PORTAGE LEARNING CHEM 210 – EXAM 1:
Bonding, Structure & Alkanes Actual Exam
UPDATED | Complete Q&A with Rationales – Pass
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Part 1 – Foundational Concepts & Naming (10 questions)
Q1: Consider the molecule nitrogen (N₂). Based on electronegativity trends, how would
you describe the bond between the two nitrogen atoms, and what type of hybridization
does each nitrogen atom utilize?
A. Nonpolar covalent bond, sp³ hybridized
B. Polar covalent bond, sp² hybridized
C. Nonpolar covalent bond, sp hybridized [CORRECT]
D. Ionic bond, sp hybridized
Correct Answer: C
Rationale: The bond is nonpolar covalent because the electronegativity difference is
zero between two identical nitrogen atoms. They are triple-bonded, which requires sp
hybridization to form one sigma and two pi bonds, leaving two lone pairs in sp orbitals.
Q2: Which of the following structures represents the most stable resonance contributor
for the nitrate ion (NO₃⁻)?
A. A structure with one N=O double bond and two N-O single bonds, where the negative
charge is localized on a single oxygen.
B. A structure where the nitrogen has a positive formal charge and all three oxygens
share a negative charge equally through resonance. [CORRECT]
C. A structure with three N=O double bonds and a negative charge on the nitrogen.
D. A structure with one N≡O triple bond and two N-O single bonds.
Correct Answer: B
Rationale: The most stable representation involves delocalization of the negative charge
over the three oxygen atoms; the actual structure is a hybrid of three resonance forms,
each placing the double bond on a different oxygen.
Q3: What is the correct IUPAC name for the following alkane:
CH₃-CH(CH₃)-CH₂-CH₂-CH₃?
A. 2-methylpentane
B. 3-methylpentane
C. 1,2-dimethylbutane
,D. Hexane
Correct Answer: A
Rationale: The longest continuous carbon chain is five carbons (pentane), and the
methyl substituent is located on carbon 2 when numbering from the end nearest the
branch.
Q4: Identify the geometry around the central carbon atom in CH₄ and the H-C-H bond
angle.
A. Trigonal planar, 120°
B. Tetrahedral, 109.5° [CORRECT]
C. Bent, 104.5°
D. Linear, 180°
Correct Answer: B
Rationale: Carbon in methane has four bonding pairs and no lone pairs, resulting in a
tetrahedral geometry with ideal bond angles of 109.5°.
Q5: Which of the following Newman projections represents the most stable
conformation of butane when looking down the C2-C3 bond?
A. The anti conformation, where the two methyl groups are 180° apart. [CORRECT]
B. The eclipsed conformation, where the two methyl groups are directly aligned.
C. The gauche conformation, where the methyl groups are 60° apart.
D. The totally eclipsed conformation.
Correct Answer: A
Rationale: The anti conformation places the bulky methyl groups opposite each other
(180°), minimizing steric strain and torsional strain, making it the lowest energy
conformation.
Q6: Calculate the formal charge on the oxygen atom in the hydroxide ion (OH⁻).
A. +1
B. 0
C. -1 [CORRECT]
D. -2
Correct Answer: C
Rationale: Oxygen has 6 valence electrons; in the hydroxide ion, it has 6 non-bonding
electrons and shares 1 bonding electron (half of 2). The calculation is 6 - (6 + 1) = -1.
Q7: In the molecule CH₂Cl₂, determine the molecular geometry and the polarity of the
molecule.
A. Tetrahedral and nonpolar
B. Tetrahedral and polar [CORRECT]
C. Trigonal pyramidal and polar
,D. Bent and nonpolar
Correct Answer: B
Rationale: The molecule is tetrahedral, but because the C-Cl bond dipoles do not cancel
out perfectly due to the different atoms attached, the molecule has a net dipole moment
and is polar.
Q8: Which statement best explains why cyclopropane is significantly more reactive than
cyclohexane?
A. Cyclopropane has angle strain due to its 60° bond angles, which deviates
significantly from the ideal tetrahedral 109.5°. [CORRECT]
B. Cyclopropane contains double bonds, whereas cyclohexane does not.
C. Cyclohexane is planar, while cyclopropane is non-planar.
D. Cyclopropane has torsional strain but no angle strain.
Correct Answer: A
Rationale: The internal bond angles in cyclopropane are forced to be 60°, creating
massive angle strain (Baeyer strain) because the sp³ hybridized orbitals cannot overlap
effectively, making the bonds weak and highly reactive.
Q9: What is the IUPAC name for the cycloalkane represented by a six-membered ring
with a methyl group on carbon 1 and an ethyl group on carbon 3?
