Organic Chemistry Solomons Fryhle Snyder 12th Ed. Ch.1-25 Test Bank Actual
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Section 1: Chapter 1-3: Bonding, Molecular Structure, Functional
Groups, and Acids/Bases (20 questions)
Q1: A carbon atom in formaldehyde (H₂C=O) exhibits which type of hybridization?
A. sp³
B. sp
C. sp² [CORRECT]
D. dsp³
Correct Answer: C
Rationale: The carbon in formaldehyde is bonded to two hydrogen atoms and
double-bonded to oxygen, giving three regions of electron density; this requires sp²
hybridization with trigonal planar geometry and a remaining unhybridized p orbital for
the π bond to oxygen.
Q2: Which of the following compounds contains a formal charge of +1 on the nitrogen
atom?
A. NH₃
,B. NH₄⁺ [CORRECT]
C. :NH₂⁻
D. N₂
Correct Answer: B
Rationale: In NH₄⁺, nitrogen has four bonds and zero lone pairs, giving it 4 valence
electrons assigned versus 5 in its neutral state; formal charge = valence electrons -
assigned electrons = 5 - 4 = +1, as confirmed by Lewis structure analysis in Solomons
12th Edition.
Q3: Which resonance structure contributes most significantly to the true hybrid
structure of the acetate ion (CH₃COO⁻)?
A. The structure with a double bond to one oxygen and single bond to the other
B. The equivalent resonance structures with delocalized negative charge equally shared
between both oxygens [CORRECT]
C. The structure with a triple bond to one oxygen
D. The structure with positive charge on the carbonyl carbon
Correct Answer: B
Rationale: The two equivalent resonance structures of acetate, each with the negative
charge on a different oxygen, contribute equally to the resonance hybrid; this
delocalization stabilizes the carboxylate anion and gives both C-O bonds identical
intermediate bond lengths.
Q4: According to VSEPR theory, what is the predicted bond angle in methane (CH₄)?
,A. 90°
B. 109.5° [CORRECT]
C. 120°
D. 180°
Correct Answer: B
Rationale: Methane has four bonding pairs around carbon with sp³ hybridization and
tetrahedral geometry; VSEPR theory predicts bond angles of 109.5° to minimize electron
pair repulsion, as described in Solomons, Fryhle, and Snyder's bonding and molecular
structure chapter.
Q5: Which intermolecular force is primarily responsible for the high boiling point of
water compared to methane?
A. London dispersion forces
B. Dipole-dipole interactions
C. Hydrogen bonding [CORRECT]
D. Ionic bonding
Correct Answer: C
Rationale: Water exhibits extensive hydrogen bonding due to the highly polar O-H bonds
and small size of oxygen, requiring significant energy to overcome; methane lacks
hydrogen bonding capability and relies only on weak London dispersion forces,
explaining its much lower boiling point.
, Q6: In the molecule CH₃CH₂O⁻, which atom is most likely to act as a Brønsted-Lowry
base?
A. The carbon atoms
B. The hydrogen atoms
C. The negatively charged oxygen [CORRECT]
D. All atoms equally
Correct Answer: C
Rationale: The ethoxide ion (CH₃CH₂O⁻) has a negative charge and lone pairs on oxygen,
making it a strong Brønsted-Lowry base (proton acceptor); the oxygen's high electron
density and negative formal charge make it the most basic site in the molecule.
Q7: Which of the following is the strongest acid?
A. CH₃CH₂OH (ethanol, pKa ≈ 16)
B. CH₃COOH (acetic acid, pKa ≈ 4.76) [CORRECT]
C. CH₃CH₂CH₃ (propane, pKa ≈ 50)
D. CH₃NH₂ (methylamine, pKa of conjugate acid ≈ 10.6)
Correct Answer: B
Rationale: Acetic acid (pKa ≈ 4.76) is the strongest acid because its conjugate base
(acetate) is stabilized by resonance delocalization of the negative charge over two
oxygen atoms; lower pKa values indicate stronger acids, and ethanol, propane, and
methylamine are all considerably weaker acids.
