Lewis Structures: Quiz
Seven questions on counting electrons, multiple bonds, formal charge, resonance and octet exceptions, with explanations.
Lewis Structures: Quiz
7 questions
2 × 1 (H) + 6 (O) = 8: two bonding pairs and two lone pairs.
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Answer: 8
2 × 1 (H) + 6 (O) = 8: two bonding pairs and two lone pairs.
8 valence electrons: three N–H bonds use 6, leaving one lone pair on N.
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Answer: 1
8 valence electrons: three N–H bonds use 6, leaving one lone pair on N.
With 16 electrons, O=C=O gives every atom an octet and all formal charges are zero.
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Answer: Two C=O double bonds
With 16 electrons, O=C=O gives every atom an octet and all formal charges are zero.
5 (N) + 3 × 6 (O) + 1 for the negative charge = 24.
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Answer: 24
5 (N) + 3 × 6 (O) + 1 for the negative charge = 24.
Carbon: 4 valence − 0 lone-pair electrons − ½ × 8 bonding electrons = 0.
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Answer: 0
Carbon: 4 valence − 0 lone-pair electrons − ½ × 8 bonding electrons = 0.
The real molecule is an average of O=O–O and O–O=O, so each bond is between a single and a double bond.
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Answer: The molecule is a resonance hybrid of two structures
The real molecule is an average of O=O–O and O–O=O, so each bond is between a single and a double bond.
Boron forms three single bonds and has no lone pairs: only 6 electrons around it.
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Answer:
Boron forms three single bonds and has no lone pairs: only 6 electrons around it.