CHEM134 C001 Final Exam | Questions and
Answers – Spring 2026 | 92% Scored.
SECTION 1: CHEMICAL BONDING & MOLECULAR GEOMETRY (8 Questions)
Q1: Which of the following molecules has a Lewis structure that violates the octet
rule?
A. CO2
B. NH3
C. PCl5
D. CH4
Correct Answer: C
Rationale: Correct because phosphorus in PCl5 has 10 valence electrons (5
bonding pairs), exceeding the octet rule; period 3 elements like P can utilize
empty d orbitals to accommodate expanded octets, unlike period 2 elements that
strictly obey the octet rule.
Q2: What is the formal charge on the central sulfur atom in the best resonance
structure of SO2?
A. +2
B. 0
C. -1
D. +1
Correct Answer: B
Rationale: Correct because formal charge = valence electrons - (nonbonding + ½
bonding) = 6 - (2 + ½(6)) = 6 - 5 = 0; the resonance structures with one double
bond and one single bond to oxygen give sulfur a formal charge of zero,
matching the lowest energy arrangement.
,Q3: According to VSEPR theory, what is the molecular geometry of SF4?
A. Tetrahedral
B. See-saw
C. Square planar
D. Trigonal bipyramidal
Correct Answer: B
Rationale: Correct because SF4 has five electron domains (four bonding pairs
and one lone pair) around sulfur, giving a trigonal bipyramidal electron geometry;
with one lone pair occupying an equatorial position to minimize repulsion, the
molecular geometry is see-saw per VSEPR theory predictions.
Q4: Which molecule is nonpolar despite containing polar bonds?
A. H2O
B. NH3
C. CCl4
D. CHCl3
Correct Answer: C
Rationale: Correct because CCl4 has four identical C-Cl bond dipoles arranged
tetrahedrally at 109.5° angles; the vector sum of these symmetrically oriented
polar bonds cancels completely, resulting in zero net dipole moment and a
nonpolar molecule despite polar individual bonds.
Q5: What is the hybridization of the central atom in XeF4?
A. sp3
B. sp3d
C. sp3d2
D. sp2
Correct Answer: C
, Rationale: Correct because xenon in XeF4 has six electron domains (four
bonding pairs and two lone pairs), requiring six hybrid orbitals; sp3d2
hybridization provides six equivalent orbitals oriented octahedrally, with lone
pairs occupying axial positions to minimize repulsion, giving square planar
molecular geometry.
Q6: How many sigma (σ) and pi (π) bonds are present in the molecule CH3C≡N?
A. 4 σ and 2 π
B. 5 σ and 2 π
C. 6 σ and 2 π
D. 5 σ and 3 π
Correct Answer: B
Rationale: Correct because CH3C≡N contains: C-H (3 σ), C-C (1 σ), C≡N (1 σ + 2
π); totaling 5 sigma bonds and 2 pi bonds; every single bond is one sigma, every
double bond adds one pi, and every triple bond adds two pi bonds to the one
sigma bond framework.
Q7: Which bond is expected to be the shortest?
A. C-C in ethane
B. C=C in ethene
C. C≡C in ethyne
D. C-O in methanol
Correct Answer: C
Rationale: Correct because bond length decreases with increasing bond order;
the carbon-carbon triple bond in ethyne has the highest bond order (3), resulting
in the strongest attraction between nuclei and the shortest internuclear distance
compared to single and double carbon-carbon bonds.
Q8: The bond angle in NH3 (107°) is less than the tetrahedral angle (109.5°)
because:
A. Nitrogen is more electronegative than hydrogen
Answers – Spring 2026 | 92% Scored.
SECTION 1: CHEMICAL BONDING & MOLECULAR GEOMETRY (8 Questions)
Q1: Which of the following molecules has a Lewis structure that violates the octet
rule?
A. CO2
B. NH3
C. PCl5
D. CH4
Correct Answer: C
Rationale: Correct because phosphorus in PCl5 has 10 valence electrons (5
bonding pairs), exceeding the octet rule; period 3 elements like P can utilize
empty d orbitals to accommodate expanded octets, unlike period 2 elements that
strictly obey the octet rule.
Q2: What is the formal charge on the central sulfur atom in the best resonance
structure of SO2?
A. +2
B. 0
C. -1
D. +1
Correct Answer: B
Rationale: Correct because formal charge = valence electrons - (nonbonding + ½
bonding) = 6 - (2 + ½(6)) = 6 - 5 = 0; the resonance structures with one double
bond and one single bond to oxygen give sulfur a formal charge of zero,
matching the lowest energy arrangement.
,Q3: According to VSEPR theory, what is the molecular geometry of SF4?
A. Tetrahedral
B. See-saw
C. Square planar
D. Trigonal bipyramidal
Correct Answer: B
Rationale: Correct because SF4 has five electron domains (four bonding pairs
and one lone pair) around sulfur, giving a trigonal bipyramidal electron geometry;
with one lone pair occupying an equatorial position to minimize repulsion, the
molecular geometry is see-saw per VSEPR theory predictions.
Q4: Which molecule is nonpolar despite containing polar bonds?
A. H2O
B. NH3
C. CCl4
D. CHCl3
Correct Answer: C
Rationale: Correct because CCl4 has four identical C-Cl bond dipoles arranged
tetrahedrally at 109.5° angles; the vector sum of these symmetrically oriented
polar bonds cancels completely, resulting in zero net dipole moment and a
nonpolar molecule despite polar individual bonds.
Q5: What is the hybridization of the central atom in XeF4?
A. sp3
B. sp3d
C. sp3d2
D. sp2
Correct Answer: C
, Rationale: Correct because xenon in XeF4 has six electron domains (four
bonding pairs and two lone pairs), requiring six hybrid orbitals; sp3d2
hybridization provides six equivalent orbitals oriented octahedrally, with lone
pairs occupying axial positions to minimize repulsion, giving square planar
molecular geometry.
Q6: How many sigma (σ) and pi (π) bonds are present in the molecule CH3C≡N?
A. 4 σ and 2 π
B. 5 σ and 2 π
C. 6 σ and 2 π
D. 5 σ and 3 π
Correct Answer: B
Rationale: Correct because CH3C≡N contains: C-H (3 σ), C-C (1 σ), C≡N (1 σ + 2
π); totaling 5 sigma bonds and 2 pi bonds; every single bond is one sigma, every
double bond adds one pi, and every triple bond adds two pi bonds to the one
sigma bond framework.
Q7: Which bond is expected to be the shortest?
A. C-C in ethane
B. C=C in ethene
C. C≡C in ethyne
D. C-O in methanol
Correct Answer: C
Rationale: Correct because bond length decreases with increasing bond order;
the carbon-carbon triple bond in ethyne has the highest bond order (3), resulting
in the strongest attraction between nuclei and the shortest internuclear distance
compared to single and double carbon-carbon bonds.
Q8: The bond angle in NH3 (107°) is less than the tetrahedral angle (109.5°)
because:
A. Nitrogen is more electronegative than hydrogen
