Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 8, Problem 8.144AP

(a)

Interpretation Introduction

Interpretation: The Lewis structure of the given species, its resonating structure, assigned formal charge and the reason for the difference in the bond length of given species of nitrogen is to be stated.

Concept introduction: The Lewis symbols of the atoms indicate the total number of valence electrons surrounding the atom or the ions.

To determine: The Lewis structure of N(NO2)2 , assigned formal charge and its resonance form.

(a)

Expert Solution
Check Mark

Answer to Problem 8.144AP

Solution

The formal charge on nitrogen 1 , nitrogen 2 , nitrogen 3 , oxygen bonded to nitrogen with single bond and oxygen bonded to nitrogen with double bond is +1,0,+1,1 and 0 .

Explanation of Solution

Explanation

The atomic number of nitrogen is 7 .

The electronic configuration of nitrogen is [He]2s22p3 . Therefore, nitrogen has five valence electrons.

The atomic number of oxygen is 8 .

The electronic configuration of oxygen is [He]2s22p4 . Therefore, oxygen has six valence electrons.

A species on gaining electron possesses a negative charge. The number of electrons gains or lost is equal to the charge raised on the structure of species.

Therefore, N(NO2)2 has an electron excess with that of the valence electrons of oxygen and nitrogen.

Thus, the total number of valence electrons in N(NO2)2 is 3×5+4×6+1=40 .

Therefore, the Lewis structure of N(NO2)2 is,

Chemistry, Chapter 8, Problem 8.144AP , additional homework tip  1

Figure 1

The above structure shows resonance as,

Chemistry, Chapter 8, Problem 8.144AP , additional homework tip  2

Figure 2

According to the above figure,

  • Nitrogen 1 has four bonds and therefore, eight bonding electrons.
  • Nitrogen 2 has six bonding electrons and two non bonding electrons.
  • Nitrogen 3 has eight bonding electrons.
  • Oxygen bonded to nitrogen with single bond has two bonding electrons and six nonbonding electrons.
  • Oxygen bonded to nitrogen with double bond has four bonding electrons and four nonbonding electrons.

The valence electrons of nitrogen are five and that of oxygen are six.

Formal charge is calculated by the formula,

Formalcharge=(Valenceelectrons(Numberofnonbondingelectrons+Numberofbondingelectrons2))

Substitute the value of valence electrons and bonding and nonbonding electrons of nitrogen 1 and 3 in formal charge formula.

Formalcharge=(5(0+82))=54=+1

Substitute the value of valence electrons and bonding and nonbonding electrons of nitrogen 2 in formal charge formula.

Formalcharge=(5(2+62))=55=0

Substitute the value of valence electrons and bonding and nonbonding electrons of oxygen bonded to nitrogen with single bond in formal charge formula.

Formalcharge=(6(6+22))=67=1

Substitute the value of valence electrons and bonding and nonbonding electrons of oxygen bonded to nitrogen with double bond in formal charge formula.

Formalcharge=(6(4+42))=66=0

Therefore, the formal charge on nitrogen 1 , nitrogen 2 , nitrogen 3 , oxygen bonded to nitrogen with single bond and oxygen bonded to nitrogen with double bond is +1,0,+1,1 and 0 .

(b)

Interpretation Introduction

To determine: The reason for the difference in the bond length of nitrogenoxygen in N(NO2)2 and in N2O .

(b)

Expert Solution
Check Mark

Answer to Problem 8.144AP

Solution

In the bond length of nitrogenoxygen in N(NO2)2 and in N2O is different due to different resonating patterns.

Explanation of Solution

Explanation

Due to the resonance effect, the nitrogenoxygen in N(NO2)2 resonates between double and single bond. Thus, the bond length of nitrogenoxygen in N(NO2)2 is between the nitrogenoxygen single bond and nitrogenoxygen double bond.

The Lewis structure of N2O is,

Chemistry, Chapter 8, Problem 8.144AP , additional homework tip  3

Figure 3

Therefore, nitrogenoxygen bond resonates between single, double and triple bond. Thus, nitrogenoxygen bond in N2O is smaller than that in N(NO2)2

(c)

Interpretation Introduction

To determine: The Lewis structure of NH4+ .

(c)

Expert Solution
Check Mark

Answer to Problem 8.144AP

Solution

The Lewis structure of NH4+ is shown below.

Explanation of Solution

Explanation

The atomic number of nitrogen is 7 .

The electronic configuration of nitrogen is [He]2s22p3 . Therefore, nitrogen has five valence electrons.

The atomic number of hydrogen is 1 .

The electronic configuration of hydrogen is 1s1 . Therefore, hydrogen has one valence electron.

A species on losing electron possesses a positive charge. The number of electrons gains or lost is equal to the charge raised on the structure of species.

Therefore, NH4+ has an electron less with that of the valence electrons of hydrogen and nitrogen.

Thus, the total number of valence electrons in NH4+ is 5+4×11=8 .

Therefore, the Lewis structure of NH4+ is,

Chemistry, Chapter 8, Problem 8.144AP , additional homework tip  4

Figure 4

Conclusion

  1. a. The formal charge on nitrogen 1 , nitrogen 2 , nitrogen 3 , oxygen bonded to nitrogen with single bond and oxygen bonded to nitrogen with double bond is +1,0,+1,1 and 0 .
  2. b. In the bond length of nitrogenoxygen in N(NO2)2 and in N2O is different due to different resonating patterns.
  3. c. The Lewis structure of NH4+ shows eight valence electrons.

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Chapter 8 Solutions

Chemistry

Ch. 8.7 - Prob. 11PECh. 8.8 - 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. 8.175APCh. 8 - Prob. 8.176AP
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