Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card)
Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card)
5th Edition
ISBN: 9780393615296
Author: Rein V. Kirss (Author), Natalie Foster (Author), Geoffrey Davies (Author) Thomas R. Gilbert (Author)
Publisher: W. W. Norton
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Chapter 8, Problem 8.109QP

(a)

Interpretation Introduction

Interpretation: The Lewis structure that contributes most to the bonding in CNO of the given Lewis structures is to be stated.

Concept introduction: The Lewis structures are diagrams that give information about the bonding electron pairs and the lone pairs of electrons in a molecule. Similar to electron dot structure in Lewis diagram the lone pair electrons are represented as dots and they also contain lines which represent bonding electron pairs in a bond.

To determine: If the given Lewis structure contributes to the bonding in CNO .

(a)

Expert Solution
Check Mark

Answer to Problem 8.109QP

Solution

The given Lewis structure does not contribute most to the bonding in CNO molecule.

Explanation of Solution

Explanation

The given Lewis structure is,

Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card), Chapter 8, Problem 8.109QP , additional homework tip  1

Figure 1

The charge present on each atom is known as formal charge which is calculated by using the formula,

Formalcharge=Valenceelectrons(Lonepairelectrons+12Bondpairelectrons)

For oxygen atom,

The number of valence electrons in oxygen atom is six, the lone pair electrons are two and the bonding electrons are six.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

Formalcharge=6(2+12×6)=6(2+3)=65=+1

For carbon atom,

The number of valence electrons in carbon atom is four, the lone pair electrons are five and the bonding electrons are two.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

Formalcharge=4(5+12×2)=4(5+1)=46=2

For nitrogen atom,

The number of valence electrons in nitrogen atom is five, the lone pair electron is zero and the bonding electrons are eight.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

Formalcharge=5(0+12×8)=5(0+4)=54=+1

In the given Lewis structure of CNO molecule, the formal charges on carbon, nitrogen and oxygen atom are 2,+1 and +1 respectively. The most preferred structure is one that contains a minimized formal charge or zero formal charge. Here, none of the atoms carry a zero formal charge. Hence, the given Lewis structure does not contribute most to the bonding in CNO molecule.

(b)

Interpretation Introduction

To determine: If the given Lewis structure contributes to the bonding in CNO .

(b)

Expert Solution
Check Mark

Answer to Problem 8.109QP

Solution

The given Lewis structure does not contribute most to the bonding in CNO molecule.

Explanation of Solution

Explanation

The given Lewis structure is,

Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card), Chapter 8, Problem 8.109QP , additional homework tip  2

Figure 2

The formal charge is calculated by using the formula,

Formalcharge=Valenceelectrons(Lonepairelectrons+12Bondpairelectrons)

For oxygen atom,

The number of valence electrons in oxygen atom is six, the lone pair electrons are four and the bonding electrons are four.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

Formalcharge=6(4+12×4)=6(4+2)=66=0

For carbon atom,

The number of valence electrons in carbon atom is four, the lone pair electrons are three and the bonding electrons are four.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

Formalcharge=4(3+12×4)=4(3+2)=45=1

For nitrogen atom,

The number of valence electrons in nitrogen atom is five, the lone pair electron is zero and the bonding electrons are eight.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

Formalcharge=5(0+12×8)=5(0+4)=54=+1

In the given Lewis structure of CNO molecule, the formal charges on carbon, nitrogen and oxygen atom are 1,+1 and 0 respectively. The most preferred structure is one that contains a minimized formal charge or zero formal charge. Here, carbon carries 1 formal charge and nitrogen carries +1 formal charge.

Thus, it is known that the nitrogen atom is more electronegative than carbon atom. Hence, the given Lewis structure does not contribute most to the bonding in CNO molecule.

(c)

Interpretation Introduction

To determine: If the given Lewis structure contributes to the bonding in CNO .

(c)

Expert Solution
Check Mark

Answer to Problem 8.109QP

Solution

The given Lewis structure does not contribute most to the bonding in CNO molecule.

Explanation of Solution

Explanation

The given Lewis structure is,

Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card), Chapter 8, Problem 8.109QP , additional homework tip  3

Figure 3

The formal charge is calculated by using the formula,

Formalcharge=Valenceelectrons(Lonepairelectrons+12Bondpairelectrons)

For oxygen atom,

The number of valence electrons in oxygen atom is six, the lone pair electrons are five and the bonding electrons are two.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

Formalcharge=6(5+12×2)=6(5+1)=66=0

For carbon atom,

The number of valence electrons in carbon atom is four, the lone pair electrons are two and the bonding electrons are six.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

Formalcharge=4(2+12×6)=4(2+3)=45=1

For nitrogen atom,

The number of valence electrons in nitrogen atom is five, the lone pair electron is zero and the bonding electrons are eight.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

Formalcharge=5(0+12×8)=5(0+4)=54=+1

In the given Lewis structure of CNO molecule, the formal charges on carbon, nitrogen and oxygen atom are 1,+1 and 0 respectively. The most preferred structure is one that contains a minimized formal charge or zero formal charge. Here, carbon carries 1 formal charge and nitrogen carries +1 formal charge. Thus, it is known that the nitrogen atom is more electronegative than carbon atom.

