1
CHEM 103 / CHEM103 Final Exam General Chemistry I
with Lab Actual Exam 2026/2027 | Complete Exam-Style
Questions | 100% Verified – Detailed Rationales – Pass
Guaranteed – A+ Graded
TABLE OF CONTENTS
Section 1 | Atomic Structure and Periodic Trends | Q1 – Q15
Section 2 | Chemical Bonding and Molecular Geometry | Q16 – Q30
Section 3 | Stoichiometry and Chemical Reactions | Q31 – Q45
Section 4 | Thermochemistry and Gases | Q46 – Q60
Section 5 | Solutions, Acids/Bases, and Lab Techniques | Q61 – Q75
SECTION 1: ATOMIC STRUCTURE AND PERIODIC TRENDS
Question 1 of 75
During a photoelectron spectroscopy experiment, a chemistry student analyzes the binding
energy spectrum of a neutral chlorine atom and observes distinct peaks for each occupied
subshell. When comparing binding energies across all occupied orbitals, which subshell
demonstrates the highest value due to its minimal shielding and maximum effective nuclear
charge?
A. The 3p subshell exhibits the highest binding energy because valence electrons experience the
full nuclear charge of 17 protons.
B. The 2s subshell shows greater binding energy than outer shells but is partially shielded by the
1s electrons.
C. The 1s subshell possesses the highest binding energy as these electrons experience the greatest
effective nuclear charge with virtually no shielding from inner electrons. ✓ CORRECT
D. The 3s subshell demonstrates the highest binding energy due to its superior penetration ability
compared to 3p orbitals.
Correct Answer: C
Rationale: The 1s electrons reside closest to the nucleus and experience an effective nuclear
charge nearly equal to the full nuclear charge of 17 protons with minimal shielding, resulting in
the highest binding energy observed in the PES spectrum. Option A represents a common
misconception that valence electrons have the highest binding energy, when in fact they are the
easiest to remove due to substantial shielding and greater distance from the nucleus.
,2
Photoelectron spectroscopy is a fundamental technique used in modern materials science to
determine elemental composition and electronic structure of surfaces.
Question 2 of 75
In a computational chemistry laboratory, researchers model the electron density maps of
isoelectronic species Na+, Mg2+, and Al3+. When comparing their ionic radii, which prediction
aligns with established periodic trends in effective nuclear charge across these isoelectronic
cations?
A. Na+ possesses the smallest radius because it has the fewest protons, creating weaker electron-
nucleus attraction.
B. Al3+ exhibits the smallest ionic radius since its greater nuclear charge of 13 protons pulls the
same 10-electron cloud inward most strongly. ✓ CORRECT
C. Mg2+ has the smallest radius as its intermediate nuclear charge achieves optimal electron
compression without excessive repulsion.
D. All three cations share identical radii because isoelectronic species always have equivalent
electron cloud distributions regardless of nuclear charge.
Correct Answer: B
Rationale: Among isoelectronic species with identical electron configurations (1s²2s²2p⁶), Al³+
has the highest nuclear charge (13 protons) and therefore exerts the strongest attraction on the
10-electron cloud, yielding the smallest ionic radius. Option D reflects the misconception that
identical electron counts guarantee identical sizes, ignoring the critical role of nuclear charge in
determining electron cloud compression. This principle explains why aluminum oxide ceramics
exhibit high density and hardness in industrial applications.
Question 3 of 75
A mass spectrometry analysis of a naturally occurring sample of magnesium yields three isotopic
peaks with the following data: 78.99% abundance for mass 23.985 u, 10.00% for 24.986 u, and
11.01% for 25.983 u. What is the calculated average atomic mass of magnesium in this sample?
A. 24.31 u ✓ CORRECT
B. 24.99 u
C. 23.99 u
D. 25.98 u
Correct Answer: A
,3
Rationale: The weighted average is computed as (0.7899 × 23.985) + (0.1000 × 24.986) +
(0.1101 × 25.983) = 18.946 + 2.499 + 2.861 = 24.31 u, which matches magnesium's accepted
atomic mass on the periodic table. Option B results from incorrectly averaging the three isotopic
masses without weighting by natural abundance, a frequent computational error in introductory
chemistry. Mass spectrometry remains the definitive analytical method for determining isotopic
composition in geochronology and forensic chemistry.
