CHEM 1020 Final Exam | 184 Questions | With Complete Solutions 2026/2027 General Chemistry II
CHEM 1020 FINAL EXAM
184 Questions | With Complete Solutions | 2026/2027
General Chemistry II
Course: CHEM 1020 - General Chemistry II Total Questions: 184
Exam Type: Final Examination (Comprehensive) Format: Multiple Choice (A-D)
Academic Year: Cognitive Mix: 25% Recall / 50% Application / 25% Analysis
Time Allotted: 180 minutes Passing Score: 60% (110/184 correct)
Topics Covered
Atomic Structure & Periodicity • Quantum Mechanics • Chemical Bonding & Molecular Geometry • Intermolecular
Forces • Thermochemistry & Thermodynamics • Chemical Kinetics • Chemical Equilibrium & Le Châtelier's
Principle • Acids, Bases & pH • Solubility Equilibria & Ksp • Electrochemistry & Redox • Nuclear Chemistry •
Organic Chemistry Basics
Section 1: Atomic Structure, Periodicity, & Quantum Mechanics
Q1: For a hydrogen atom, which set of quantum numbers (n, l, m_l, m_s) is NOT allowed?
A. (3, 2, -2, +1/2)
B. (3, 1, 0, -1/2)
C. (3, 3, 1, +1/2) *[CORRECT]*
D. (2, 0, 0, -1/2)
Correct Answer: C
Rationale: For n = 3, the allowed values of l are 0, 1, and 2 (l < n); l = 3 is forbidden. Option C violates this rule.
The other sets follow the standard quantum-number constraints taught in General Chemistry II.
Q2: Which transition in a hydrogen atom corresponds to the longest wavelength of emitted light?
A. n = 4 to n = 1
B. n = 6 to n = 3 *[CORRECT]*
C. n = 5 to n = 1
D. n = 3 to n = 1
Correct Answer: B
Rationale: Longest wavelength means smallest energy difference (E = hc/lambda). The n = 6 to n = 3 transition in
the Paschen series has the smallest energy gap among the listed options, producing the longest wavelength (1094
nm, infrared).
Q3: The electron configuration [Xe] 4f14 5d10 6s2 6p3 corresponds to which element?
A. Lead (Pb)
B. Bismuth (Bi) *[CORRECT]*
C. Polonium (Po)
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,CHEM 1020 Final Exam | 184 Questions | With Complete Solutions 2026/2027 General Chemistry II
D. Astatine (At)
Correct Answer: B
Rationale: Counting electrons: [Xe] = 54, plus 4f14 (14) + 5d10 (10) + 6s2 (2) + 6p3 (3) = 83 total, which is
Bismuth (Bi, Z = 83). Lead has 82, Polonium has 84, and Astatine has 85 electrons.
Q4: How many orbitals are present in the n = 3 shell of a hydrogen atom?
A. 3
B. 6
C. 9 *[CORRECT]*
D. 18
Correct Answer: C
Rationale: The n = 3 shell contains subshells l = 0 (1 orbital, 3s), l = 1 (3 orbitals, 3p), and l = 2 (5 orbitals, 3d),
giving 1 + 3 + 5 = 9 orbitals total. Each orbital holds max 2 electrons, so n = 3 holds up to 18 electrons.
Q5: What is the maximum number of electrons that can occupy a single molecular orbital?
A. 1
B. 2 *[CORRECT]*
C. 4
D. 6
Correct Answer: B
Rationale: Following the Pauli exclusion principle, a single molecular orbital can hold a maximum of two
electrons with opposite spins. This applies to both bonding and antibonding MOs in molecular orbital theory.
Q6: Which statement about the four quantum numbers is correct?
A. ml can be any integer from 0 to +l
B. ms can only be 0 or 1
C. l can take integer values from 0 to n-1 *[CORRECT]*
D. n determines the shape of the orbital
Correct Answer: C
Rationale: The azimuthal quantum number l takes integer values from 0 to n - 1. Option A is wrong (ml ranges
from -l to +l), Option B is wrong (ms = +/-1/2), and Option D is wrong (n determines the energy/size, l determines
shape).
Q7: Which element has the highest first ionization energy?