Exam 2026/2027 – Comprehensive Exam with Detailed Rationales | 100% Verified |
Pass Guaranteed – A+ Graded
Section 1: Chapter 1-3: Bonding, Molecular Structure, Functional
Groups, and Acids/Bases (20 questions)
Q1: A carbon atom in formaldehyde (H₂C=O) exhibits which type of hybridization?
A. sp³
B. sp
C. sp² [CORRECT]
D. dsp³
Correct Answer: C
Rationale: The carbon in formaldehyde is bonded to two hydrogen atoms and
double-bonded to oxygen, giving three regions of electron density; this requires sp²
hybridization with trigonal planar geometry and a remaining unhybridized p orbital for
the π bond to oxygen.
Q2: Which of the following compounds contains a formal charge of +1 on the nitrogen
atom?
A. NH₃
,B. NH₄⁺ [CORRECT]
C. :NH₂⁻
D. N₂
Correct Answer: B
Rationale: In NH₄⁺, nitrogen has four bonds and zero lone pairs, giving it 4 valence
electrons assigned versus 5 in its neutral state; formal charge = valence electrons -
assigned electrons = 5 - 4 = +1, as confirmed by Lewis structure analysis in Solomons
12th Edition.
Q3: Which resonance structure contributes most significantly to the true hybrid
structure of the acetate ion (CH₃COO⁻)?
A. The structure with a double bond to one oxygen and single bond to the other
B. The equivalent resonance structures with delocalized negative charge equally shared
between both oxygens [CORRECT]
C. The structure with a triple bond to one oxygen
D. The structure with positive charge on the carbonyl carbon
Correct Answer: B
Rationale: The two equivalent resonance structures of acetate, each with the negative
charge on a different oxygen, contribute equally to the resonance hybrid; this
delocalization stabilizes the carboxylate anion and gives both C-O bonds identical
intermediate bond lengths.
Q4: According to VSEPR theory, what is the predicted bond angle in methane (CH₄)?
,A. 90°
B. 109.5° [CORRECT]
C. 120°
D. 180°
Correct Answer: B
Rationale: Methane has four bonding pairs around carbon with sp³ hybridization and
tetrahedral geometry; VSEPR theory predicts bond angles of 109.5° to minimize electron
pair repulsion, as described in Solomons, Fryhle, and Snyder's bonding and molecular
structure chapter.
Q5: Which intermolecular force is primarily responsible for the high boiling point of
water compared to methane?
A. London dispersion forces
B. Dipole-dipole interactions
C. Hydrogen bonding [CORRECT]
D. Ionic bonding
Correct Answer: C
Rationale: Water exhibits extensive hydrogen bonding due to the highly polar O-H bonds
and small size of oxygen, requiring significant energy to overcome; methane lacks
hydrogen bonding capability and relies only on weak London dispersion forces,
explaining its much lower boiling point.
, Q6: In the molecule CH₃CH₂O⁻, which atom is most likely to act as a Brønsted-Lowry
base?
A. The carbon atoms
B. The hydrogen atoms
C. The negatively charged oxygen [CORRECT]
D. All atoms equally
Correct Answer: C
Rationale: The ethoxide ion (CH₃CH₂O⁻) has a negative charge and lone pairs on oxygen,
making it a strong Brønsted-Lowry base (proton acceptor); the oxygen's high electron
density and negative formal charge make it the most basic site in the molecule.
Q7: Which of the following is the strongest acid?
A. CH₃CH₂OH (ethanol, pKa ≈ 16)
B. CH₃COOH (acetic acid, pKa ≈ 4.76) [CORRECT]
C. CH₃CH₂CH₃ (propane, pKa ≈ 50)
D. CH₃NH₂ (methylamine, pKa of conjugate acid ≈ 10.6)
Correct Answer: B
Rationale: Acetic acid (pKa ≈ 4.76) is the strongest acid because its conjugate base
(acetate) is stabilized by resonance delocalization of the negative charge over two
oxygen atoms; lower pKa values indicate stronger acids, and ethanol, propane, and
methylamine are all considerably weaker acids.
