,Verified CHEM 219 Module 1 Exam Principles of
Organic Chemistry w/Lab | Portage (2026–2027)
1. Considering the periodic trend of atomic radius, which of the following sequences correctly
arranges the elements in order of increasing atomic radius?
A) F < O < C < B
B) B < C < O < F
C) O < F < B < C
D) C < B < F < O
Correct Answer: F < O < C < B
Rationale: Atomic radius increases as you move down a group and decreases as you move from left to
right across a period. Fluorine (F) is the smallest, followed by oxygen (O), then carbon (C), and boron
(B) is the largest among these. The other sequences incorrectly order the elements based on their
positions in the periodic table.
2. The Lewis structure of a molecule shows a central atom with three bonding domains and one lone
pair of electrons. According to VSEPR theory, what is the molecular geometry of this molecule?
A) Trigonal planar
B) Tetrahedral
C) Trigonal pyramidal
D) Bent
Correct Answer: Trigonal pyramidal
Rationale: Four electron domains (three bonds + one lone pair) correspond to a tetrahedral electron-
group geometry. However, the molecular geometry is described by the positions of the atoms only,
which forms a trigonal pyramidal shape. A trigonal planar geometry results from three bonding
domains and no lone pairs.
,3. In the formation of a carbon-carbon double bond, what type of orbital overlap is responsible for the
pi (π) bond?
A) End-to-end overlap of sp² hybrid orbitals
B) Side-to-side overlap of unhybridized p orbitals
C) End-to-end overlap of sp³ hybrid orbitals
D) Side-to-side overlap of sp hybrid orbitals
Correct Answer: Side-to-side overlap of unhybridized p orbitals
Rationale: A pi bond is formed by the side-to-side overlap of parallel, unhybridized p orbitals. The
sigma bond is formed by the end-to-end overlap of hybrid orbitals. The other options describe sigma
bond formation or incorrect orbital types for a pi bond.
4. Which of the following compounds exhibits cis-trans isomerism?
A) 1-butene
B) 2-butene
C) 2-methylpropene
D) Cyclohexane
Correct Answer: 2-butene
Rationale: Cis-trans (geometric) isomerism requires a carbon-carbon double bond where each carbon
in the double bond has two different substituents. In 2-butene (CH₃-CH=CH-CH₃), each alkene carbon
has a hydrogen and a methyl group. 1-butene has a terminal double bond, and 2-methylpropene has
identical methyl groups on one carbon.
5. A scientist compares the boiling points of methane (CH₄) and water (H₂O). Water has a significantly
higher boiling point. What is the primary intermolecular force responsible for this difference?
A) London dispersion forces
B) Dipole-dipole interactions
C) Hydrogen bonding
, D) Ionic bonding
Correct Answer: Hydrogen bonding
Rationale: Water exhibits strong hydrogen bonding due to the highly polar O-H bond, while methane
only has weak London dispersion forces. Hydrogen bonding requires a significant amount of energy to
overcome, leading to a much higher boiling point. Dipole-dipole interactions are weaker than
hydrogen bonds.
6. A molecule has the formula C₂H₆O. Which of the following statements correctly distinguishes its
two possible constitutional isomers?
A) One is an alcohol and the other is an ether.
B) One is an alkane and the other is an alkene.
C) One is a ketone and the other is an aldehyde.
D) One is a carboxylic acid and the other is an ester.
Correct Answer: One is an alcohol and the other is an ether.
Rationale: The formula C₂H₆O corresponds to two constitutional isomers: ethanol (an alcohol,
CH₃CH₂OH) and dimethyl ether (an ether, CH₃OCH₃). The other options describe functional groups that
require different atomic compositions or numbers of carbon atoms.
7. A student calculates the formal charge on the oxygen atom in a molecule and obtains a value of +1.
Which of the following Lewis structures is most consistent with this formal charge?
A) An oxygen atom with two lone pairs and two single bonds
B) An oxygen atom with one lone pair and three single bonds
C) An oxygen atom with three lone pairs and one single bond
D) An oxygen atom with one lone pair and one double bond
Correct Answer: An oxygen atom with one lone pair and three single bonds
Organic Chemistry w/Lab | Portage (2026–2027)
1. Considering the periodic trend of atomic radius, which of the following sequences correctly
arranges the elements in order of increasing atomic radius?