Question 4 of 75
An electron in a ground-state phosphorus atom occupies an orbital described by the quantum
numbers n = 3, l = 1, m_l = 0. Which statement accurately describes the spatial characteristics
and electron capacity of this specific orbital?
A. This orbital is spherical in shape and can hold a maximum of 10 electrons when completely
filled.
B. The orbital is one of five degenerate orbitals in the 3d subshell, each capable of holding two
electrons.
C. The described orbital is a 3s orbital with a radial node and capacity for two electrons with
opposite spins.
D. The quantum numbers correspond to one of three degenerate 3p orbitals, which has a
dumbbell shape and holds up to two electrons. ✓ CORRECT
Correct Answer: D
Rationale: The angular momentum quantum number l = 1 corresponds to a p orbital, and with n
= 3, this identifies a 3p orbital; each p subshell contains three degenerate orbitals (m_l = -1, 0,
+1) that each accommodate two electrons. Option B incorrectly assigns l = 1 to the d subshell,
when d orbitals require l = 2 and five values of m_l, reflecting a common confusion about
quantum number-orbital correspondence. Understanding quantum numbers is essential for
interpreting electron configurations in atomic spectroscopy and laser technology.
Question 5 of 75
In an advanced materials research facility, engineers select elements for semiconductor doping
based on electronegativity differences with silicon (χ = 1.90). When comparing boron (χ = 2.04),
aluminum (χ = 1.61), phosphorus (χ = 2.19), and gallium (χ = 1.81), which element produces the
smallest electronegativity gap with silicon while remaining distinct enough to alter charge carrier
density?
A. Boron creates the smallest gap with a difference of only 0.14, making it nearly
indistinguishable from silicon.
, 4
B. Aluminum provides a moderate gap of 0.29, offering sufficient distinction for p-type doping
applications.
C. Gallium exhibits the smallest meaningful gap at 0.09, allowing controlled doping while
minimizing interfacial strain. ✓ CORRECT
D. Phosphorus shows a gap of 0.29, which is ideal for creating n-type semiconductors with high
carrier mobility.
Correct Answer: C
Rationale: The electronegativity difference between gallium (1.81) and silicon (1.90) is |1.81 -
1.90| = 0.09, which is the smallest among the options while still providing distinct electronic
properties for controlled doping. Option A miscalculates or misidentifies the smallest gap, as
boron's 0.14 difference exceeds gallium's 0.09 difference. Electronegativity matching is critical
in semiconductor fabrication to ensure minimal lattice mismatch at atomic interfaces.
Question 6 of 75
A chemistry instructor demonstrates the photoelectric effect using cesium metal, which has a
work function of 2.14 eV. When irradiated with photons of wavelength 400 nm, what observable
outcome occurs regarding electron emission, and which calculation supports this prediction? (h =
4.136 × 10⁻¹⁵ eV·s, c = 3.00 × 10⁸ m/s)
A. No electrons are emitted because the photon energy of 1.86 eV falls below the work function
threshold.
B. Electrons are emitted with excess kinetic energy since the 400 nm photons carry 3.10 eV,
exceeding the 2.14 eV work function. ✓ CORRECT
C. Electrons are emitted only if the light intensity increases sufficiently to compensate for the
low photon energy.
D. The metal requires photons with wavelength longer than 400 nm to overcome the binding
energy of cesium atoms.
Correct Answer: B
Rationale: The photon energy is calculated as E = hc/λ = (4.136 × 10⁻¹⁵ eV·s)(3.00 × 10⁸
m/s)/(400 × 10⁻⁹ m) = 3.10 eV, which exceeds cesium's 2.14 eV work function, enabling electron
emission with 0.96 eV of kinetic energy. Option A incorrectly computes the photon energy as
1.86 eV, likely by using an erroneous wavelength value or arithmetic mistake, when the correct
calculation yields 3.10 eV. The photoelectric effect underlies the operation of photovoltaic cells
and photomultiplier tubes in analytical instrumentation.
