Resonance structures of propenal are given below. Which of these resonating structures is more stable? Give re

NCERT Exemplar · Q47

Q. Resonance structures of propenal are given below. Which of these resonating structures is more stable? Give reason for your answer. I. CH2=CH—CH=O II. ^+CH2—CH=CH—O^-

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Concept understanding — Resonance Structures Drawing

Resonance Structures: What They Are and How to Draw Them

Let's start with a simple question. When you draw a molecule like ozone ( $O_3$ ), you might put a double bond between the central oxygen and one of the end oxygens, and a single bond to the other. But experiments show both O–O bonds are identical — same length, same strength. So which drawing is correct?

Neither single drawing is correct. The real molecule is a hybrid of both possibilities. That's the core idea of resonance.

The Intuition: A Musical Analogy

Think of a chord played on a piano. A C major chord is made of three notes: C, E, G. No single note is the chord — the chord is the blend of all three. Similarly, a resonance hybrid is the blend of all valid Lewis structures (called resonance contributors or canonical forms) for a molecule. The real molecule is not flipping between these forms; it exists as a single, stable average.

Note

Resonance structures are not in equilibrium. The molecule does not switch from one form to another. It is a single structure that is the weighted average of all contributors.

The Precise Definition

Resonance structures are two or more Lewis structures that differ only in the placement of electrons (pi bonds and lone pairs), never in the positions of atoms. The real molecule is described by a resonance hybrid — a superposition of all contributors.

Rules for Valid Resonance Structures

Atoms never move. Only electrons (pi bonds, lone pairs, and sometimes sigma bonds in special cases) change positions.
The total number of electrons stays the same. You are just redistributing them.
Each structure must obey the octet rule (for second-period elements) and have valid formal charges.
All structures must have the same net charge and the same number of unpaired electrons (if any).

How to Draw Resonance Structures: A Step-by-Step Method

Let's use the carbonate ion ( $CO_3^{2-}$ ) as our example.

Step 1: Draw the best Lewis structure

Start with the skeleton: carbon in the center, three oxygens around it. Count valence electrons: C has 4, each O has 6, plus 2 for the charge = $4 + 18 + 2 = 24$ electrons. Place bonds and lone pairs to satisfy octets. You'll get one structure with a C=O double bond and two C–O single bonds, each single-bonded oxygen carrying a negative charge.

Step 2: Identify movable electrons

Look for pi bonds (double or triple bonds) and lone pairs that are adjacent to pi bonds or to atoms with an empty p orbital. In carbonate, the C=O pi bond and the lone pairs on the negatively charged oxygens are the movable parts.

Step 3: Push electrons using curved arrows

An arrow starts at the electron source (a pi bond or lone pair) and points to where the electrons go (to form a new pi bond or to become a lone pair). In carbonate:

Take the pi bond from C=O and push it to become a lone pair on that oxygen.
Simultaneously, take a lone pair from a negatively charged oxygen and push it to form a new C=O pi bond.

Step 4: Draw the new structure

After pushing, you get a second structure where the double bond is on a different oxygen. Repeat to get the third structure (all three oxygens take turns being double-bonded).

Tip

Always check that the total number of electrons and the net charge remain unchanged after each arrow push. A common mistake is to accidentally add or remove electrons.

Common Patterns to Recognize

Pattern	Example	What moves
Allylic system	$CH_2=CH-CH_2^+$	Pi bond shifts, positive charge moves
Conjugated diene	$CH_2=CH-CH=CH_2$	Pi bonds shift (less common in neutral molecules)
Carbonyl group	$R_2C=O$	Lone pair from O forms pi bond, pi bond becomes lone pair
Benzene ring	$C_6H_6$	Alternating double bonds shift around the ring

The Most Common Mistake Beginners Make

Breaking sigma bonds. Remember: sigma bonds (single bonds between atoms) never break in resonance. Only pi bonds and lone pairs move. If you find yourself moving an atom or breaking a single bond, you are drawing a different molecule (a constitutional isomer), not a resonance structure. …

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Structure I is more stable because all atoms (except hydrogen) have complete octets and there is no charge separation, unlike Structure II which has an incomplete octet on carbon and separated charges. The more stable structure is I. CH $_2$ =CH—CH=O.

