AP Chemical Bonding and Molecular Structure 2

Question 1

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

Accepted Answer

Seesaw (sawhorse) geometry, ~102° and ~173°

Answer

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.

Question 2

Which of the following molecules has a dipole moment of zero?

Accepted Answer

CO₂

Answer

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.

Question 3

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

Accepted Answer

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

Answer

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.

Question 4

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

Accepted Answer

Bond order = 2, paramagnetic (2 unpaired electrons)

Answer

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.

Question 5

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

Accepted Answer

C–H < C≡C < C=C < C–C

Answer

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.

AP Practice Test

AP Practice Test

AP Chemical Bonding and Molecular Structure 2