A. 1-ethyl-3-methylcyclohexane
B. 1-methyl-3-ethylcyclohexane [CORRECT]
C. 3-ethyl-1-methylcyclohexane
D. 1,3-diethylcyclohexane
Correct Answer: B
Rationale: Substituents are listed alphabetically (ethyl before methyl), and the
numbering is assigned to give the lowest set of locants (1,3-), though in a symmetric
ring like this, starting anywhere yields the same numbers relative to each other.
Q10: Identify the hybridization state of the carbon atoms in an alkyne triple bond.
A. sp³
B. sp²
C. sp [CORRECT]
D. d
Correct Answer: C
Rationale: A carbon participating in a triple bond has two regions of electron density
(one single bond and one triple bond, or two triple bonds), requiring sp hybridization
which yields linear geometry with 180° bond angles.
Part 2 – Structure & Stability (10 questions)
, Q11: When drawing the chair conformation of cyclohexane, which type of hydrogen
positions (axial or equatorial) experience the most steric hindrance when bulky
substituents are attached?
A. Equatorial
B. Axial [CORRECT]
C. Both equally
D. Neither, due to ring flipping
Correct Answer: B
Rationale: Axial bonds are oriented perpendicular to the ring and point straight up or
down, causing 1,3-diaxial interactions with other axial hydrogens/substituents, leading
to significant steric strain.
Q12: Which of the following constitutional isomers of pentane has the lowest boiling
point?
A. n-Pentane
B. Isopentane (2-methylbutane)
C. Neopentane (2,2-dimethylpropane) [CORRECT]
D. They all have the same boiling point.
Correct Answer: C
Rationale: Neopentane is the most compact and spherical shape, resulting in the
smallest surface area and the weakest London dispersion forces, which leads to the
lowest boiling point.
Q13: Which factor primarily determines the stability of a carbocation?
A. Inductive effects from electron-donating groups [CORRECT]
B. The number of hydrogen atoms attached
C. Electronegativity of the adjacent atoms
D. Hybridization of the carbon with the positive charge
Correct Answer: A
Rationale: Alkyl groups are electron-donating via inductive effects and
hyperconjugation, which help stabilize the positive charge on the carbon; therefore,
tertiary carbocations are more stable than primary ones.
Q14: In the context of VSEPR theory, why is the H₂O molecule bent rather than linear?
A. Oxygen forms double bonds.
B. The two lone pairs on oxygen exert greater repulsion than the bonding pairs,
compressing the H-O-H bond angle. [CORRECT]
C. Hydrogen is more electronegative than oxygen.
D. The oxygen atom is sp hybridized.
Correct Answer: B
Bonding, Structure & Alkanes Actual Exam
UPDATED | Complete Q&A with Rationales – Pass
Guaranteed - A+ Graded
Part 1 – Foundational Concepts & Naming (10 questions)
Q1: Consider the molecule nitrogen (N₂). Based on electronegativity trends, how would
you describe the bond between the two nitrogen atoms, and what type of hybridization
does each nitrogen atom utilize?
A. Nonpolar covalent bond, sp³ hybridized
B. Polar covalent bond, sp² hybridized
C. Nonpolar covalent bond, sp hybridized [CORRECT]
D. Ionic bond, sp hybridized
Correct Answer: C
Rationale: The bond is nonpolar covalent because the electronegativity difference is
zero between two identical nitrogen atoms. They are triple-bonded, which requires sp
hybridization to form one sigma and two pi bonds, leaving two lone pairs in sp orbitals.
Q2: Which of the following structures represents the most stable resonance contributor
for the nitrate ion (NO₃⁻)?
A. A structure with one N=O double bond and two N-O single bonds, where the negative
charge is localized on a single oxygen.
B. A structure where the nitrogen has a positive formal charge and all three oxygens
share a negative charge equally through resonance. [CORRECT]
C. A structure with three N=O double bonds and a negative charge on the nitrogen.
D. A structure with one N≡O triple bond and two N-O single bonds.
Correct Answer: B
Rationale: The most stable representation involves delocalization of the negative charge
over the three oxygen atoms; the actual structure is a hybrid of three resonance forms,
each placing the double bond on a different oxygen.
Q3: What is the correct IUPAC name for the following alkane:
CH₃-CH(CH₃)-CH₂-CH₂-CH₃?
A. 2-methylpentane
B. 3-methylpentane
C. 1,2-dimethylbutane
,D. Hexane
Correct Answer: A
Rationale: The longest continuous carbon chain is five carbons (pentane), and the
methyl substituent is located on carbon 2 when numbering from the end nearest the
branch.
Q4: Identify the geometry around the central carbon atom in CH₄ and the H-C-H bond
angle.
A. Trigonal planar, 120°
B. Tetrahedral, 109.5° [CORRECT]
C. Bent, 104.5°
D. Linear, 180°
Correct Answer: B
Rationale: Carbon in methane has four bonding pairs and no lone pairs, resulting in a
tetrahedral geometry with ideal bond angles of 109.5°.
