EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
7th Edition
ISBN: 8220106637203
Author: Chang
Publisher: YUZU
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Chapter 9, Problem 9.54QP
Interpretation Introduction

Interpretation:

The three resonance structures for the given molecule OCS should be drawn and the formal charges should be indicated.

Concept Introduction:

Octet rule: According to octet rule, when an atom contains eight electrons in its valence shell (outermost shell), it will be stable. Most of the atom will donate or accept electrons to its outer most orbital to satisfy the octet rule.

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.

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.

All the possible resonance structures are imaginary whereas the resonance hybrid is real.

Resonance is a method used to describe about delocalized electrons inside certain molecules or polyatomic ions since the Lewis structure can’t express it. A molecule or ion containing delocalized electrons can be represented by using several similar structures such structures are called as resonance structures or canonical structures.

The general rules to be followed for drawing resonance structures are given below:

  1. 1. The position of the atoms does not change only the pi electrons and a lone pair of electrons can change their positions.
  2. 2. The total number of valence electrons in all the resonance structures remains the same.
  3. 3. Octet rule must not be violated. No atom can have electrons more than 8 in its valence shell except sulphur.
  4. 4. Transfer of electrons between the bonds is shown by curved arrows.

Formal charge of an atom can be determined by the given formula.

Formalcharge(FC)=(numberofvalenceelectroninatom)12(numberofbondingelectrons)(numberofnon-bondingelectrons)

Expert Solution & Answer
Check Mark

Answer to Problem 9.54QP

The resonance structure can be drawn as,

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO, Chapter 9, Problem 9.54QP , additional homework tip  1

Explanation of Solution

Total electrons in OCS is,

(valence e on O )+(valence e on C)+ (valence e on S) = 6+4+6 = 16

Accordingly Lewis structure of OCS is drawn which shows 16 electrons in the molecule, including bonding and non-bonding electrons.

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO, Chapter 9, Problem 9.54QP , additional homework tip  2

The resonance structures of molecule is as follows,

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO, Chapter 9, Problem 9.54QP , additional homework tip  3

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO, Chapter 9, Problem 9.54QP , additional homework tip  4

  • Oxygen atom

Numberofvalenceelectron=6Numberofbondingelectron=4Numberofnon-bondingelectron=4

Substituting these values to the equation,

FC=6-(12×4)-4=0

  • Carbon atom

Numberofvalenceelectron=4Numberofbondingelectron=8Numberofnon-bondingelectron=0

Substituting these values to the equation,

FC=4-(12×8)-0=0

  • Sulphur atom

Numberofvalenceelectron=6Numberofbondingelectron=4Numberofnon-bondingelectron=4

Substituting these values to the equation,

FC=6-(12×4)-4=0

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO, Chapter 9, Problem 9.54QP , additional homework tip  5

  • Oxygen atom

Numberofvalenceelectron=6Numberofbondingelectron=2Numberofnon-bondingelectron=6

Substituting these values to the equation,

FC=6-(12×2)-6=-1

  • Carbon atom

Numberofvalenceelectron=4Numberofbondingelectron=8Numberofnon-bondingelectron=0

Substituting these values to the equation,

FC=4-(12×8)-0=0

  • Sulphur atom

Numberofvalenceelectron=6Numberofbondingelectron=6Numberofnon-bondingelectron=2

Substituting these values to the equation,

FC=6-(12×6)-2=+1

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO, Chapter 9, Problem 9.54QP , additional homework tip  6

  • Oxygen atom

Numberofvalenceelectron=6Numberofbondingelectron=6Numberofnon-bondingelectron=2

Substituting these values to the equation,

FC=6-(12×6)-2=+1

  • Carbon atom

Numberofvalenceelectron=4Numberofbondingelectron=8Numberofnon-bondingelectron=0

Substituting these values to the equation,

FC=4-(12×8)-0=0

  • Sulphur atom

Numberofvalenceelectron=6Numberofbondingelectron=2Numberofnon-bondingelectron=6

Substituting these values to the equation,

FC=6-(12×2)-6=-1

Conclusion

The three resonance structures for the given molecule OCS were drawn and the formal charges were indicated.