A. Na
B. Mg
C. Al
D. Si *[CORRECT]*
Correct Answer: D
Rationale: Across period 3, ionization energy generally increases from left to right due to increasing effective
nuclear charge. Silicon (Si), being farthest right among the options, has the highest first ionization energy (786
kJ/mol).
Q8: Which of the following atoms has the largest atomic radius?
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,CHEM 1020 Final Exam | 184 Questions | With Complete Solutions 2026/2027 General Chemistry II
A. F
B. Cl
C. Br
D. I *[CORRECT]*
Correct Answer: D
Rationale: Atomic radius increases down a group as additional principal energy levels are added. Iodine (I), being
in period 5, has the largest atomic radius among the halogens listed (≈imately 133 pm covalent radius).
Q9: The electron configuration of Cr (Z = 24) is:
A. [Ar] 4s2 3d4
B. [Ar] 4s1 3d5 *[CORRECT]*
C. [Ar] 4s2 3d6
D. [Ar] 4s1 3d4
Correct Answer: B
Rationale: Chromium is an exception to the Aufbau principle. The half-filled d subshell ([Ar] 4s1 3d5) provides
extra stability from exchange energy. The same exception occurs for Mo, Cu, Ag, and Au.
Q10: Which set of isoelectronic species is correctly ordered by increasing ionic radius?
A. O2- < F- < Na+ < Mg2+
B. Mg2+ < Na+ < F- < O2- *[CORRECT]*
C. Na+ < F- < O2- < Mg2+
D. F- < O2- < Mg2+ < Na+
Correct Answer: B
Rationale: All four species have 10 electrons. With constant electron count, fewer protons mean larger radius (less
nuclear pull). Mg2+ (12p) < Na+ (11p) < F- (9p) < O2- (8p), so Mg2+ is smallest and O2- is largest.
Q11: Which of the following has the most unpaired electrons in its ground state?
A. O
B. N *[CORRECT]*
C. C
D. F
Correct Answer: B
Rationale: Nitrogen (1s2 2s2 2p3) has three unpaired electrons (Hund's rule in 2p). Oxygen has 2, carbon has 2,
and fluorine has 1. Thus N has the most unpaired electrons.
Q12: The de Broglie wavelength of a particle is given by:
A. lambda = h/p *[CORRECT]*
B. lambda = hv
C. lambda = c/v
D. lambda = E/h
Correct Answer: A
Rationale: de Broglie's equation, lambda = h/p (where p = momentum = mv), describes the wave-particle duality
of matter. This fundamental relationship bridges quantum mechanics and classical momentum.
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, CHEM 1020 Final Exam | 184 Questions | With Complete Solutions 2026/2027 General Chemistry II
Q13: Which of the following transitions in a hydrogen atom emits the highest-energy photon?
A. n = 2 to n = 1 *[CORRECT]*
B. n = 3 to n = 2
C. n = 4 to n = 3
D. n = 5 to n = 4
Correct Answer: A
Rationale: Energy of an emitted photon is largest when the difference between energy levels is largest. The n = 2 to
n = 1 transition (Lyman series) releases the most energy among the listed options (10.2 eV, UV region).
Q14: Which statement about periodic trends is correct?
A. Electronegativity decreases across a period
B. Atomic radius increases across a period
C. Metallic character decreases across a period *[CORRECT]*
D. Ionization energy decreases across a period
Correct Answer: C
Rationale: Metallic character decreases across a period (left to right) because elements become more nonmetallic
as effective nuclear charge increases. The other options state trends opposite to what is observed.
Q15: The shape of an s orbital is best described as:
A. dumbbell
B. spherical *[CORRECT]*
C. cloverleaf
D. torus
Correct Answer: B
Rationale: An s orbital (l = 0) has spherical symmetry, meaning the probability of finding an electron depends only
on distance from the nucleus, not direction. p orbitals are dumbbell-shaped, and d orbitals have more complex
cloverleaf shapes.
Q16: Which element has the electron configuration [Ar] 3d10 4s2 4p5?
A. Br *[CORRECT]*
B. Kr
C. Se
D. As
Correct Answer: A
Rationale: Counting: [Ar] = 18, plus 3d10 (10) + 4s2 (2) + 4p5 (5) = 35, which is Bromine (Br, Z = 35). This
places it in Group 17 (halogens) of period 4.
Q17: How many electrons can occupy the 4f subshell?
