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CHEM 120 WEEK 8 FINAL EXAM 2026/2027 | General
Chemistry Atomic Structure Stoichiometry
Thermochemistry Kinetics Equilibrium Acids/Bases | 100%
Correct Solutions Graded A | Pass Guaranteed
Section 1: Atomic Structure & Periodic Trends (Questions 1–12)
Q1. In Rutherford's gold foil experiment, most alpha particles passed straight
through the foil, but a small fraction were deflected at large angles or even bounced
back. What conclusion did Rutherford draw from these observations?
A. Atoms are solid, indivisible spheres with uniform density throughout
B. Atoms contain a small, dense, positively charged nucleus surrounded by mostly
empty space
C. Electrons are embedded in a positively charged "plum pudding" matrix
D. Atoms absorb and emit energy only at specific discrete wavelengths
Correct Answer: B
Rationale: The large-angle deflections indicated that alpha particles encountered a
small, massive, positively charged region—the nucleus—while most particles passed
through empty space. This disproved Thomson's plum pudding model and Dalton's
solid sphere model. The discrete energy emission conclusion belongs to Bohr, not
Rutherford. [CORRECT]
Q2. Which of the following sets of quantum numbers (n, ℓ, mℓ, ms) is valid for an
electron in a 4d orbital?
A. 4, 3, 2, +½
B. 4, 2, 3, −½
C. 4, 2, −2, +½
D. 4, 1, 0, +½
,2
Correct Answer: C
Rationale: For a 4d orbital, n = 4 and ℓ = 2 (since d corresponds to ℓ = 2). The
magnetic quantum number mℓ must range from −ℓ to +ℓ, so mℓ = −2 is valid. Choice
A has ℓ = 3 (f orbital); choice B has mℓ = 3 which exceeds ℓ = 2; choice D has ℓ = 1 (p
orbital). [CORRECT]
Q3. What is the ground-state electron configuration of copper (Cu, Z = 29)?
A. [Ar] 4s² 3d⁹
B. [Ar] 4s¹ 3d¹⁰
C. [Ar] 4s² 3d¹⁰
D. [Ar] 3d¹⁰ 4p¹
Correct Answer: B
Rationale: Copper is an exception to the Aufbau principle due to the extra stability
of a completely filled d subshell. The configuration [Ar] 4s¹ 3d¹⁰ is more stable than
[Ar] 4s² 3d⁹ because a filled 3d¹⁰ subshell minimizes electron-electron repulsion.
Choice A is the expected but incorrect Aufbau configuration; choice C is zinc.
[CORRECT]
Q4. Which element has the highest first ionization energy?
A. Beryllium (Be)
B. Boron (B)
C. Carbon (C)
D. Nitrogen (N)
Correct Answer: D
Rationale: First ionization energy generally increases left to right across a period, but
there is an exception between Be (2s², fully filled subshell) and B (2p¹, less stable).
Nitrogen has a half-filled 2p³ subshell, which provides extra stability, giving it the
highest first ionization energy among these elements. [CORRECT]
,3
Q5. Arrange the following species in order of increasing ionic radius: Mg²⁺, Na⁺, F⁻,
O²⁻.
A. Mg²⁺ < Na⁺ < F⁻ < O²⁻
B. O²⁻ < F⁻ < Na⁺ < Mg²⁺
C. Mg²⁺ < O²⁻ < Na⁺ < F⁻
D. Na⁺ < Mg²⁺ < F⁻ < O²⁻
Correct Answer: A
Rationale: All four species are isoelectronic with neon (10 electrons). Ionic radius
decreases as nuclear charge increases because the same number of electrons is
pulled closer to a larger nucleus. The nuclear charges are: Mg²⁺ (12), Na⁺ (11), F⁻ (9),
O²⁻ (8). Thus Mg²⁺ is smallest and O²⁻ is largest. [CORRECT]
Q6. Which statement about effective nuclear charge (Z_eff) is correct?
A. Z_eff increases down a group because the principal quantum number increases
B. Z_eff decreases across a period because more electrons are added
C. Z_eff increases across a period due to poor shielding by electrons in the same shell
D. Z_eff is constant for all elements in the same period
Correct Answer: C
Rationale: Effective nuclear charge increases across a period because electrons
added to the same principal shell shield each other poorly, while proton number
increases. This pulls valence electrons closer to the nucleus, explaining decreasing
atomic radius across a period. Down a group, Z_eff remains relatively constant.
