CHEM 219 PRINCIPLES OF ORGANIC CHEMISTRY
EXAM 2026/2027 | Portage Learning Complete
Cumulative Test | Pass Guaranteed - A+ Graded
Module 1: Bonding, Structure & Functional Groups (Q1-15)
Q1. What is the hybridization state of the carbon atom in formaldehyde (H₂C=O)?
A. sp³
B. sp²
C. sp
D. dsp³
Rationale: The carbon in formaldehyde is bonded to two hydrogen atoms via single
bonds and to oxygen via a double bond, giving three regions of electron density.
Three regions require sp² hybridization (one s + two p orbitals), leaving one
unhybridized p orbital for the C=O π bond. sp³ would give four regions; sp gives two
regions; dsp³ is not applicable to carbon.
Correct Answer: B
Q2. Which molecule exhibits the strongest hydrogen bonding in the pure liquid
state?
A. CH₃CH₂OCH₂CH₃ (diethyl ether)
B. CH₃CH₂CH₂CH₃ (butane)
C. CH₃CH₂OH (ethanol)
D. CH₃COCH₃ (acetone)
Rationale: Ethanol contains an O–H bond, allowing it to act as both a hydrogen
bond donor and acceptor, producing strong intermolecular hydrogen bonding.
Diethyl ether and acetone can only accept hydrogen bonds (no O–H or N–H); butane
is nonpolar with only London dispersion forces.
Correct Answer: C
,Q3. Which of the following represents a valid resonance structure for the acetate ion
(CH₃COO⁻)?
A. A structure with a C=C double bond and both oxygens neutral
B. A structure with one C=O double bond and the other C–O single bond with
negative charge on the single-bonded oxygen
C. A structure with a positive charge on carbon and negative charges on both
oxygens
D. A structure with a C≡O triple bond
Rationale: The acetate ion has two equivalent resonance structures where the
negative charge is delocalized over both oxygen atoms via the C=O π bond. Each
structure shows one C=O double bond and one C–O single bond bearing the
negative charge; the true structure is a hybrid with equal C–O bond lengths. A C=C
or C≡O violates valence rules; placing positive charge on carbon is incorrect.
Correct Answer: B
Q4. What is the molecular geometry around the nitrogen atom in ammonia (NH₃)
according to VSEPR theory?
A. Trigonal planar
B. Tetrahedral
C. Trigonal pyramidal
D. Bent
Rationale: Ammonia has four electron domains (three N–H bonds and one lone
pair). The electron geometry is tetrahedral, but the molecular geometry—describing
only atom positions—is trigonal pyramidal. Trigonal planar applies to three domains
with no lone pair (e.g., BH₃); bent applies to two bonds with lone pairs (e.g., H₂O).
Correct Answer: C
,Q5. Which compound contains a carbon atom with sp hybridization?
A. Ethene (C₂H₄)
B. Ethane (C₂H₆)
C. Ethyne (C₂H₂)
D. Cyclohexane
Rationale: Ethyne (acetylene) has a carbon–carbon triple bond. Each carbon is
bonded to one hydrogen and one carbon via one sigma bond each, giving two
regions of electron density—requiring sp hybridization (one s + one p orbital), with
two unhybridized p orbitals forming the two π bonds. Ethene carbons are sp²; ethane
and cyclohexane carbons are sp³.
Correct Answer: C
Q6. Which functional group is present in the compound CH₃CH₂COOH?
A. Ester
B. Ketone
C. Carboxylic acid
D. Aldehyde
Rationale: The –COOH group (carbonyl carbon bonded to OH) defines a carboxylic
acid. An ester would be –COOR; a ketone would have C=O between two carbons (–
CO–); an aldehyde would have –CHO.
Correct Answer: C
Q7. Arrange the following bonds in order of increasing polarity: C–H, C–O, O–H, C–N.
A. C–H < C–N < C–O < O–H
B. O–H < C–O < C–N < C–H
C. C–H < C–O < C–N < O–H
D. C–N < C–H < O–H < C–O
, Rationale: Bond polarity depends on electronegativity difference (ΔEN). Using
Pauling values: C–H (ΔEN = 0.4), C–N (ΔEN = 0.5), C–O (ΔEN = 1.0), O–H (ΔEN = 1.4).
Thus the order is C–H < C–N < C–O < O–H.
Correct Answer: A
Q8. Which statement about resonance is correct?
A. Resonance structures are in rapid equilibrium with each other
B. The actual molecule is a weighted average of all contributing resonance structures
C. Resonance structures must have different numbers of electrons
D. Resonance only occurs in molecules with double bonds
Rationale: The true structure of a resonance-stabilized molecule is a hybrid
(weighted average) of all valid contributing structures, not an equilibrium mixture. All
resonance structures must have the same number of electrons and the same atom
positions; only electron distribution differs. Resonance can involve lone pairs and
empty p orbitals without formal double bonds (e.g., allyl cation).
Correct Answer: B
Q9. What is the formal charge on the oxygen atom in the Lewis structure of
methanol (CH₃OH)?
A. +1
B. 0
C. –1
D. –2
Rationale: Formal charge = valence electrons – (nonbonding electrons + ½ bonding
electrons). For oxygen in methanol: 6 – (4 + ½×4) = 6 – 6 = 0. Oxygen has two lone
pairs and two single bonds (to carbon and hydrogen), giving a neutral formal charge.