Hence, the given Lewis structure does not contribute most to the bonding in CNO molecule.

(d)

Interpretation Introduction

To determine: If the given Lewis structure contributes to the bonding in CNO .

(d)

Expert Solution
Check Mark

Answer to Problem 8.109QP

Solution

The given Lewis structure contributes most to the bonding in CNO molecule.

Explanation of Solution

Explanation

The given Lewis structure is,

Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card), Chapter 8, Problem 8.109QP , additional homework tip  4

Figure 4

The formal charge is calculated by using the formula,

Formalcharge=Valenceelectrons(Lonepairelectrons+12Bondpairelectrons)

For oxygen atom,

The number of valence electrons in oxygen atom is six, the lone pair electrons are six and the bonding electrons are two.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

Formalcharge=6(6+12×2)=6(6+1)=67=1

For carbon atom,

The number of valence electrons in carbon atom is four, the lone pair electron is one and the bonding electrons are six.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

Formalcharge=4(1+12×6)=4(1+3)=44=0

For nitrogen atom,

The number of valence electrons in nitrogen atom is five, the lone pair electron is zero and the bonding electrons are eight.

Substitute the value of valence electrons, lone pair of electrons and bond pair of electrons in the above formula to calculate the formal charge.

Formalcharge=5(0+12×8)=5(0+4)=54=+1

In the given Lewis structure of CNO molecule, the formal charges on carbon, nitrogen and oxygen atom are 0,+1 and 1 respectively. The most preferred structure is one that contains a minimized formal charge or zero formal charge. Here, the formal charges are in accordance to the electronegativity values. Therefore, the given Lewis structure contributes the most to the bonding in the CNO molecule.

Conclusion

The Lewis structure that contributes most to the bonding in the CNO molecule is option (d).

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Chapter 8 Solutions

Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card)

