Chemistry

Resonance structures: electron movement

Step-by-step chemistry solution: Resonance structures: electron movement

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1. What the problem is asking

The student is given a set of resonance‑type structures (the first and the last of the series) and is asked to:

1. Draw every reasonable resonance contributor for each of those two molecules.
2. Indicate the electron‑flow with curved‑arrow notation (starting point = lone‑pair or π‑bond; arrow head = where the electron pair ends up).
3. Assign the correct formal charges on every atom after the electron movement.

In short – start from the “parent” Lewis structure, move electrons according to the rules of resonance, and write the resulting structures, making sure that the total number of electrons is unchanged and that each atom obeys the octet (or duet for H) rule.

Below I will walk through the complete reasoning for two very common textbook examples that are often placed at the beginning and the end of a resonance‑exercise worksheet:

• Molecule 1 – an amide fragment: R‑C(=O)‑NR₂ (e.g. N,N‑dimethylformamide).
• Molecule 2 – a nitro‑substituted aromatic fragment: Ph‑NO₂ (nitrobenzene).

The same logic applies to any other first‑/last‑molecule you may have in the worksheet; just replace the substituents (R‑groups) with those shown in your figure.

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2. Step‑by‑step construction of the resonance forms

2.1 General rules for drawing resonance structures

Rule	What it means
Only π‑bonds or lone‑pairs move	σ‑bonds (single bonds that are part of the skeletal framework) never shift in a resonance step.
Arrow starts on the electron source	A lone‑pair or a π‑bond is the donor; the arrow points to the acceptor (usually an adjacent atom or a π‑bond).
The total number of valence electrons stays the same	You never create or destroy electrons; you only redistribute them.
Octet (or duet) rule must be satisfied	After the electron shift, each atom (except H) should have 8 electrons around it (or the appropriate expanded‑octet for third‑row elements).
Formal charge is calculated as (\displaystyle FC = V - (L + \tfrac{1}{2}B))	V = valence electrons of the free atom, L = non‑bonding electrons, B = bonding electrons.
Only the most stable contributors are kept	The “best” resonance forms have (i) the smallest absolute formal charges, (ii) negative charge on the more electronegative atom, (iii) full octets, (iv) charge separation minimized.

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2.2 Molecule 1 – The amide fragment R‑C(=O)‑NR₂

2.2.1 Write the initial Lewis structure

   R
   |
   C
  ║
  O
   |
   N
  / \
 R   R

• The carbonyl C is double‑bonded to O and single‑bonded to N.
• O carries two lone pairs, N carries one lone pair.
• Formal charges in the parent structure: all atoms are neutral.

2.2.2 Identify the possible electron donors

• Lone pair on the nitrogen (N has a lone pair that can be donated).
• π‑bond of the C=O carbonyl (the carbon‑oxygen double bond can shift).

Only one of these can move at a time in a single resonance step.

2.2.3 First resonance move – nitrogen donates its lone pair to the carbonyl

Electron flow

N:  →  C=O   (arrow from N lone pair to the carbonyl carbon)

What happens

• The N→C donation creates a C–N double bond.
• The C=O double bond becomes a C–O single bond, and the oxygen receives the two electrons from the former π‑bond → oxygen now bears a negative formal charge.
• Because nitrogen now has four bonds (R–C, two R‑substituents, and the new C=N double bond) it carries a positive formal charge.

Resulting resonance form

   R
   |
   C
  ║
  O⁻
   |
   N⁺
  / \
 R   R

• Curved arrows:

  • Lone‑pair on N → C (forming C=N)
  • One of the C=O π‑bond electrons → O (giving O⁻)

• Formal charges: O⁻, N⁺, all others neutral.

2.2.4 Second resonance move – carbonyl π‑bond donates to nitrogen

The opposite direction of the first move gives back the original structure, so the only two significant contributors for an amide are:

1. Neutral form (C=O, N with lone pair) – the major contributor because it has no formal charges.
2. Charge‑separated form (C–O⁻, N⁺=C) – contributes because it allows delocalisation of the lone pair onto the carbonyl carbon, stabilising the C‑N bond.

Both obey the octet rule (the carbonyl carbon has a full octet in both forms; nitrogen expands its octet in the charged form, which is allowed for a third‑row element).

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2.3 Molecule 2 – Nitrobenzene Ph‑NO₂

2.3.1 Draw the initial Lewis structure

   O
   ║
Ph–N–O
   ║
   O⁻   (actually the conventional representation is N attached to two O atoms, one double‑bonded, one single‑bonded with a negative charge; the N carries a +1 formal charge)

A more explicit drawing:

   O
   ║
Ph–N⁺–O⁻
   |
   O

• One oxygen is double‑bonded to N, the other is single‑bonded and carries a negative charge.
• Nitrogen carries a positive charge (formal charge +1).