Question 7 of 75
CHEM 103 / CHEM103 Final Exam General Chemistry I
with Lab Actual Exam 2026/2027 | Complete Exam-Style
Questions | 100% Verified – Detailed Rationales – Pass
Guaranteed – A+ Graded
TABLE OF CONTENTS
Section 1 | Atomic Structure and Periodic Trends | Q1 – Q15
Section 2 | Chemical Bonding and Molecular Geometry | Q16 – Q30
Section 3 | Stoichiometry and Chemical Reactions | Q31 – Q45
Section 4 | Thermochemistry and Gases | Q46 – Q60
Section 5 | Solutions, Acids/Bases, and Lab Techniques | Q61 – Q75
SECTION 1: ATOMIC STRUCTURE AND PERIODIC TRENDS
Question 1 of 75
During a photoelectron spectroscopy experiment, a chemistry student analyzes the binding
energy spectrum of a neutral chlorine atom and observes distinct peaks for each occupied
subshell. When comparing binding energies across all occupied orbitals, which subshell
demonstrates the highest value due to its minimal shielding and maximum effective nuclear
charge?
A. The 3p subshell exhibits the highest binding energy because valence electrons experience the
full nuclear charge of 17 protons.
B. The 2s subshell shows greater binding energy than outer shells but is partially shielded by the
1s electrons.
C. The 1s subshell possesses the highest binding energy as these electrons experience the greatest
effective nuclear charge with virtually no shielding from inner electrons. ✓ CORRECT
D. The 3s subshell demonstrates the highest binding energy due to its superior penetration ability
compared to 3p orbitals.
Correct Answer: C
Rationale: The 1s electrons reside closest to the nucleus and experience an effective nuclear
charge nearly equal to the full nuclear charge of 17 protons with minimal shielding, resulting in
the highest binding energy observed in the PES spectrum. Option A represents a common
misconception that valence electrons have the highest binding energy, when in fact they are the
easiest to remove due to substantial shielding and greater distance from the nucleus.
,2
Photoelectron spectroscopy is a fundamental technique used in modern materials science to
determine elemental composition and electronic structure of surfaces.
Question 2 of 75
In a computational chemistry laboratory, researchers model the electron density maps of
isoelectronic species Na+, Mg2+, and Al3+. When comparing their ionic radii, which prediction
aligns with established periodic trends in effective nuclear charge across these isoelectronic
cations?
A. Na+ possesses the smallest radius because it has the fewest protons, creating weaker electron-
nucleus attraction.
B. Al3+ exhibits the smallest ionic radius since its greater nuclear charge of 13 protons pulls the
same 10-electron cloud inward most strongly. ✓ CORRECT
C. Mg2+ has the smallest radius as its intermediate nuclear charge achieves optimal electron
compression without excessive repulsion.
D. All three cations share identical radii because isoelectronic species always have equivalent
electron cloud distributions regardless of nuclear charge.
Correct Answer: B
Rationale: Among isoelectronic species with identical electron configurations (1s²2s²2p⁶), Al³+
has the highest nuclear charge (13 protons) and therefore exerts the strongest attraction on the
10-electron cloud, yielding the smallest ionic radius. Option D reflects the misconception that
identical electron counts guarantee identical sizes, ignoring the critical role of nuclear charge in
determining electron cloud compression. This principle explains why aluminum oxide ceramics
exhibit high density and hardness in industrial applications.
Question 3 of 75
A mass spectrometry analysis of a naturally occurring sample of magnesium yields three isotopic
peaks with the following data: 78.99% abundance for mass 23.985 u, 10.00% for 24.986 u, and
11.01% for 25.983 u. What is the calculated average atomic mass of magnesium in this sample?
A. 24.31 u ✓ CORRECT
B. 24.99 u
C. 23.99 u
D. 25.98 u
Correct Answer: A
,3
Rationale: The weighted average is computed as (0.7899 × 23.985) + (0.1000 × 24.986) +
(0.1101 × 25.983) = 18.946 + 2.499 + 2.861 = 24.31 u, which matches magnesium's accepted
atomic mass on the periodic table. Option B results from incorrectly averaging the three isotopic
masses without weighting by natural abundance, a frequent computational error in introductory
chemistry. Mass spectrometry remains the definitive analytical method for determining isotopic
composition in geochronology and forensic chemistry.