Correct Answer: B
EXAM 2026/2027 | Portage Learning Complete
Cumulative Test | Pass Guaranteed - A+ Graded
Module 1: Bonding, Structure & Functional Groups (Q1-15)
Q1. What is the hybridization state of the carbon atom in formaldehyde (H₂C=O)?
A. sp³
B. sp²
C. sp
D. dsp³
Rationale: The carbon in formaldehyde is bonded to two hydrogen atoms via single
bonds and to oxygen via a double bond, giving three regions of electron density.
Three regions require sp² hybridization (one s + two p orbitals), leaving one
unhybridized p orbital for the C=O π bond. sp³ would give four regions; sp gives two
regions; dsp³ is not applicable to carbon.
Correct Answer: B
Q2. Which molecule exhibits the strongest hydrogen bonding in the pure liquid
state?
A. CH₃CH₂OCH₂CH₃ (diethyl ether)
B. CH₃CH₂CH₂CH₃ (butane)
C. CH₃CH₂OH (ethanol)
D. CH₃COCH₃ (acetone)
Rationale: Ethanol contains an O–H bond, allowing it to act as both a hydrogen
bond donor and acceptor, producing strong intermolecular hydrogen bonding.
Diethyl ether and acetone can only accept hydrogen bonds (no O–H or N–H); butane
is nonpolar with only London dispersion forces.
Correct Answer: C
,Q3. Which of the following represents a valid resonance structure for the acetate ion
(CH₃COO⁻)?
A. A structure with a C=C double bond and both oxygens neutral
B. A structure with one C=O double bond and the other C–O single bond with
negative charge on the single-bonded oxygen
C. A structure with a positive charge on carbon and negative charges on both
oxygens
D. A structure with a C≡O triple bond
Rationale: The acetate ion has two equivalent resonance structures where the
negative charge is delocalized over both oxygen atoms via the C=O π bond. Each
structure shows one C=O double bond and one C–O single bond bearing the
negative charge; the true structure is a hybrid with equal C–O bond lengths. A C=C
or C≡O violates valence rules; placing positive charge on carbon is incorrect.
Correct Answer: B
Q4. What is the molecular geometry around the nitrogen atom in ammonia (NH₃)
according to VSEPR theory?
A. Trigonal planar
B. Tetrahedral
C. Trigonal pyramidal
D. Bent
Rationale: Ammonia has four electron domains (three N–H bonds and one lone
pair). The electron geometry is tetrahedral, but the molecular geometry—describing
only atom positions—is trigonal pyramidal. Trigonal planar applies to three domains
with no lone pair (e.g., BH₃); bent applies to two bonds with lone pairs (e.g., H₂O).
Correct Answer: C
,Q5. Which compound contains a carbon atom with sp hybridization?
A. Ethene (C₂H₄)
B. Ethane (C₂H₆)
C. Ethyne (C₂H₂)
D. Cyclohexane
Rationale: Ethyne (acetylene) has a carbon–carbon triple bond. Each carbon is
bonded to one hydrogen and one carbon via one sigma bond each, giving two
regions of electron density—requiring sp hybridization (one s + one p orbital), with
two unhybridized p orbitals forming the two π bonds. Ethene carbons are sp²; ethane
and cyclohexane carbons are sp³.
Correct Answer: C
Q6. Which functional group is present in the compound CH₃CH₂COOH?
A. Ester
B. Ketone
C. Carboxylic acid
D. Aldehyde
Rationale: The –COOH group (carbonyl carbon bonded to OH) defines a carboxylic
acid. An ester would be –COOR; a ketone would have C=O between two carbons (–
CO–); an aldehyde would have –CHO.
Correct Answer: C
Q7. Arrange the following bonds in order of increasing polarity: C–H, C–O, O–H, C–N.
A. C–H < C–N < C–O < O–H
B. O–H < C–O < C–N < C–H
C. C–H < C–O < C–N < O–H
D. C–N < C–H < O–H < C–O
, Rationale: Bond polarity depends on electronegativity difference (ΔEN). Using
Pauling values: C–H (ΔEN = 0.4), C–N (ΔEN = 0.5), C–O (ΔEN = 1.0), O–H (ΔEN = 1.4).
Thus the order is C–H < C–N < C–O < O–H.
Correct Answer: A
Q8. Which statement about resonance is correct?
A. Resonance structures are in rapid equilibrium with each other
B. The actual molecule is a weighted average of all contributing resonance structures
C. Resonance structures must have different numbers of electrons
D. Resonance only occurs in molecules with double bonds
Rationale: The true structure of a resonance-stabilized molecule is a hybrid
(weighted average) of all valid contributing structures, not an equilibrium mixture. All
resonance structures must have the same number of electrons and the same atom
positions; only electron distribution differs. Resonance can involve lone pairs and
empty p orbitals without formal double bonds (e.g., allyl cation).
Correct Answer: B
Q9. What is the formal charge on the oxygen atom in the Lewis structure of
methanol (CH₃OH)?
A. +1
B. 0
C. –1
D. –2
Rationale: Formal charge = valence electrons – (nonbonding electrons + ½ bonding
electrons). For oxygen in methanol: 6 – (4 + ½×4) = 6 – 6 = 0. Oxygen has two lone
pairs and two single bonds (to carbon and hydrogen), giving a neutral formal charge.
Correct Answer: B
