AP Chemical Bonding and Molecular Structure 2 — Questions and Answers

Question 1: Using VSEPR theory, what is the molecular geometry and approximate bond angle in SF₄?

Seesaw (sawhorse) geometry, ~102° and ~173° (Correct answer)
Tetrahedral geometry, 109.5°
Trigonal pyramidal geometry, ~107°
Square planar geometry, 90°

Correct answer: Seesaw (sawhorse) geometry, ~102° and ~173°

SF₄ has 5 electron domains around S (4 bonding + 1 lone pair), giving a trigonal bipyramidal electron geometry. With one lone pair in an equatorial position, the molecular geometry is seesaw with axial (~173°) and equatorial (~102°) bond angles.

VSEPR analysis of SF₄: S has 6 valence electrons. Each of 4 F atoms contributes 1 bond. Lewis structure: 4 S–F bonds + 1 lone pair on S = 5 electron domains. 5 electron domains → trigonal bipyramidal electron geometry. The lone pair preferentially occupies an equatorial position (120° spacing = less repulsion than the 90° spacing of an axial position). With the lone pair equatorial, the 4 F atoms occupy 2 equatorial and 2 axial positions → seesaw molecular geometry. Bond angles: equatorial F–S–F ≈ 102° (compressed from 120° by lone pair repulsion); axial F–S–F ≈ 173° (compressed from 180° by lone pair repulsion). SF₄ is an important industrial fluorinating agent. Understanding its geometry is key to predicting its reactivity and dipole moment (the molecule is polar despite having 4 identical bonds, due to asymmetric geometry).

Question 2: Which of the following molecules has a dipole moment of zero?

CO₂ (Correct answer)
H₂O
NH₃
SO₂

Correct answer: CO₂

CO₂ has a linear geometry (O=C=O), so the two equal and opposite C=O bond dipoles cancel exactly, giving a net dipole moment of zero. The other molecules have bent or pyramidal geometries with nonzero dipole moments.

Molecular polarity depends on both the polarity of individual bonds and the molecular geometry. CO₂: linear geometry (sp hybridization on C). The two C=O bond dipoles point in exactly opposite directions (180° apart) and are equal in magnitude, so they cancel. Net μ = 0. H₂O: bent geometry (~104.5°). The two O–H bond dipoles point toward the electronegative oxygen with a net downward vector. Net μ = 1.85 D (quite polar). NH₃: trigonal pyramidal. Three N–H bond dipoles plus the lone pair's electron density all point in the same general direction. Net μ = 1.47 D. SO₂: bent geometry (similar to H₂O). The two S=O bond dipoles do not cancel. Net μ ≠ 0. Note that CCl₄ (tetrahedral) and BF₃ (trigonal planar) also have zero dipole moments due to symmetric geometry, even though individual bonds are polar. Geometry is crucial — XeF₄ (square planar) is nonpolar, while SF₄ (seesaw) is polar.

Question 3: What is the hybridization of the central carbon in ketene, H₂C=C=O?

The terminal CH₂ carbon is sp² and the central carbon bonded to O is sp (Correct answer)
Both carbons are sp²
Both carbons are sp
The terminal CH₂ is sp³ and the central carbon is sp²

Correct answer: The terminal CH₂ carbon is sp² and the central carbon bonded to O is sp

In H₂C=C=O: the left carbon (CH₂) has 3 electron domains (2 H + 1 double bond) → sp². The central carbon has 2 electron domains (2 double bonds) → sp hybridized.

Ketene (H₂C=C=O) is a cumulated diene system with consecutive double bonds sharing the middle carbon. The terminal carbon (CH₂=): bonded to 2 H atoms and double-bonded to the central C. Total sigma bonds = 3 (2 to H + 1 to C) + 0 lone pairs = 3 electron domains → sp² hybridized. It has 1 remaining p orbital for π bonding. The central carbon (=C=): double-bonded to CH₂ on one side and double-bonded to O on the other. Total sigma bonds = 2 + 0 lone pairs = 2 electron domains → sp hybridized. It has 2 remaining p orbitals at 90° to each other for the two perpendicular π bonds. This means the π bonds in ketene are perpendicular to each other — one π bond is in the horizontal plane (C–C π bond) and the other is in the vertical plane (C–O π bond). Ketene's unusual structure makes it highly reactive as a carbonyl-like electrophile, and it is used industrially in acetylation reactions. Cumulated diene systems always have sp hybridization at the central atom.

Question 4: In molecular orbital theory, which of the following correctly describes the bond order and magnetic properties of O₂?