Ch. 8.6 - Prob. 11PECh. 8.7 - Prob. 12PECh. 8 - Prob. 8.1VPCh. 8 - Prob. 8.2VPCh. 8 - Prob. 8.3VPCh. 8 - Prob. 8.4VPCh. 8 - Prob. 8.5VPCh. 8 - Prob. 8.6VPCh. 8 - Prob. 8.7VPCh. 8 - Prob. 8.8VPCh. 8 - Prob. 8.9VPCh. 8 - Prob. 8.10VPCh. 8 - Prob. 8.11VPCh. 8 - Prob. 8.12VPCh. 8 - Prob. 8.13VPCh. 8 - Prob. 8.14VPCh. 8 - Prob. 8.15VPCh. 8 - Prob. 8.16VPCh. 8 - Prob. 8.17VPCh. 8 - Prob. 8.18VPCh. 8 - Prob. 8.19QPCh. 8 - Prob. 8.20QPCh. 8 - Prob. 8.21QPCh. 8 - Prob. 8.22QPCh. 8 - Prob. 8.23QPCh. 8 - Prob. 8.24QPCh. 8 - Prob. 8.25QPCh. 8 - Prob. 8.26QPCh. 8 - Prob. 8.27QPCh. 8 - Prob. 8.28QPCh. 8 - Prob. 8.29QPCh. 8 - Prob. 8.30QPCh. 8 - Prob. 8.31QPCh. 8 - Prob. 8.32QPCh. 8 - Prob. 8.33QPCh. 8 - Prob. 8.34QPCh. 8 - Prob. 8.35QPCh. 8 - Prob. 8.36QPCh. 8 - Prob. 8.37QPCh. 8 - Prob. 8.38QPCh. 8 - Prob. 8.39QPCh. 8 - Prob. 8.40QPCh. 8 - Prob. 8.41QPCh. 8 - Prob. 8.42QPCh. 8 - Prob. 8.43QPCh. 8 - Prob. 8.44QPCh. 8 - Prob. 8.45QPCh. 8 - Prob. 8.46QPCh. 8 - Prob. 8.47QPCh. 8 - Prob. 8.48QPCh. 8 - Prob. 8.49QPCh. 8 - Prob. 8.50QPCh. 8 - Prob. 8.51QPCh. 8 - Prob. 8.52QPCh. 8 - Prob. 8.53QPCh. 8 - Prob. 8.54QPCh. 8 - Prob. 8.55QPCh. 8 - Prob. 8.56QPCh. 8 - Prob. 8.57QPCh. 8 - Prob. 8.58QPCh. 8 - Prob. 8.59QPCh. 8 - Prob. 8.60QPCh. 8 - Prob. 8.61QPCh. 8 - Prob. 8.62QPCh. 8 - Prob. 8.63QPCh. 8 - Prob. 8.64QPCh. 8 - Prob. 8.65QPCh. 8 - Prob. 8.66QPCh. 8 - Prob. 8.67QPCh. 8 - Prob. 8.68QPCh. 8 - Prob. 8.69QPCh. 8 - Prob. 8.70QPCh. 8 - Prob. 8.71QPCh. 8 - Prob. 8.72QPCh. 8 - Prob. 8.73QPCh. 8 - Prob. 8.74QPCh. 8 - Prob. 8.75QPCh. 8 - Prob. 8.76QPCh. 8 - Prob. 8.77QPCh. 8 - Prob. 8.78QPCh. 8 - Prob. 8.79QPCh. 8 - Prob. 8.80QPCh. 8 - Prob. 8.81QPCh. 8 - Prob. 8.82QPCh. 8 - Prob. 8.83QPCh. 8 - Prob. 8.84QPCh. 8 - Prob. 8.85QPCh. 8 - Prob. 8.86QPCh. 8 - Prob. 8.87QPCh. 8 - Prob. 8.88QPCh. 8 - Prob. 8.89QPCh. 8 - Prob. 8.90QPCh. 8 - Prob. 8.91QPCh. 8 - Prob. 8.92QPCh. 8 - Prob. 8.93QPCh. 8 - Prob. 8.94QPCh. 8 - Prob. 8.95QPCh. 8 - Prob. 8.96QPCh. 8 - Prob. 8.97QPCh. 8 - Prob. 8.98QPCh. 8 - Prob. 8.99QPCh. 8 - Prob. 8.100QPCh. 8 - Prob. 8.101QPCh. 8 - Prob. 8.102QPCh. 8 - Prob. 8.103QPCh. 8 - Prob. 8.104QPCh. 8 - Prob. 8.105QPCh. 8 - Prob. 8.106QPCh. 8 - Prob. 8.107QPCh. 8 - Prob. 8.108QPCh. 8 - Prob. 8.109QPCh. 8 - Prob. 8.110QPCh. 8 - Prob. 8.111QPCh. 8 - Prob. 8.112QPCh. 8 - Prob. 8.113QPCh. 8 - Prob. 8.114QPCh. 8 - Prob. 8.115QPCh. 8 - Prob. 8.116QPCh. 8 - Prob. 8.117QPCh. 8 - Prob. 8.118QPCh. 8 - Prob. 8.119QPCh. 8 - Prob. 8.120QPCh. 8 - Prob. 8.121QPCh. 8 - Prob. 8.122QPCh. 8 - Prob. 8.123QPCh. 8 - Prob. 8.124QPCh. 8 - Prob. 8.125QPCh. 8 - Prob. 8.126QPCh. 8 - Prob. 8.127QPCh. 8 - Prob. 8.128QPCh. 8 - Prob. 8.129QPCh. 8 - Prob. 8.130QPCh. 8 - Prob. 8.131QPCh. 8 - Prob. 8.132QPCh. 8 - Prob. 8.133QPCh. 8 - Prob. 8.134QPCh. 8 - Prob. 8.135QPCh. 8 - Prob. 8.136QPCh. 8 - Prob. 8.137QPCh. 8 - Prob. 8.138QPCh. 8 - Prob. 8.139APCh. 8 - Prob. 8.140APCh. 8 - Prob. 8.141APCh. 8 - Prob. 8.142APCh. 8 - Prob. 8.143APCh. 8 - Prob. 8.144APCh. 8 - Prob. 8.145APCh. 8 - Prob. 8.146APCh. 8 - Prob. 8.147APCh. 8 - Prob. 8.148APCh. 8 - Prob. 8.149APCh. 8 - Prob. 8.150APCh. 8 - Prob. 8.151APCh. 8 - Prob. 8.152APCh. 8 - Prob. 8.153APCh. 8 - Prob. 8.154APCh. 8 - Prob. 8.155APCh. 8 - Prob. 8.156APCh. 8 - Prob. 8.157APCh. 8 - Prob. 8.158APCh. 8 - Prob. 8.159APCh. 8 - Prob. 8.160APCh. 8 - Prob. 8.161APCh. 8 - Prob. 8.162APCh. 8 - Prob. 8.163APCh. 8 - Prob. 8.164APCh. 8 - Prob. 8.165APCh. 8 - Prob. 8.166APCh. 8 - Prob. 8.167APCh. 8 - Prob. 8.168APCh. 8 - Prob. 8.169APCh. 8 - Prob. 8.170APCh. 8 - Prob. 8.171APCh. 8 - Prob. 8.172APCh. 8 - Prob. 8.173APCh. 8 - Prob. 8.174APCh. 8 - Prob. 8.175AP
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