Chapter 9, Problem 9.54QP

Draw two resonance structures for diazomethane, CH₂N₂. Show formal charges. The skeletal structure of the molecule is

Interpretation Introduction

Interpretation: The resonance structure of diazomethane should be found.

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.

In some molecules, there is possibility of more than one Lewis structure where all the structures are equally acceptable. One of the acceptable Lewis structures of these molecules is called resonance structure.

All the possible resonance structures are imaginary whereas the resonance hybrid is real.

Any of the possible structure does not exist as such like a stable real molecule. So it is not possible to isolate one resonance structure.

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

Structure with greater number of covalent bonds are more stable comparing to that with lower number of covalent bonds.

Structure which does not involve charge separation is more stable when comparing with structure having positive and negative charge separation.

While drawing resonance structure of a molecule some rules should be followed where the position, over whole charge and chemical framework remains intact. Also only π and nonbonding electron has been moved in all the three resonance structures

Answer to Problem 9.54QP

Explanation of Solution

The given atomic arrangement of diazomethane is ${\text{CH}}_{\text{2}}{\text{N}}_{\text{2}}$.

Resonance structure of hydrazoic acid is drawn below.

In the case of diazomethane, 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 the diazomethane molecule were drawn.

Interpretation Introduction

Interpretation: The formal charge of diazomethane molecule should be found.

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.

The Lewis structure with formal charge on each of the atoms close to zero is taken as the most plausible structure.

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 9.54QP

Resonance structure of diazomethane is given below.

Explanation of Solution

The formal charge of the given compound is calculated,

• First hydrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=1}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}0\end{array}$

Substituting these values to the equation,

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

• Second hydrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=1}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}0\end{array}$

Substituting these values to the equation,

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

• Carbon atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=4}\\ \text{Number}\text{of}\text{bonding}\text{electron=}8\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}0\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}4-\left(\frac{\text{1}}{\text{2}}\text{×8}\right)\\ \text{=0}\end{array}$

• First 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,

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

• Second nitrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=5}\\ \text{Number}\text{of}\text{bonding}\text{electron=}4\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}4\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}5-\left(\frac{\text{1}}{\text{2}}\text{×4}\right)-4\\ \text{=}-1\end{array}$

Resonance structure of diazomethane is given below.

The formal charge of the given compound is calculated,

• First hydrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=1}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}0\end{array}$

Substituting these values to the equation,

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

• Second hydrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=1}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}0\end{array}$

Substituting these values to the equation,

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

• Carbon atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=4}\\ \text{Number}\text{of}\text{bonding}\text{electron=}6\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}2\end{array}$

Substituting these values to the equation,

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

• First 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,

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

• Second nitrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=5}\\ \text{Number}\text{of}\text{bonding}\text{electron=}6\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}2\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}5-\left(\frac{\text{1}}{\text{2}}\text{×6}\right)-2\\ \text{=}0\end{array}$

Resonance structure of the diazomethane is given below.

The formal charge of the given compound is calculated,

• First hydrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=1}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}0\end{array}$

Substituting these values to the equation,

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

• Second hydrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=1}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}0\end{array}$

Substituting these values to the equation,

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

• Carbon atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=4}\\ \text{Number}\text{of}\text{bonding}\text{electron=}6\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}2\end{array}$

Substituting these values to the equation,

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

• First nitrogen atom,

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=5}\\ \text{Number}\text{of}\text{bonding}\text{electron=}6\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}2\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}5-\left(\frac{\text{1}}{\text{2}}\text{×6}\right)-2\\ \text{=}0\end{array}$

• Second nitrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=5}\\ \text{Number}\text{of}\text{bonding}\text{electron=}4\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}2\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}5-\left(\frac{\text{1}}{\text{2}}\text{×4}\right)-2\\ \text{=}+1\end{array}$

Conclusion

The formal charges of diazomethane molecule were shown.