Hybridization and Sigma and Pi Bonds: Quiz
Seven questions on hybridization, electron domains and sigma and pi bonds, with explanations.
Hybridization and Sigma and Pi Bonds: Quiz
7 questions
Carbon has 4 bonded atoms and no lone pairs: 4 electron domains, so sp³, tetrahedral, 109.5°.
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Answer: sp³
Carbon has 4 bonded atoms and no lone pairs: 4 electron domains, so sp³, tetrahedral, 109.5°.
The first bond between two atoms is always σ (head-on); the second is a π bond (side-on overlap of p orbitals).
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Answer: one σ and one π bond
The first bond between two atoms is always σ (head-on); the second is a π bond (side-on overlap of p orbitals).
Carbon is bonded to two O atoms with no lone pairs: 2 domains (each double bond counts once), so sp and linear.
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Answer: sp
Carbon is bonded to two O atoms with no lone pairs: 2 domains (each double bond counts once), so sp and linear.
Three sp² hybrids lie in a plane, as far apart as possible: 120° (trigonal planar).
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Answer: 120°
Three sp² hybrids lie in a plane, as far apart as possible: 120° (trigonal planar).
Two C–H bonds (2 σ) plus the C≡C (1 σ + 2 π): 3 σ and 2 π in total.
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Answer: 3 σ, 2 π
Two C–H bonds (2 σ) plus the C≡C (1 σ + 2 π): 3 σ and 2 π in total.
O has 2 bonds and 2 lone pairs: 4 domains, so sp³. Lone pairs count as domains.
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Answer: sp³
O has 2 bonds and 2 lone pairs: 4 domains, so sp³. Lone pairs count as domains.
Both bonds hold two electrons, but side-on overlap is less effective, so the π bond is weaker: C=C (614 kJ/mol) is less than twice C–C (348 kJ/mol).
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Answer: Side-on overlap of p orbitals is smaller than head-on overlap
Both bonds hold two electrons, but side-on overlap is less effective, so the π bond is weaker: C=C (614 kJ/mol) is less than twice C–C (348 kJ/mol).