Resonance structures are

Answer: Multiple Lewis structures that contribute to the actual bonding (a single, delocalized structure) The true structure is a hybrid (weighted average) of all valid resonance forms. e.g., O₃ has two resonance structures averaging to equal O-O bonds.

The actual nitrite ion has:

Answer: Two equivalent N-O bonds with bond order 1.5 Resonance averages the structures: both N-O bonds are equivalent with bond order 1.5.

For SO₄²⁻, the best Lewis structure (minimizing formal charges) has S with

Answer: Two single bonds and two double bonds (expanded octet, all formal charges minimized) With expanded octet, S can form 2 single + 2 double bonds; the extra electrons reduce formal charges to a minimum (S = 0, single-bonded O = -1, double-bonded O = 0).

AP Chemistry · Topic 2.6

Resonance and Formal Charge Practice

Part of Compound Structure and Properties.(SAP-4.B)

Practice questions

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Sample questions

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Sample 1difficulty 3/5
$Resonance structures are$
- A
  $Always equivalent$
- B
  $Errors$
- C
  $Different molecules$
- D
  $Multiple Lewis structures that contribute to the actual bonding (a single, delocalized structure)$
  check_circle
Why
$The true structure is a hybrid (weighted average) of all valid resonance forms. e.g., O₃ has two resonance structures averaging to equal O-O bonds.$
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Sample 2difficulty 3/5
$The nitrite ion (NO2-) is shown in two resonance structures. In each, one N-O bond is single and the other is double.$
$The actual nitrite ion has:$
- A
  $Two equivalent N-O bonds with bond order 1.5$
  check_circle
- B
  $Two different N-O bond lengths$
- C
  $Three equivalent bonds of order 1.33$
- D
  $One single and one double bond, frozen in place$
Why
$Resonance averages the structures: both N-O bonds are equivalent with bond order 1.5.$
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Sample 3difficulty 3/5
$For SO₄²⁻, the best Lewis structure (minimizing formal charges) has S with$
- A
  $No bonds$
- B
  $Two single bonds and two double bonds (expanded octet, all formal charges minimized)$
  check_circle
- C
  $All triple bonds$
- D
  $Four single bonds (octet on S)$
Why
$With expanded octet, S can form 2 single + 2 double bonds; the extra electrons reduce formal charges to a minimum (S = 0, single-bonded O = -1, double-bonded O = 0).$
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Sample 4difficulty 3/5
$A Lewis structure of formate (HCO2-) places a double bond on one O and a single bond plus three lone pairs on the other O.$
$The formal charge on the singly-bonded oxygen is:$
- A
  $-1$
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- B
  $-2$
- C
  $0$
- D
  $+1$
Why
$FC = 6 - 6 (lone pair e-) - 1 (half of bonding e-) = -1.$
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Sample 5difficulty 3/5
$Formal charge of an atom in a Lewis structure is calculated as$
- A
  $Atomic number$
- B
  $Number of valence electrons - (lone-pair electrons + ½ \times shared electrons)$
  check_circle
- C
  $Total electrons$
- D
  $Number of bonds$
Why
$FC = V - (L + S/2). The "best" Lewis structures minimize formal charges and place negatives on more electronegative atoms.$
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