Bond order = 2, paramagnetic (2 unpaired electrons) (Correct answer)
Bond order = 3, diamagnetic
Bond order = 2, diamagnetic
Bond order = 1, paramagnetic

Correct answer: Bond order = 2, paramagnetic (2 unpaired electrons)

O₂ MO configuration: (σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)². Bond order = (8−4)/2 = 2. The 2 electrons in degenerate π*2p orbitals are unpaired (Hund's rule), making O₂ paramagnetic.

Molecular orbital theory is essential for correctly predicting O₂ properties — a case where Lewis structures fail (they predict a diamagnetic double bond). MO filling order for O₂ (16 electrons total, 8 per atom): (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π2p)²(π*2p)¹(π*2p)¹ Using only the valence (n=2) electrons: bonding electrons = 8, antibonding electrons = 4. Bond order = (bonding − antibonding)/2 = (8 − 4)/2 = 2 → double bond. ✓ The two degenerate π*2p orbitals each receive one electron (Hund's rule). These electrons have parallel spins → paramagnetic (attracted to magnetic fields). This was experimentally confirmed by liquid oxygen being attracted to magnets. This is a landmark success of MO theory and one reason it is preferred over valence bond theory for diatomic molecules. Bond order 2 is consistent with the O=O double bond in Lewis structures, but the paramagnetic character can only be explained by MO theory.

Question 5: Arrange these bonds in order of increasing bond length: C–C single, C=C double, C≡C triple, C–H.

C≡C < C=C < C–C < C–H
C–H < C≡C < C=C < C–C (Correct answer)
C≡C < C–H < C=C < C–C
C–C < C=C < C≡C < C–H

Correct answer: C–H < C≡C < C=C < C–C

C–H (109 pm) < C≡C (120 pm) < C=C (134 pm) < C–C (154 pm). More bonds between C atoms shorten the C–C distance, and C–H is shorter than any C–C bond due to hydrogen's small atomic radius.

Bond length is inversely related to bond order and is influenced by atomic radii. For carbon-carbon bonds: - C–C (single, bond order 1): ~154 pm — long, weaker - C=C (double, bond order 2): ~134 pm — more electron density, shorter - C≡C (triple, bond order 3): ~120 pm — maximum electron density, shortest For C–H: ~109 pm. Hydrogen has a very small atomic radius (covalent radius ~31 pm), much smaller than carbon's (~77 pm). The resulting C–H bond is shorter than even the C≡C bond. The trend in bond strength parallels the trend in shortness: C–H (413 kJ/mol), C–C (347), C=C (614), C≡C (839 kJ/mol). Longer bonds are generally weaker. This ordering is fundamental to organic chemistry and spectroscopy. IR stretching frequencies also follow this pattern: higher frequencies for shorter, stronger bonds (ν: C≡C > C=C > C–C).

Question 6: Which of the following best explains why the H–N–H bond angle in NH₃ (~107°) is smaller than the H–C–H bond angle in CH₄ (109.5°)?

N is more electronegative than C, pulling bonding electrons toward it and increasing angles
The lone pair on N exerts greater repulsion on the bonding pairs than a bonding pair would, compressing the H–N–H angles (Correct answer)
NH₃ is a polar molecule while CH₄ is nonpolar, and polarity decreases bond angles
N–H bonds are shorter than C–H bonds, bringing H atoms closer together

Correct answer: The lone pair on N exerts greater repulsion on the bonding pairs than a bonding pair would, compressing the H–N–H angles

In NH₃, the lone pair on N repels the three bonding pairs more strongly than a fourth bonding pair would in a tetrahedral arrangement, pushing the N–H bonds together and decreasing the H–N–H angle below 109.5°.

VSEPR theory ranks electron pair repulsions: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair. In CH₄: 4 bonding pairs, no lone pairs. Symmetrical tetrahedral arrangement, all H–C–H angles = 109.5°. In NH₃: 3 bonding pairs + 1 lone pair. The lone pair occupies more space than a bonding pair (lone pair electrons are held by one nucleus, spreading out; bonding electrons are attracted by two nuclei and pulled inward). This greater lone pair repulsion pushes the three N–H bonding pairs closer together, compressing H–N–H angles to ~107°. In H₂O: 2 bonding pairs + 2 lone pairs. The even greater lone pair repulsion compresses H–O–H to ~104.5°. The progression CH₄ (109.5°) → NH₃ (107°) → H₂O (104.5°) beautifully illustrates the increasing influence of lone pairs on bond angles. This is one of VSEPR's most powerful predictive tools.

AP Chemical Bonding and Molecular Structure 2

Using VSEPR theory, what is the molecular geometry and approximate bond angle in SF₄?