 6 - Prob. 6.23QPCh. 6 - Prob. 6.24QPCh. 6 - Prob. 6.25QPCh. 6 - Prob. 6.26QPCh. 6 - Prob. 6.27QPCh. 6 - Prob. 6.28QPCh. 6 - Prob. 6.29QPCh. 6 - Prob. 6.30QPCh. 6 - Prob. 6.31QPCh. 6 - Prob. 6.32QPCh. 6 - Prob. 6.33QPCh. 6 - Prob. 6.34QPCh. 6 - Draw all of the resonance structures for the...Ch. 6 - Prob. 6.36QPCh. 6 - Prob. 6.37QPCh. 6 - Draw three resonance structures for the molecule...Ch. 6 - Draw three reasonable resonance structures for the...Ch. 6 - Indicate which of the following are resonance...Ch. 6 - Prob. 6.41QPCh. 6 - Prob. 6.42QPCh. 6 - Draw a resonance structure of the guanine molecule...Ch. 6 - Prob. 6.44QPCh. 6 - Give three examples of compounds that do not...Ch. 6 - Prob. 6.46QPCh. 6 - Prob. 6.47QPCh. 6 - Prob. 6.48QPCh. 6 - Prob. 6.49QPCh. 6 - Prob. 6.50QPCh. 6 - Prob. 6.51QPCh. 6 - Prob. 6.52QPCh. 6 - Prob. 6.53QPCh. 6 - Draw Lewis structures for the radical species ClF2...Ch. 6 - Prob. 6.55QPCh. 6 - Prob. 6.56QPCh. 6 - Prob. 6.57QPCh. 6 - Prob. 6.58QPCh. 6 - Prob. 6.59QPCh. 6 - Prob. 6.60QPCh. 6 - Give an example of an ion or molecule containing...Ch. 6 - Prob. 6.62QPCh. 6 - Prob. 6.63QPCh. 6 - Prob. 6.64QPCh. 6 - Are the following statements true or false? (a)...Ch. 6 - Prob. 6.66QPCh. 6 - Prob. 6.67QPCh. 6 - Most organic acids can be represented as RCOOH,...Ch. 6 - Prob. 6.69QPCh. 6 - Prob. 6.70QPCh. 6 - Prob. 6.71QPCh. 6 - The following species have been detected in...Ch. 6 - Prob. 6.73QPCh. 6 - Prob. 6.74QPCh. 6 - The triiodide ion (I3) in which the I atoms are...Ch. 6 - Prob. 6.76QPCh. 6 - Prob. 6.77QPCh. 6 - The chlorine nitrate (ClONO2) molecule is believed...Ch. 6 - Prob. 6.79QPCh. 6 - For each of the following organic molecules draw a...Ch. 6 - Prob. 6.81QPCh. 6 - Draw Lewis structures for the following organic...Ch. 6 - Draw Lewis structures for the following four...Ch. 6 - Prob. 6.84QPCh. 6 - Prob. 6.85QPCh. 6 - Draw three resonance structures for (a) the...Ch. 6 - Prob. 6.87QPCh. 6 - Prob. 6.88QPCh. 6 - Prob. 6.89QPCh. 6 - Draw a Lewis structure for nitrogen pentoxide...Ch. 6 - Prob. 6.91QPCh. 6 - Nitrogen dioxide (NO2) is a stable compound....Ch. 6 - Prob. 6.93QPCh. 6 - Vinyl chloride (C2H3Cl) differs from ethylene...Ch. 6 - Prob. 6.95QPCh. 6 - Prob. 6.96QPCh. 6 - In 1999 an unusual cation containing only nitrogen...Ch. 6 - Prob. 6.98QPCh. 6 - Prob. 6.99QPCh. 6 - Electrostatic potential maps for three compounds...Ch. 6 - Which of the following atoms must always obey the...Ch. 6 - Prob. 6.2KSPCh. 6 - Prob. 6.3KSPCh. 6 - How many lone pairs are on the central atom in the...
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