Propose three resonance forms (that respect the octet rule and energetically nonequivalent) contributing to the structures for each of the following molecule or ion. (No pts for first correct Lewis structure) – (3 pts for the second Lewis that is resonance form to the first Lewis structure, if any) – (3 pts for the third Lewis that is resonance form to the first or second Lewis structures) (ii) Show formal charges where they exist. (Do not show zero FC) (1 pt). (a) C3H4NO^-
The Correct Answer and Explanation is:
To determine three resonance structures for C₃H₄NO⁻, we must draw structures that follow the octet rule, show delocalization of electrons, and differ in the location of pi bonds or lone pairs, while keeping the atom arrangement the same.
Step 1: Determine the structure’s skeleton
We interpret C₃H₄NO⁻ as likely derived from a propynamide or related structure. One likely backbone is:
CH₂=C=N⁻−CH₃
This contains:
- A carbon chain of three atoms (C₁−C₂−C₃),
- A nitrogen atom double bonded to the central carbon,
- An oxygen atom bonded to the same carbon.
A typical form is related to propiolamide (or its anion), where the negative charge may localize on O or N.
Let’s consider this backbone: CH₂−C(=O)−NH⁻ (with one hydrogen lost from the NH₂ group to give the negative charge). The correct molecular formula from this structure is C₃H₄NO⁻.
Resonance Form 1 (Main contributor)
mathematicaCopyEditH₂C−C(=O)−NH⁻
||
O
- C₂ is double bonded to O
- N has a single bond to C and carries a negative charge
- Octet rule is satisfied for all atoms
- Formal charges:
- Nitrogen: −1 (has 6 valence electrons instead of 5)
- Oxygen: 0
Resonance Form 2
mathematicaCopyEditH₂C−C(−O⁻)=NH
- The double bond shifts from C=O to C=N
- Oxygen now has three lone pairs, giving it the −1 charge
- Nitrogen now has a formal charge of 0
Formal charges:
- Oxygen: −1
- Nitrogen: 0
This structure is valid, follows the octet rule, and represents electron delocalization from the nitrogen to oxygen.
Resonance Form 3
mathematicaCopyEditH₂C=C⁻−O−NH₂
- Move the negative charge to the terminal carbon and form a double bond between C₁=C₂
- C₂ is now singly bonded to both O and N
- O is neutral, N has a lone pair
Formal charges:
- Carbon (C₁): −1
- Other atoms: neutral
This form is less stable but still follows the octet rule and illustrates resonance with different electron distribution.
Summary of Formal Charges:
Resonance Form | Negative Charge Location |
---|---|
Form 1 | Nitrogen (N) |
Form 2 | Oxygen (O) |
Form 3 | Terminal Carbon (C₁) |
Explanation
Resonance structures depict the delocalization of electrons in a molecule where a single Lewis structure cannot adequately represent electron distribution. For the molecule C₃H₄NO⁻, three resonance structures can be drawn that follow the octet rule and differ in their distribution of formal charges and pi bonds.
The first resonance form places the negative charge on nitrogen, following typical amide behavior where nitrogen can retain a lone pair. This form shows a carbon double bonded to oxygen and singly bonded to nitrogen, giving nitrogen a negative charge due to one more electron than usual.
In the second resonance form, the lone pair on nitrogen forms a pi bond with carbon, while the pi electrons between carbon and oxygen shift to oxygen. This pushes the negative charge onto oxygen. This resonance form is stabilized by the high electronegativity of oxygen, which can accommodate a negative charge better than nitrogen.
The third resonance form involves shifting electrons further, with a negative charge on the terminal carbon. This form is energetically less favorable but still valid. It shows how even the carbon framework can participate in resonance if conjugated with adjacent atoms.
All three forms satisfy the octet rule and highlight the molecule’s potential delocalization. Such resonance contributes to overall molecular stability by spreading out electron density and lowering potential energy. Understanding these resonance contributors helps predict reactivity, acidity, and other chemical behaviors of the molecule.
