AP - Advanced Placement Chemical Bonding and Molecular Structure 1 — Questions and Answers
Question 1: What is the molecular geometry and approximate bond angle of XeF₄?
- Tetrahedral, 109.5°
- See-saw, ~102°
- Square planar, 90° (Correct answer)
- Square pyramidal, 90°
Correct answer: Square planar, 90°
XeF₄ has six electron domains around xenon (4 bonding pairs + 2 lone pairs), producing an octahedral electron geometry. The two lone pairs occupy opposite axial positions, placing all four F atoms in a plane at 90° to one another — a square planar molecular geometry.
Question 2: Which of the following C–O bonds has the shortest bond length?
- C–O in CH₃OH (bond order 1)
- C–O in CO₃²⁻ (bond order 1.33)
- C=O in CO₂ (bond order 2)
- C≡O in CO (bond order 3) (Correct answer)
Correct answer: C≡O in CO (bond order 3)
Bond length decreases as bond order increases between the same pair of atoms. CO has a triple bond (bond order 3), giving it the highest bond order among these species and therefore the shortest C–O distance (~113 pm), shorter than CO₂ (~116 pm), CO₃²⁻ (~129 pm), or CH₃OH (~143 pm).
Question 3: According to molecular orbital theory, what is the bond order of NO and what are its magnetic properties?
- Bond order = 2, diamagnetic
- Bond order = 2.5, paramagnetic (Correct answer)
- Bond order = 3, diamagnetic
- Bond order = 1.5, paramagnetic
Correct answer: Bond order = 2.5, paramagnetic
NO has 11 valence electrons. Filling the MO diagram yields σ2s², σ*2s², σ2p², π2p⁴, π*2p¹. Bond order = (8 bonding − 3 antibonding) ÷ 2 = 2.5. The single unpaired electron in the π*2p orbital makes NO paramagnetic.
Question 4: What is the hybridization of phosphorus in PCl₅ and what molecular geometry results?
- sp²; trigonal planar
- sp³; tetrahedral
- sp³d; trigonal bipyramidal (Correct answer)
- sp³d²; octahedral
Correct answer: sp³d; trigonal bipyramidal
PCl₅ has five bonding pairs and no lone pairs around P, requiring five equivalent hybrid orbitals. Mixing one s, three p, and one d orbital produces sp³d hybridization. This gives a trigonal bipyramidal geometry with 120° equatorial and 180° axial bond angles.
Question 5: Which of the following molecules has a non-zero dipole moment?
- BF₃
- XeF₄
- PF₅
- ClF₃ (Correct answer)
Correct answer: ClF₃
ClF₃ adopts a T-shaped geometry with two equatorial lone pairs, creating an asymmetric distribution that prevents bond dipoles from canceling, producing a net dipole. BF₃ (trigonal planar), XeF₄ (square planar), and PF₅ (trigonal bipyramidal) are all highly symmetric, so their bond dipoles cancel exactly.
Question 6: In the Lewis structure of carbon monoxide written as :C≡O:, what are the formal charges on C and O respectively?
- C = 0, O = 0
- C = +1, O = −1
- C = −1, O = +1 (Correct answer)
- C = −2, O = +2
Correct answer: C = −1, O = +1
Using FC = (valence e⁻) − (nonbonding e⁻) − ½(bonding e⁻): carbon has FC = 4 − 2 − ½(6) = −1, and oxygen has FC = 6 − 2 − ½(6) = +1. Although oxygen is more electronegative, C carries the negative formal charge because the lone pair on C is counted against it in this calculation.
AP - Advanced Placement Chemical Bonding and Molecular Structure 1
What is the molecular geometry and approximate bond angle of XeF₄?