A. 6
B. 10
C. 14 *[CORRECT]*
D. 2
Correct Answer: C
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CHEM 1020 FINAL EXAM
184 Questions | With Complete Solutions | 2026/2027
General Chemistry II
Course: CHEM 1020 - General Chemistry II Total Questions: 184
Exam Type: Final Examination (Comprehensive) Format: Multiple Choice (A-D)
Academic Year: Cognitive Mix: 25% Recall / 50% Application / 25% Analysis
Time Allotted: 180 minutes Passing Score: 60% (110/184 correct)
Topics Covered
Atomic Structure & Periodicity • Quantum Mechanics • Chemical Bonding & Molecular Geometry • Intermolecular
Forces • Thermochemistry & Thermodynamics • Chemical Kinetics • Chemical Equilibrium & Le Châtelier's
Principle • Acids, Bases & pH • Solubility Equilibria & Ksp • Electrochemistry & Redox • Nuclear Chemistry •
Organic Chemistry Basics
Section 1: Atomic Structure, Periodicity, & Quantum Mechanics
Q1: For a hydrogen atom, which set of quantum numbers (n, l, m_l, m_s) is NOT allowed?
A. (3, 2, -2, +1/2)
B. (3, 1, 0, -1/2)
C. (3, 3, 1, +1/2) *[CORRECT]*
D. (2, 0, 0, -1/2)
Correct Answer: C
Rationale: For n = 3, the allowed values of l are 0, 1, and 2 (l < n); l = 3 is forbidden. Option C violates this rule.
The other sets follow the standard quantum-number constraints taught in General Chemistry II.
Q2: Which transition in a hydrogen atom corresponds to the longest wavelength of emitted light?
A. n = 4 to n = 1
B. n = 6 to n = 3 *[CORRECT]*
C. n = 5 to n = 1
D. n = 3 to n = 1
Correct Answer: B
Rationale: Longest wavelength means smallest energy difference (E = hc/lambda). The n = 6 to n = 3 transition in
the Paschen series has the smallest energy gap among the listed options, producing the longest wavelength (1094
nm, infrared).
Q3: The electron configuration [Xe] 4f14 5d10 6s2 6p3 corresponds to which element?
A. Lead (Pb)
B. Bismuth (Bi) *[CORRECT]*
C. Polonium (Po)
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,CHEM 1020 Final Exam | 184 Questions | With Complete Solutions 2026/2027 General Chemistry II
D. Astatine (At)
Correct Answer: B
Rationale: Counting electrons: [Xe] = 54, plus 4f14 (14) + 5d10 (10) + 6s2 (2) + 6p3 (3) = 83 total, which is
Bismuth (Bi, Z = 83). Lead has 82, Polonium has 84, and Astatine has 85 electrons.
Q4: How many orbitals are present in the n = 3 shell of a hydrogen atom?
A. 3
B. 6
C. 9 *[CORRECT]*
D. 18
Correct Answer: C
Rationale: The n = 3 shell contains subshells l = 0 (1 orbital, 3s), l = 1 (3 orbitals, 3p), and l = 2 (5 orbitals, 3d),
giving 1 + 3 + 5 = 9 orbitals total. Each orbital holds max 2 electrons, so n = 3 holds up to 18 electrons.
Q5: What is the maximum number of electrons that can occupy a single molecular orbital?
A. 1
B. 2 *[CORRECT]*
C. 4
D. 6
Correct Answer: B
Rationale: Following the Pauli exclusion principle, a single molecular orbital can hold a maximum of two
electrons with opposite spins. This applies to both bonding and antibonding MOs in molecular orbital theory.
Q6: Which statement about the four quantum numbers is correct?
A. ml can be any integer from 0 to +l
B. ms can only be 0 or 1
C. l can take integer values from 0 to n-1 *[CORRECT]*
D. n determines the shape of the orbital
Correct Answer: C
Rationale: The azimuthal quantum number l takes integer values from 0 to n - 1. Option A is wrong (ml ranges
from -l to +l), Option B is wrong (ms = +/-1/2), and Option D is wrong (n determines the energy/size, l determines
shape).
Q7: Which element has the highest first ionization energy?
A. Na
B. Mg
C. Al
D. Si *[CORRECT]*
Correct Answer: D
Rationale: Across period 3, ionization energy generally increases from left to right due to increasing effective
nuclear charge. Silicon (Si), being farthest right among the options, has the highest first ionization energy (786
kJ/mol).