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Chapter 9 Solutions

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO

Ch. 9.6 - Prob. 1RCCh. 9.7 - Prob. 1PECh. 9.7 - Prob. 2PECh. 9.7 - Prob. 1RCCh. 9.8 - Prob. 1PECh. 9.8 - Prob. 1RCCh. 9.9 - Practice Exercise Draw the Lewis structure for...Ch. 9.9 - Prob. 2PECh. 9.9 - Prob. 3PECh. 9.9 - Prob. 4PECh. 9.9 - Prob. 1RCCh. 9.10 - Prob. 1PECh. 9.10 - Prob. 1RCCh. 9 - Prob. 9.1QPCh. 9 - 9.2 Use the second member of each group from Group...Ch. 9 - Prob. 9.3QPCh. 9 - Prob. 9.4QPCh. 9 - Prob. 9.5QPCh. 9 - Prob. 9.6QPCh. 9 - Prob. 9.7QPCh. 9 - Prob. 9.8QPCh. 9 - Prob. 9.9QPCh. 9 - Prob. 9.10QPCh. 9 - Prob. 9.11QPCh. 9 - Prob. 9.13QPCh. 9 - Prob. 9.14QPCh. 9 - Prob. 9.15QPCh. 9 - Prob. 9.16QPCh. 9 - Prob. 9.17QPCh. 9 - Prob. 9.18QPCh. 9 - Prob. 9.19QPCh. 9 - Prob. 9.20QPCh. 9 - Prob. 9.21QPCh. 9 - 9.22 Explain how the lattice energy of an ionic...Ch. 9 - Prob. 9.23QPCh. 9 - Prob. 9.24QPCh. 9 - Prob. 9.25QPCh. 9 - Prob. 9.26QPCh. 9 - Prob. 9.27QPCh. 9 - Prob. 9.28QPCh. 9 - Prob. 9.29QPCh. 9 - Prob. 9.30QPCh. 9 - Prob. 9.31QPCh. 9 - Prob. 9.33QPCh. 9 - 9.34 Arrange these bonds in order of increasing...Ch. 9 - Prob. 9.35QPCh. 9 - Prob. 9.36QPCh. 9 - Prob. 9.37QPCh. 9 - Prob. 9.38QPCh. 9 - Prob. 9.39QPCh. 9 - Prob. 9.40QPCh. 9 - Prob. 9.41QPCh. 9 - Prob. 9.42QPCh. 9 - Prob. 9.43QPCh. 9 - Prob. 9.44QPCh. 9 - Prob. 9.45QPCh. 9 - Prob. 9.46QPCh. 9 - Prob. 9.47QPCh. 9 - Prob. 9.48QPCh. 9 - Prob. 9.49QPCh. 9 - Prob. 9.50QPCh. 9 - Prob. 9.51QPCh. 9 - Prob. 9.52QPCh. 9 - Prob. 9.53QPCh. 9 - Prob. 9.54QPCh. 9 - Prob. 9.55QPCh. 9 - Prob. 9.56QPCh. 9 - Prob. 9.57QPCh. 9 - Prob. 9.58QPCh. 9 - Prob. 9.59QPCh. 9 - Prob. 9.60QPCh. 9 - Prob. 9.61QPCh. 9 - Prob. 9.62QPCh. 9 - Prob. 9.63QPCh. 9 - Prob. 9.64QPCh. 9 - Prob. 9.65QPCh. 9 - Prob. 9.66QPCh. 9 - Prob. 9.67QPCh. 9 - Prob. 9.68QPCh. 9 - Prob. 9.69QPCh. 9 - Prob. 9.70QPCh. 9 - Prob. 9.71QPCh. 9 - Prob. 9.72QPCh. 9 - Prob. 9.73QPCh. 9 - Prob. 9.74QPCh. 9 - Prob. 9.75QPCh. 9 - Prob. 9.76QPCh. 9 - Prob. 9.77QPCh. 9 - Prob. 9.78QPCh. 9 - Prob. 9.79QPCh. 9 - Prob. 9.80QPCh. 9 - 9.81 Draw reasonable resonance structures for...Ch. 9 - Prob. 9.82QPCh. 9 - Prob. 9.83QPCh. 9 - Prob. 9.84QPCh. 9 - Prob. 9.85QPCh. 9 - Prob. 9.86QPCh. 9 - Prob. 9.87QPCh. 9 - Prob. 9.88QPCh. 9 - Prob. 9.89QPCh. 9 - Prob. 9.90QPCh. 9 - Prob. 9.91QPCh. 9 - Prob. 9.92QPCh. 9 - Prob. 9.93QPCh. 9 - Prob. 9.94QPCh. 9 - Prob. 9.95QPCh. 9 - Prob. 9.96QPCh. 9 - Prob. 9.97QPCh. 9 - Prob. 9.98QPCh. 9 - Prob. 9.99QPCh. 9 - Prob. 9.100QPCh. 9 - Prob. 9.101QPCh. 9 - Prob. 9.102QPCh. 9 - Prob. 9.103QPCh. 9 - Prob. 9.104QPCh. 9 - Prob. 9.105QPCh. 9 - Prob. 9.106QPCh. 9 - Prob. 9.107QPCh. 9 - Prob. 9.108QPCh. 9 - 9.109 Among the common inhaled anesthetics...Ch. 9 - 9.110 Industrially, ammonia is synthesized by the...Ch. 9 - Prob. 9.111QPCh. 9 - Prob. 9.112QPCh. 9 - Prob. 9.113SPCh. 9 - Prob. 9.114SPCh. 9 - Prob. 9.115SPCh. 9 - Prob. 9.116SPCh. 9 - Prob. 9.117SPCh. 9 - Prob. 9.118SP
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