Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 9, Problem 9.57QP

(a)

Interpretation Introduction

Interpretation: The hybridization of the nitrogen atom in the given compounds is to be identified.

Concept introduction: Hybridization is a process in which two or more atomic orbital of nearly similar energy combine together to form the equal number of hybrid orbital of similar energy.

To determine: The hybridization of the nitrogen atom in the given compound.

(a)

Expert Solution
Check Mark

Answer to Problem 9.57QP

Solution

The hybridization of carbon atom in NO2+ is sp .

Explanation of Solution

Explanation

Hybridization is a process in which two or more atomic orbital of nearly similar energy combine together to form the equal number of hybrid orbital of similar energy.

The Lewis structure of NO2+ is,

Chemistry: The Science in Context (Fifth Edition), Chapter 9, Problem 9.57QP , additional homework tip  1

Figure 1

The two unpaired electrons of nitrogen atom take part in sigma bond formation with two oxygen atoms. One s and one p orbital of nitrogen atom is used for the formation of two sp hybrid orbital and these two hybrid orbital are taking part in the formation of two sigma bonds.

The remaining two unhybrid p orbital undergoes collateral overlapping for the formation of pi bonds.

Therefore, the hybridization of carbon atom in NO2+ is sp .

(b)

Interpretation Introduction

To determine: The hybridization of the nitrogen atom in the given compound.

(b)

Expert Solution
Check Mark

Answer to Problem 9.57QP

Solution:

The hybridization of nitrogen atom in NO2 is sp2 .

Explanation of Solution

The Lewis structure of NO2 is,

Chemistry: The Science in Context (Fifth Edition), Chapter 9, Problem 9.57QP , additional homework tip  2

Figure 2

The two unpaired electrons of nitrogen atom take part in sigma bond formation with two oxygen atoms. One p orbital of nitrogen atom is used for the formation of one sigma bonds.

The remaining one unhybrid p orbital undergoes collateral overlapping for the formation of pi bond.

Therefore, the hybridization of nitrogen atom in NO2 is sp2 .

(c)

Interpretation Introduction

To determine: The hybridization of the nitrogen atom in the given compound.

(c)

Expert Solution
Check Mark

Answer to Problem 9.57QP

Solution:

The hybridization of nitrogen 1 atom in N2O is sp .

The hybridization of nitrogen 2 atom in N2O is sp .

Explanation of Solution

The Lewis structure of N2O is,

Chemistry: The Science in Context (Fifth Edition), Chapter 9, Problem 9.57QP , additional homework tip  3

Figure 3

The two unpaired electrons of nitrogen 1 atom take part in sigma bond formation with nitrogen and oxygen atoms. One s and one p orbital of nitrogen atom is used for the formation of two sigma bonds. The remaining two unhybrid p orbital undergoes collateral overlapping for the formation of pi bond.

Therefore, the hybridization of nitrogen 1 atom in N2O is sp .

The one unpaired electrons of nitrogen 2 atom take part in sigma bond formation with nitrogen atom. One p orbital of nitrogen atom is used for the formation of one sigma bond. The two non bonding pair of electrons are present in s and p orbitals. The one unhybrid p orbital undergoes collateral overlapping for the formation of pi bond.

Therefore, the hybridization of nitrogen 2 atom in N2O is sp .

(d)

Interpretation Introduction

To determine: The hybridization of the nitrogen atom in the given compound.

(d)

Expert Solution
Check Mark

Answer to Problem 9.57QP

Solution:

The hybridization of nitrogen 1 atom in N2O3 is sp3 .

The hybridization of nitrogen 2 atom in N2O3 is sp3 .

Explanation of Solution

The Lewis structure of N2O3 is,

Chemistry: The Science in Context (Fifth Edition), Chapter 9, Problem 9.57QP , additional homework tip  4

Figure 4

The two unpaired electrons of nitrogen 1 atom take part in sigma bond formation with nitrogen and oxygen atoms. Two p orbital of nitrogen atom is used for the formation of two sigma bonds. The one non bonding pair of electrons are present in s orbital. The remaining one unhybrid p orbital undergoes collateral overlapping for the formation of pi bond.

Therefore, the hybridization of nitrogen 1 atom in N2O3 is sp3 .