Q8: Which of the following atoms has the largest atomic radius?
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,CHEM 1020 Final Exam | 184 Questions | With Complete Solutions 2026/2027 General Chemistry II
A. F
B. Cl
C. Br
D. I *[CORRECT]*
Correct Answer: D
Rationale: Atomic radius increases down a group as additional principal energy levels are added. Iodine (I), being
in period 5, has the largest atomic radius among the halogens listed (≈imately 133 pm covalent radius).
Q9: The electron configuration of Cr (Z = 24) is:
A. [Ar] 4s2 3d4
B. [Ar] 4s1 3d5 *[CORRECT]*
C. [Ar] 4s2 3d6
D. [Ar] 4s1 3d4
Correct Answer: B
Rationale: Chromium is an exception to the Aufbau principle. The half-filled d subshell ([Ar] 4s1 3d5) provides
extra stability from exchange energy. The same exception occurs for Mo, Cu, Ag, and Au.
Q10: Which set of isoelectronic species is correctly ordered by increasing ionic radius?
A. O2- < F- < Na+ < Mg2+
B. Mg2+ < Na+ < F- < O2- *[CORRECT]*
C. Na+ < F- < O2- < Mg2+
D. F- < O2- < Mg2+ < Na+
Correct Answer: B
Rationale: All four species have 10 electrons. With constant electron count, fewer protons mean larger radius (less
nuclear pull). Mg2+ (12p) < Na+ (11p) < F- (9p) < O2- (8p), so Mg2+ is smallest and O2- is largest.
Q11: Which of the following has the most unpaired electrons in its ground state?
A. O
B. N *[CORRECT]*
C. C
D. F
Correct Answer: B
Rationale: Nitrogen (1s2 2s2 2p3) has three unpaired electrons (Hund's rule in 2p). Oxygen has 2, carbon has 2,
and fluorine has 1. Thus N has the most unpaired electrons.
Q12: The de Broglie wavelength of a particle is given by:
A. lambda = h/p *[CORRECT]*
B. lambda = hv
C. lambda = c/v
D. lambda = E/h
Correct Answer: A
Rationale: de Broglie's equation, lambda = h/p (where p = momentum = mv), describes the wave-particle duality
of matter. This fundamental relationship bridges quantum mechanics and classical momentum.
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, CHEM 1020 Final Exam | 184 Questions | With Complete Solutions 2026/2027 General Chemistry II
Q13: Which of the following transitions in a hydrogen atom emits the highest-energy photon?
A. n = 2 to n = 1 *[CORRECT]*
B. n = 3 to n = 2
C. n = 4 to n = 3
D. n = 5 to n = 4
Correct Answer: A
Rationale: Energy of an emitted photon is largest when the difference between energy levels is largest. The n = 2 to
n = 1 transition (Lyman series) releases the most energy among the listed options (10.2 eV, UV region).
Q14: Which statement about periodic trends is correct?
A. Electronegativity decreases across a period
B. Atomic radius increases across a period
C. Metallic character decreases across a period *[CORRECT]*
D. Ionization energy decreases across a period
Correct Answer: C
Rationale: Metallic character decreases across a period (left to right) because elements become more nonmetallic
as effective nuclear charge increases. The other options state trends opposite to what is observed.
Q15: The shape of an s orbital is best described as:
A. dumbbell
B. spherical *[CORRECT]*
C. cloverleaf
D. torus
Correct Answer: B
Rationale: An s orbital (l = 0) has spherical symmetry, meaning the probability of finding an electron depends only
on distance from the nucleus, not direction. p orbitals are dumbbell-shaped, and d orbitals have more complex
cloverleaf shapes.
Q16: Which element has the electron configuration [Ar] 3d10 4s2 4p5?
A. Br *[CORRECT]*
B. Kr
C. Se
D. As
Correct Answer: A
Rationale: Counting: [Ar] = 18, plus 3d10 (10) + 4s2 (2) + 4p5 (5) = 35, which is Bromine (Br, Z = 35). This
places it in Group 17 (halogens) of period 4.
Q17: How many electrons can occupy the 4f subshell?
A. 6
B. 10
C. 14 *[CORRECT]*
D. 2
Correct Answer: C
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