Question 4 of 75
An electron in a ground-state phosphorus atom occupies an orbital described by the quantum
numbers n = 3, l = 1, m_l = 0. Which statement accurately describes the spatial characteristics
and electron capacity of this specific orbital?
A. This orbital is spherical in shape and can hold a maximum of 10 electrons when completely
filled.
B. The orbital is one of five degenerate orbitals in the 3d subshell, each capable of holding two
electrons.
C. The described orbital is a 3s orbital with a radial node and capacity for two electrons with
opposite spins.
D. The quantum numbers correspond to one of three degenerate 3p orbitals, which has a
dumbbell shape and holds up to two electrons. ✓ CORRECT
Correct Answer: D
Rationale: The angular momentum quantum number l = 1 corresponds to a p orbital, and with n
= 3, this identifies a 3p orbital; each p subshell contains three degenerate orbitals (m_l = -1, 0,
+1) that each accommodate two electrons. Option B incorrectly assigns l = 1 to the d subshell,
when d orbitals require l = 2 and five values of m_l, reflecting a common confusion about
quantum number-orbital correspondence. Understanding quantum numbers is essential for
interpreting electron configurations in atomic spectroscopy and laser technology.
Question 5 of 75
In an advanced materials research facility, engineers select elements for semiconductor doping
based on electronegativity differences with silicon (χ = 1.90). When comparing boron (χ = 2.04),
aluminum (χ = 1.61), phosphorus (χ = 2.19), and gallium (χ = 1.81), which element produces the
smallest electronegativity gap with silicon while remaining distinct enough to alter charge carrier
density?
A. Boron creates the smallest gap with a difference of only 0.14, making it nearly
indistinguishable from silicon.
, 4
B. Aluminum provides a moderate gap of 0.29, offering sufficient distinction for p-type doping
applications.
C. Gallium exhibits the smallest meaningful gap at 0.09, allowing controlled doping while
minimizing interfacial strain. ✓ CORRECT
D. Phosphorus shows a gap of 0.29, which is ideal for creating n-type semiconductors with high
carrier mobility.
Correct Answer: C
Rationale: The electronegativity difference between gallium (1.81) and silicon (1.90) is |1.81 -
1.90| = 0.09, which is the smallest among the options while still providing distinct electronic
properties for controlled doping. Option A miscalculates or misidentifies the smallest gap, as
boron's 0.14 difference exceeds gallium's 0.09 difference. Electronegativity matching is critical
in semiconductor fabrication to ensure minimal lattice mismatch at atomic interfaces.
Question 6 of 75
A chemistry instructor demonstrates the photoelectric effect using cesium metal, which has a
work function of 2.14 eV. When irradiated with photons of wavelength 400 nm, what observable
outcome occurs regarding electron emission, and which calculation supports this prediction? (h =
4.136 × 10⁻¹⁵ eV·s, c = 3.00 × 10⁸ m/s)
A. No electrons are emitted because the photon energy of 1.86 eV falls below the work function
threshold.
B. Electrons are emitted with excess kinetic energy since the 400 nm photons carry 3.10 eV,
exceeding the 2.14 eV work function. ✓ CORRECT
C. Electrons are emitted only if the light intensity increases sufficiently to compensate for the
low photon energy.
D. The metal requires photons with wavelength longer than 400 nm to overcome the binding
energy of cesium atoms.
Correct Answer: B
Rationale: The photon energy is calculated as E = hc/λ = (4.136 × 10⁻¹⁵ eV·s)(3.00 × 10⁸
m/s)/(400 × 10⁻⁹ m) = 3.10 eV, which exceeds cesium's 2.14 eV work function, enabling electron
emission with 0.96 eV of kinetic energy. Option A incorrectly computes the photon energy as
1.86 eV, likely by using an erroneous wavelength value or arithmetic mistake, when the correct
calculation yields 3.10 eV. The photoelectric effect underlies the operation of photovoltaic cells
and photomultiplier tubes in analytical instrumentation.
Question 7 of 75