This is the standard resonance form that people first draw.

2.3.2 Identify electron donors

• Lone pair on the negatively charged O⁻ (the single‑bonded oxygen).
• π‑bond of the N=O double bond (the double bond can shift).

2.3.3 Resonance move #1 – Lone pair on O⁻ forms a N=O double bond

Electron flow

O⁻ :  →  N (arrow from O⁻ lone pair to N)

Simultaneously the N=O double bond breaks, sending its π‑electrons onto the other oxygen (the one that was double‑bonded).

Resulting structure

   O⁻
   |
Ph–N=O
   |
   O⁺

• Curved arrows:

  • Lone pair on O⁻ → N (forming N–O single bond that becomes N=O)
  • π electrons of the original N=O double bond → the other O (giving it a positive formal charge).

• Formal charges: O⁻ (the one that donated its pair) becomes neutral, the former double‑bonded O becomes O⁺. Nitrogen returns to neutral (it now has three bonds: to the phenyl ring, to the newly formed N=O, and to the O⁻).

Thus the new resonance contributor is Ph–N(=O)–O⁺ (often drawn as Ph–N–O⁺=O⁻ but the arrow direction shows the charge movement).

2.3.4 Resonance move #2 – The opposite direction (back to the original)

If you start from the second form and move the electrons in the reverse direction (π‑bond of the newly formed N=O moves back, and the lone pair on the O⁺ goes to N), you recover the original Ph–N⁺(=O)–O⁻ structure.

2.3.5 Complete resonance picture

Therefore, nitrobenzene has two major resonance contributors:

1. Structure A (canonical) – ( \displaystyle \text{Ph–N}^{+}=O\; \text{O}^{-})
2. Structure B – ( \displaystyle \text{Ph–N}=O\; \text{O}^{+})

Both obey the octet rule; the charge‑separated form (A) places the negative charge on the more electronegative oxygen, which makes it slightly more important, but the contribution of B is essential for explaining the equal N–O bond lengths observed experimentally (the real molecule is a hybrid of the two).

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3. Final answer – the resonance contributors (with arrows)

3.1 Amide fragment

Resonance form	Curved‑arrow description	Formal charges
(1) Neutral carbonyl (major)	– (no movement)	all atoms neutral
(2) Charge‑separated	1️⃣ N lone‑pair → C (forms C=N) 2️⃣ C=O π‑electrons → O (gives O⁻)	N⁺, O⁻

   R                 R
   |                 |
   C=O   ⇌   C–O⁻
   |          ║   |
   N:          N⁺
  / \        / \
 R   R      R   R

3.2 Nitrobenzene

Resonance form	Curved‑arrow description	Formal charges
(A) Conventional (major)	– (no movement)	N⁺, O⁻
(B) Alternate	1️⃣ O⁻ lone‑pair → N (forming N=O) 2️⃣ π‑electrons of N=O → the other O (giving O⁺)	N neutral, O⁻ becomes neutral, O⁺ appears

   O⁻               O
   ║                ||
Ph–N⁺–O   ⇌   Ph–N=O
                 |
                O⁺

Both sets of structures satisfy the octet rule and together describe the true electronic distribution of the molecules.

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4. Common Mistakes (and how to avoid them)

Mistake	Why it’s wrong	How to fix it
Moving σ‑bonds (e.g., breaking a C–C single bond)	σ‑bonds are not part of resonance; only π‑bonds or lone pairs move.	Look for double bonds or lone pairs that are adjacent to the atom that will accept the electrons.
Creating or destroying electrons (ending up with more/less than the original count)	Resonance conserves the total number of valence electrons.	After each arrow movement, recount the electrons on each atom; the sum must equal the original total.
Violating the octet rule (e.g., giving a second‑row atom 10 electrons)	Most stable resonance forms keep all second‑row atoms with 8 electrons.	If a move would give an atom >8 electrons, that form is disfavored – look for an alternative electron source.
Assigning formal charges incorrectly	Forgetting to count half of the bonding electrons.	Use the formula (FC = V - (L + \tfrac12 B)) for every atom after each step.
Leaving charge separation without justification (e.g., putting a negative charge on carbon)	Negative charge prefers the more electronegative atom (O, N, halogen).	When you generate a charge‑separated form, put the negative charge on the most electronegative atom

Original question: Resonance structures: electron movement on Chemistry Stack Exchange, licensed CC BY-SA.

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