A) F < O < C < B
B) B < C < O < F
C) O < F < B < C
D) C < B < F < O
Correct Answer: F < O < C < B
Rationale: Atomic radius increases as you move down a group and decreases as you move from left to
right across a period. Fluorine (F) is the smallest, followed by oxygen (O), then carbon (C), and boron
(B) is the largest among these. The other sequences incorrectly order the elements based on their
positions in the periodic table.
2. The Lewis structure of a molecule shows a central atom with three bonding domains and one lone
pair of electrons. According to VSEPR theory, what is the molecular geometry of this molecule?
A) Trigonal planar
B) Tetrahedral
C) Trigonal pyramidal
D) Bent
Correct Answer: Trigonal pyramidal
Rationale: Four electron domains (three bonds + one lone pair) correspond to a tetrahedral electron-
group geometry. However, the molecular geometry is described by the positions of the atoms only,
which forms a trigonal pyramidal shape. A trigonal planar geometry results from three bonding
domains and no lone pairs.
,3. In the formation of a carbon-carbon double bond, what type of orbital overlap is responsible for the
pi (π) bond?
A) End-to-end overlap of sp² hybrid orbitals
B) Side-to-side overlap of unhybridized p orbitals
C) End-to-end overlap of sp³ hybrid orbitals
D) Side-to-side overlap of sp hybrid orbitals
Correct Answer: Side-to-side overlap of unhybridized p orbitals
Rationale: A pi bond is formed by the side-to-side overlap of parallel, unhybridized p orbitals. The
sigma bond is formed by the end-to-end overlap of hybrid orbitals. The other options describe sigma
bond formation or incorrect orbital types for a pi bond.
4. Which of the following compounds exhibits cis-trans isomerism?
A) 1-butene
B) 2-butene
C) 2-methylpropene
D) Cyclohexane
Correct Answer: 2-butene
Rationale: Cis-trans (geometric) isomerism requires a carbon-carbon double bond where each carbon
in the double bond has two different substituents. In 2-butene (CH₃-CH=CH-CH₃), each alkene carbon
has a hydrogen and a methyl group. 1-butene has a terminal double bond, and 2-methylpropene has
identical methyl groups on one carbon.
5. A scientist compares the boiling points of methane (CH₄) and water (H₂O). Water has a significantly
higher boiling point. What is the primary intermolecular force responsible for this difference?
A) London dispersion forces
B) Dipole-dipole interactions
C) Hydrogen bonding
, D) Ionic bonding
Correct Answer: Hydrogen bonding
Rationale: Water exhibits strong hydrogen bonding due to the highly polar O-H bond, while methane
only has weak London dispersion forces. Hydrogen bonding requires a significant amount of energy to
overcome, leading to a much higher boiling point. Dipole-dipole interactions are weaker than
hydrogen bonds.
6. A molecule has the formula C₂H₆O. Which of the following statements correctly distinguishes its
two possible constitutional isomers?
A) One is an alcohol and the other is an ether.
B) One is an alkane and the other is an alkene.
C) One is a ketone and the other is an aldehyde.
D) One is a carboxylic acid and the other is an ester.
Correct Answer: One is an alcohol and the other is an ether.
Rationale: The formula C₂H₆O corresponds to two constitutional isomers: ethanol (an alcohol,
CH₃CH₂OH) and dimethyl ether (an ether, CH₃OCH₃). The other options describe functional groups that
require different atomic compositions or numbers of carbon atoms.
7. A student calculates the formal charge on the oxygen atom in a molecule and obtains a value of +1.
Which of the following Lewis structures is most consistent with this formal charge?
A) An oxygen atom with two lone pairs and two single bonds
B) An oxygen atom with one lone pair and three single bonds
C) An oxygen atom with three lone pairs and one single bond
D) An oxygen atom with one lone pair and one double bond
Correct Answer: An oxygen atom with one lone pair and three single bonds