The three unpaired electrons of nitrogen 2 atom take part in sigma bond formation with two oxygen and nitrogen atoms. One s and two p orbital of nitrogen atom is used for the formation of three sigma bond. The unhybrid p orbital undergoes collateral overlapping for the formation of pi bond.

Therefore, the hybridization of nitrogen 2 atom in N2O3 is sp3 .

(e)

Interpretation Introduction

To determine: The hybridization of the nitrogen atom in the given compound.

(e)

Expert Solution
Check Mark

Answer to Problem 9.57QP

Solution:

The hybridization of nitrogen 1 atom in N2O5 is sp2 .

The hybridization of nitrogen 2 atom in N2O5 is sp2 .

Explanation of Solution

The Lewis structure of N2O5 is,

Chemistry: The Science in Context (Fifth Edition), Chapter 9, Problem 9.57QP , additional homework tip  5

Figure 5

The three unpaired electrons of nitrogen 1 atom take part in sigma bond formation with oxygen atoms. One s and two p orbital of nitrogen atom is used for the formation of two sigma bonds. The one unhybrid p orbital undergoes collateral overlapping for the formation of pi bond.

Therefore, the hybridization of nitrogen 1 atom in N2O5 is sp2 .

The three unpaired electrons of nitrogen 2 atom take part in sigma bond formation with three oxygen atoms. One s and two p orbital of nitrogen atom is used for the formation of three sigma bond. The one unhybrid p orbital undergoes collateral overlapping for the formation of pi bond.

Therefore, the hybridization of nitrogen 2 atom in N2O5 is sp2 .

Conclusion

  1. a. The hybridization of carbon atom in NO2+ is sp .
  2. b. The hybridization of nitrogen atom in NO2 is sp2 .
  3. c. The hybridization of nitrogen 1 atom in N2O is sp .

    The hybridization of nitrogen 2 atom in N2O is sp .

  4. d. The hybridization of nitrogen 1 atom in N2O3 is sp3 .

    The hybridization of nitrogen 2 atom in N2O3 is sp3 .

  5. e. The hybridization of nitrogen 1 atom in N2O5 is sp2 .

    The hybridization of nitrogen 2 atom in N2O5 is sp2 .

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Chapter 9 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 9.7 - Prob. 11PECh. 9.7 - Prob. 12PECh. 9.7 - Prob. 13PECh. 9 - Prob. 9.1VPCh. 9 - Prob. 9.2VPCh. 9 - Prob. 9.3VPCh. 9 - Prob. 9.4VPCh. 9 - Prob. 9.5VPCh. 9 - Prob. 9.6VPCh. 9 - Prob. 9.7VPCh. 9 - Prob. 9.8VPCh. 9 - Prob. 9.9QPCh. 9 - Prob. 9.10QPCh. 9 - Prob. 9.11QPCh. 9 - Prob. 9.12QPCh. 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 - Prob. 9.22QPCh. 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. 9.32QPCh. 9 - Prob. 9.33QPCh. 9 - Prob. 9.34QPCh. 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 - Prob. 9.81QPCh. 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 - Prob. 9.109QPCh. 9 - Prob. 9.110QPCh. 9 - Prob. 9.111QPCh. 9 - Prob. 9.112QPCh. 9 - Prob. 9.113QPCh. 9 - Prob. 9.114QPCh. 9 - Prob. 9.115APCh. 9 - Prob. 9.116APCh. 9 - Prob. 9.117APCh. 9 - Prob. 9.118APCh. 9 - Prob. 9.119APCh. 9 - Prob. 9.120APCh. 9 - Prob. 9.121APCh. 9 - Prob. 9.122APCh. 9 - Prob. 9.123APCh. 9 - Prob. 9.124APCh. 9 - Prob. 9.125APCh. 9 - Prob. 9.126APCh. 9 - Prob. 9.127APCh. 9 - Prob. 9.128APCh. 9 - Prob. 9.129APCh. 9 - Prob. 9.130APCh. 9 - Prob. 9.131APCh. 9 - Prob. 9.132APCh. 9 - Prob. 9.133APCh. 9 - Prob. 9.134APCh. 9 - Prob. 9.135APCh. 9 - Prob. 9.136APCh. 9 - Prob. 9.137APCh. 9 - Prob. 9.138APCh. 9 - Prob. 9.139APCh. 9 - Prob. 9.140APCh. 9 - Prob. 9.141APCh. 9 - Prob. 9.142APCh. 9 - Prob. 9.143AP
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