Chapter 6, Problem 6.92QP

Nitrogen dioxide (NO₂) is a stable compound. Explain why there is a tendency for two such molecules to combine to form dinitrogen tetroxide (N₂O₄). Draw four resonance structures of N₂O₄, showing formal charges.

Interpretation Introduction

Interpretation: The resonance structure of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ should be drawn. The tendency for two ${\text{NO}}_{\text{2}}$ molecules to combine to form dinitrogen tetroxide should be explained

Concept Introduction:

• 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.
• These structures will differ only in the arrangement of the electrons not in the relative position of the atomic nuclei.

To find: The resonance structure of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ and the explanation for combination of two ${\text{NO}}_{\text{2}}$ molecules results in the formation of dinitrogen tetroxide.

Answer to Problem 6.92QP

Combination of two ${\text{NO}}_{\text{2}}$ molecules is explained as below,

${\text{O}}_{\text{2}}\text{N}·\text{+}·{\text{NO}}_{\text{2}}\stackrel{}{\to }{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$

The single electron of one ${\text{NO}}_{\text{2}}$ reacts with the single electron of other ${\text{NO}}_{\text{2}}$ molecule to form a bond. The covalent bond formed by these shared electrons is more stable since octet is complete in ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$.

Explanation of Solution

• To know the tendency that how two ${\text{NO}}_{\text{2}}$ molecules combine to form dinitrogen tetroxide.

${\text{O}}_{\text{2}}\text{N}·\text{+}·{\text{NO}}_{\text{2}}\stackrel{}{\to }{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$

The ${\text{NO}}_{\text{2}}$ molecule contains a total of 17 valence electrons since nitrogen contributes 5 electrons and each of the oxygen contributes 6 electrons. It has odd number of electrons. It does not obey octet rule and is a stable free radical. The single electron of one ${\text{NO}}_{\text{2}}$ reacts with the single electron of other ${\text{NO}}_{\text{2}}$ molecule to form a bond. The covalent bond formed by these shared electrons is more stable since octet is complete in ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$.

• Resonance structure for ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is drawn below.

In the case of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$, the chemical bonding of a molecule cannot be represented using a single Lewis structure. The chemical bonding are described by delocalization of electrons forming 3 possible resonance structures. In all the 3 resonance structures the position, over whole charge and chemical framework remains intact.

Conclusion

The resonance structures of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ were drawn and combination of two ${\text{NO}}_{\text{2}}$ molecules were explained.

Interpretation Introduction

Interpretation: The formal charges of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ should be drawn. The tendency for two ${\text{NO}}_{\text{2}}$ molecules to combine to form dinitrogen tetroxide should be explained

Concept Introduction

• A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.
• This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.
• Formal charge of an atom can be determined by the given formula.
• $\text{Formal}\text{charge}\left(\text{F}\text{C}\right)\text{=}\left(\begin{array}{l}\text{no}\text{.of}\text{valence}\\ \text{electron}\text{in}\text{atom}\end{array}\right)-\frac{\text{1}}{\text{2}}\left(\begin{array}{l}\text{no}\text{.of}\text{bonding}\\ \text{electrons}\end{array}\right)-\left(\begin{array}{l}\text{no}\text{.of}\text{non-bonding}\\ \text{electrons}\end{array}\right)$

Answer to Problem 6.92QP

Explanation of Solution

The formal charge for the first given resonance structure is depicted below.

The formal charge of the given compound is calculated,

• Nitrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=5}\\ \text{Number}\text{of}\text{bonding}\text{electron=}8\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}0\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}5-\left(\frac{\text{1}}{\text{2}}\text{×8}\right)\\ \text{=}+1\end{array}$

• Both nitrogen atoms are similar in ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule thus have similar formal charges.
• Oxygen atom having single bond with nitrogen

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}-\text{1}\end{array}$

• Both oxygen atom having single bond with nitrogen is similar in ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule thus have similar formal charges.
• Oxygen atom having double bond with nitrogen

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}4\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}4\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}\text{6}-\left(\frac{\text{1}}{\text{2}}\text{×4}\right)-\text{4}\\ \text{=}0\end{array}$

• Both oxygen atom having double bond with nitrogen is similar in ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule thus have similar formal charges.

The formal charge for the second given resonance structure is given below.

The formal charge of the given compound is calculated,

• Nitrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=5}\\ \text{Number}\text{of}\text{bonding}\text{electron=}8\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}0\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}5-\left(\frac{\text{1}}{\text{2}}\text{×8}\right)\\ \text{=}+1\end{array}$

• Both nitrogen atoms are similar in ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule thus have similar formal charges.
• Oxygen atom having single bond with nitrogen

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}-\text{1}\end{array}$

• Both oxygen atom having single bond with nitrogen is similar in ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule thus have similar formal charges.
• Oxygen atom having double bond with nitrogen

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}4\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}4\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}\text{6}-\left(\frac{\text{1}}{\text{2}}\text{×4}\right)-\text{4}\\ \text{=}0\end{array}$

• Both oxygen atom having double bond with nitrogen is similar in ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule thus have similar formal charges.

The formal charge for the third given resonance structure is given below.

The formal charge of the given compound is calculated,

• Nitrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=5}\\ \text{Number}\text{of}\text{bonding}\text{electron=}8\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}0\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}5-\left(\frac{\text{1}}{\text{2}}\text{×8}\right)\\ \text{=}+1\end{array}$

• Both nitrogen atoms are similar in ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule thus have similar formal charges.
• Oxygen atom having single bond with nitrogen

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}-\text{1}\end{array}$

• Both oxygen atom having single bond with nitrogen is similar in ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule thus have similar formal charges.
• Oxygen atom having double bond with nitrogen

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}4\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}4\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}\text{6}-\left(\frac{\text{1}}{\text{2}}\text{×4}\right)-\text{4}\\ \text{=}0\end{array}$

• Both oxygen atom having double bond with nitrogen is similar in ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule thus have similar formal charges.

The formal charge for the fourth given resonance structure is given below.

The formal charge of the given compound is calculated,

• Nitrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=5}\\ \text{Number}\text{of}\text{bonding}\text{electron=}8\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}0\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}5-\left(\frac{\text{1}}{\text{2}}\text{×8}\right)\\ \text{=}+1\end{array}$

• Both nitrogen atoms are similar in ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule thus have similar formal charges.
• Oxygen atom having single bond with nitrogen

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}-\text{1}\end{array}$

• Both oxygen atom having single bond with nitrogen is similar in ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule thus have similar formal charges.
• Oxygen atom having double bond with nitrogen

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}4\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}4\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}\text{6}-\left(\frac{\text{1}}{\text{2}}\text{×4}\right)-\text{4}\\ \text{=}0\end{array}$

• Both oxygen atom having double bond with nitrogen is similar in ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule thus have similar formal charges.

Conclusion

The resonance structure of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ and its corresponding formal charges were drawn. The tendency for two ${\text{NO}}_{\text{2}}$ molecules to combine to form dinitrogen tetroxide was explained