When comparing the stability of different resonance structures, we follow a set of rules, often in a hierarchical order. The goal is to identify the structure that contributes most significantly to the overall resonance hybrid, meaning it is the most stable representation.

Here are the key principles for determining the relative stability of resonance structures:

Octet Rule: Structures where all atoms (except hydrogen, which needs a duet) have a complete octet are generally more stable. This is the most important rule.
Number of Covalent Bonds: Structures with a greater number of covalent bonds are more stable. More bonds mean more energy released, leading to a lower energy state.
Charge Separation: Structures with less charge separation are more stable. Separating charges requires energy, making structures with separated charges less stable than neutral ones or those with minimal charge separation.
Placement of Charges:
- If charge separation is unavoidable, structures with negative charges on more electronegative atoms (like O, N, F) and positive charges on less electronegative atoms (like C) are more stable.
- Adjacent like charges (e.g., two positive charges next to each other) are highly destabilising.

Let's apply these rules to the given resonance structures of propenal:

Analyze Structure I: CH $_2$ =CH—CH=O
- Let's count the valence electrons around each non-hydrogen atom:
  - Carbon 1 (CH $_2$ ): Forms two single bonds with H and one double bond with C. Total bonds = 4. Total valence electrons = $2 \times 2 (\text{from H}) + 4 (\text{from C-C double bond}) = 8$ . Complete octet.
  - Carbon 2 (CH): Forms one single bond with H, one double bond with C1, and one single bond with C3. Total bonds = 4. Total valence electrons = $2 (\text{from H}) + 4 (\text{from C-C double bond}) + 2 (\text{from C-C single bond}) = 8$ . Complete octet.
  - Carbon 3 (CH): Forms one single bond with H, one single bond with C2, and one double bond with O. Total bonds = 4. Total valence electrons = $2 (\text{from H}) + 2 (\text{from C-C single bond}) + 4 (\text{from C-O double bond}) = 8$ . Complete octet.
  - Oxygen (O): Forms one double bond with C3 and has two lone pairs. Total bonds = 2. Total valence electrons = $4 (\text{from C-O double bond}) + 4 (\text{from 2 lone pairs}) = 8$ . Complete octet.
- Formal Charges: All atoms in Structure I have a formal charge of zero.
- Conclusion for I: All atoms have complete octets, and there is no charge separation.
Analyze Structure II: $^+$ CH $_2$ —CH=CH—O $^-$
- Let's count the valence electrons around each non-hydrogen atom:
  - Carbon 1 ( $^+$ CH $_2$ ): Forms two single bonds with H and one single bond with C2. Total bonds = 3. Total valence electrons = $2 \times 2 (\text{from H}) + 2 (\text{from C-C single bond}) = 6$ . Incomplete octet.
  - Carbon 2 (CH): Forms one single bond with H, one single bond with C1, and one double bond with C3. Total bonds = 4. Total valence electrons = $2 (\text{from H}) + 2 (\text{from C-C single bond}) + 4 (\text{from C-C double bond}) = 8$ . Complete octet.
  - Carbon 3 (CH): Forms one single bond with H, one double bond with C2, and one single bond with O. Total bonds = 4. Total valence electrons = $2 (\text{from H}) + 4 (\text{from C-C double bond}) + 2 (\text{from C-O single bond}) = 8$ . Complete octet. …

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Q.Resonance structures of propenal are given below. Which of these resonating structures is more stable? Give reason for your answer. I. CH2=CH—CH=O II. ^+CH2—CH=CH—O^-

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Q.Resonance structures of propenal are given below. Which of these resonating structures is more stable? Give reason for your answer.
I. CH2=CH—CH=O
II. ^+CH2—CH=CH—O^-

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