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.

Answer to Problem 9.57QP

Solution

The hybridization of carbon atom in ${\text{NO}}_{\text{2}}{}^{+}$ is $\text{sp}$.

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 ${\text{NO}}_{\text{2}}{}^{+}$ is,

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 ${\text{NO}}_{\text{2}}{}^{+}$ is $\text{sp}$.

(b)

Interpretation Introduction

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

Answer to Problem 9.57QP

Solution:

The hybridization of nitrogen atom in ${\text{NO}}_{\text{2}}{}^{-}$ is ${\text{sp}}^2$.

Explanation of Solution

The Lewis structure of ${\text{NO}}_{\text{2}}{}^{-}$ is,

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 ${\text{NO}}_{\text{2}}{}^{-}$ is ${\text{sp}}^2$.

(c)

Interpretation Introduction

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

Answer to Problem 9.57QP

Solution:

The hybridization of nitrogen 1 atom in ${\text{N}}_2{\text{O}}^{-}$ is $\text{sp}$.

The hybridization of nitrogen 2 atom in ${\text{N}}_2{\text{O}}^{-}$ is $\text{sp}$.

Explanation of Solution

The Lewis structure of ${\text{N}}_2{\text{O}}^{-}$ is,

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 ${\text{N}}_2{\text{O}}^{-}$ is $\text{sp}$.

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 ${\text{N}}_2{\text{O}}^{-}$ is $\text{sp}$.

(d)

Interpretation Introduction

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

Answer to Problem 9.57QP

Solution:

The hybridization of nitrogen 1 atom in ${\text{N}}_2{\text{O}}_3$ is ${\text{sp}}^3$.

The hybridization of nitrogen 2 atom in ${\text{N}}_2{\text{O}}_3$ is ${\text{sp}}^3$.

Explanation of Solution

The Lewis structure of ${\text{N}}_2{\text{O}}_3$ is,

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 ${\text{N}}_2{\text{O}}_3$ is ${\text{sp}}^3$.

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 ${\text{N}}_2{\text{O}}_3$ is ${\text{sp}}^3$.

(e)

Interpretation Introduction

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

Answer to Problem 9.57QP

Solution:

The hybridization of nitrogen 1 atom in ${\text{N}}_2{\text{O}}_5$ is ${\text{sp}}^2$.

The hybridization of nitrogen 2 atom in ${\text{N}}_2{\text{O}}_5$ is ${\text{sp}}^2$.

Explanation of Solution

The Lewis structure of ${\text{N}}_2{\text{O}}_5$ is,

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 ${\text{N}}_2{\text{O}}_5$ is ${\text{sp}}^2$.

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 ${\text{N}}_2{\text{O}}_5$ is ${\text{sp}}^2$.

Conclusion

1. a. The hybridization of carbon atom in ${\text{NO}}_{\text{2}}{}^{+}$ is $\text{sp}$.
2. b. The hybridization of nitrogen atom in ${\text{NO}}_{\text{2}}{}^{-}$ is ${\text{sp}}^2$.
3. c. The hybridization of nitrogen 1 atom in ${\text{N}}_2{\text{O}}^{-}$ is $\text{sp}$.

   The hybridization of nitrogen 2 atom in ${\text{N}}_2{\text{O}}^{-}$ is $\text{sp}$.

4. d. The hybridization of nitrogen 1 atom in ${\text{N}}_2{\text{O}}_3$ is ${\text{sp}}^3$.

   The hybridization of nitrogen 2 atom in ${\text{N}}_2{\text{O}}_3$ is ${\text{sp}}^3$.

5. e. The hybridization of nitrogen 1 atom in ${\text{N}}_2{\text{O}}_5$ is ${\text{sp}}^2$.

   The hybridization of nitrogen 2 atom in ${\text{N}}_2{\text{O}}_5$ is ${\text{sp}}^2$.