Q5: Which of the following Newman projections represents the most stable
conformation of butane when looking down the C2-C3 bond?
A. The anti conformation, where the two methyl groups are 180° apart. [CORRECT]
B. The eclipsed conformation, where the two methyl groups are directly aligned.
C. The gauche conformation, where the methyl groups are 60° apart.
D. The totally eclipsed conformation.
Correct Answer: A
Rationale: The anti conformation places the bulky methyl groups opposite each other
(180°), minimizing steric strain and torsional strain, making it the lowest energy
conformation.
Q6: Calculate the formal charge on the oxygen atom in the hydroxide ion (OH⁻).
A. +1
B. 0
C. -1 [CORRECT]
D. -2
Correct Answer: C
Rationale: Oxygen has 6 valence electrons; in the hydroxide ion, it has 6 non-bonding
electrons and shares 1 bonding electron (half of 2). The calculation is 6 - (6 + 1) = -1.
Q7: In the molecule CH₂Cl₂, determine the molecular geometry and the polarity of the
molecule.
A. Tetrahedral and nonpolar
B. Tetrahedral and polar [CORRECT]
C. Trigonal pyramidal and polar
,D. Bent and nonpolar
Correct Answer: B
Rationale: The molecule is tetrahedral, but because the C-Cl bond dipoles do not cancel
out perfectly due to the different atoms attached, the molecule has a net dipole moment
and is polar.
Q8: Which statement best explains why cyclopropane is significantly more reactive than
cyclohexane?
A. Cyclopropane has angle strain due to its 60° bond angles, which deviates
significantly from the ideal tetrahedral 109.5°. [CORRECT]
B. Cyclopropane contains double bonds, whereas cyclohexane does not.
C. Cyclohexane is planar, while cyclopropane is non-planar.
D. Cyclopropane has torsional strain but no angle strain.
Correct Answer: A
Rationale: The internal bond angles in cyclopropane are forced to be 60°, creating
massive angle strain (Baeyer strain) because the sp³ hybridized orbitals cannot overlap
effectively, making the bonds weak and highly reactive.
Q9: What is the IUPAC name for the cycloalkane represented by a six-membered ring
with a methyl group on carbon 1 and an ethyl group on carbon 3?
A. 1-ethyl-3-methylcyclohexane
B. 1-methyl-3-ethylcyclohexane [CORRECT]
C. 3-ethyl-1-methylcyclohexane
D. 1,3-diethylcyclohexane
Correct Answer: B
Rationale: Substituents are listed alphabetically (ethyl before methyl), and the
numbering is assigned to give the lowest set of locants (1,3-), though in a symmetric
ring like this, starting anywhere yields the same numbers relative to each other.
Q10: Identify the hybridization state of the carbon atoms in an alkyne triple bond.
A. sp³
B. sp²
C. sp [CORRECT]
D. d
Correct Answer: C
Rationale: A carbon participating in a triple bond has two regions of electron density
(one single bond and one triple bond, or two triple bonds), requiring sp hybridization
which yields linear geometry with 180° bond angles.
Part 2 – Structure & Stability (10 questions)
, Q11: When drawing the chair conformation of cyclohexane, which type of hydrogen
positions (axial or equatorial) experience the most steric hindrance when bulky
substituents are attached?
A. Equatorial
B. Axial [CORRECT]
C. Both equally
D. Neither, due to ring flipping
Correct Answer: B
Rationale: Axial bonds are oriented perpendicular to the ring and point straight up or
down, causing 1,3-diaxial interactions with other axial hydrogens/substituents, leading
to significant steric strain.
Q12: Which of the following constitutional isomers of pentane has the lowest boiling
point?
A. n-Pentane
B. Isopentane (2-methylbutane)
C. Neopentane (2,2-dimethylpropane) [CORRECT]
D. They all have the same boiling point.
Correct Answer: C
Rationale: Neopentane is the most compact and spherical shape, resulting in the
smallest surface area and the weakest London dispersion forces, which leads to the
lowest boiling point.
Q13: Which factor primarily determines the stability of a carbocation?
A. Inductive effects from electron-donating groups [CORRECT]
B. The number of hydrogen atoms attached
C. Electronegativity of the adjacent atoms
D. Hybridization of the carbon with the positive charge
Correct Answer: A
Rationale: Alkyl groups are electron-donating via inductive effects and
hyperconjugation, which help stabilize the positive charge on the carbon; therefore,
tertiary carbocations are more stable than primary ones.
Q14: In the context of VSEPR theory, why is the H₂O molecule bent rather than linear?
A. Oxygen forms double bonds.
B. The two lone pairs on oxygen exert greater repulsion than the bonding pairs,
compressing the H-O-H bond angle. [CORRECT]
C. Hydrogen is more electronegative than oxygen.
D. The oxygen atom is sp hybridized.
Correct Answer: B