[CORRECT]
Q7. Which of the following is the correct electron configuration for the Fe³⁺ ion?
A. [Ar] 4s² 3d³
B. [Ar] 4s² 3d⁶
, 4
C. [Ar] 3d⁵
D. [Ar] 3d⁶
Correct Answer: C
Rationale: Neutral Fe is [Ar] 4s² 3d⁶. Transition metals lose electrons from the s
orbital before the d orbital. Fe³⁺ loses all three valence electrons (two 4s and one 3d),
leaving [Ar] 3d⁵. Choice A incorrectly removes two 3d and one 4s electron; choice B is
Fe²⁺. [CORRECT]
Q8. The Heisenberg uncertainty principle states that:
A. It is impossible to know both the exact position and exact momentum of an
electron simultaneously
B. Electrons occupy discrete energy levels in the atom
C. No two electrons in an atom can have the same set of four quantum numbers
D. Electrons fill degenerate orbitals singly before pairing
Correct Answer: A
Rationale: The Heisenberg uncertainty principle mathematically states that Δx · Δp ≥
h/4π, meaning the more precisely position is known, the less precisely momentum
can be known, and vice versa. Choice B describes Bohr's postulate; choice C is the
Pauli exclusion principle; choice D is Hund's rule. [CORRECT]
Q9. Which element has the most negative electron affinity?
A. Sodium (Na)
B. Magnesium (Mg)
C. Chlorine (Cl)
D. Argon (Ar)
Correct Answer: C
Rationale: Electron affinity is most negative (most energy released) for halogens
because adding one electron gives them a stable noble gas configuration. Chlorine
has a very high electron affinity because the added electron enters the 3p subshell
CHEM 120 WEEK 8 FINAL EXAM 2026/2027 | General
Chemistry Atomic Structure Stoichiometry
Thermochemistry Kinetics Equilibrium Acids/Bases | 100%
Correct Solutions Graded A | Pass Guaranteed
Section 1: Atomic Structure & Periodic Trends (Questions 1–12)
Q1. In Rutherford's gold foil experiment, most alpha particles passed straight
through the foil, but a small fraction were deflected at large angles or even bounced
back. What conclusion did Rutherford draw from these observations?
A. Atoms are solid, indivisible spheres with uniform density throughout
B. Atoms contain a small, dense, positively charged nucleus surrounded by mostly
empty space
C. Electrons are embedded in a positively charged "plum pudding" matrix
D. Atoms absorb and emit energy only at specific discrete wavelengths
Correct Answer: B
Rationale: The large-angle deflections indicated that alpha particles encountered a
small, massive, positively charged region—the nucleus—while most particles passed
through empty space. This disproved Thomson's plum pudding model and Dalton's
solid sphere model. The discrete energy emission conclusion belongs to Bohr, not
Rutherford. [CORRECT]
Q2. Which of the following sets of quantum numbers (n, ℓ, mℓ, ms) is valid for an
electron in a 4d orbital?
A. 4, 3, 2, +½
B. 4, 2, 3, −½
C. 4, 2, −2, +½
D. 4, 1, 0, +½
,2
Correct Answer: C
Rationale: For a 4d orbital, n = 4 and ℓ = 2 (since d corresponds to ℓ = 2). The
magnetic quantum number mℓ must range from −ℓ to +ℓ, so mℓ = −2 is valid. Choice
A has ℓ = 3 (f orbital); choice B has mℓ = 3 which exceeds ℓ = 2; choice D has ℓ = 1 (p
orbital). [CORRECT]
Q3. What is the ground-state electron configuration of copper (Cu, Z = 29)?
A. [Ar] 4s² 3d⁹
B. [Ar] 4s¹ 3d¹⁰
C. [Ar] 4s² 3d¹⁰
D. [Ar] 3d¹⁰ 4p¹
Correct Answer: B
Rationale: Copper is an exception to the Aufbau principle due to the extra stability
of a completely filled d subshell. The configuration [Ar] 4s¹ 3d¹⁰ is more stable than
[Ar] 4s² 3d⁹ because a filled 3d¹⁰ subshell minimizes electron-electron repulsion.
Choice A is the expected but incorrect Aufbau configuration; choice C is zinc.
[CORRECT]
Q4. Which element has the highest first ionization energy?
A. Beryllium (Be)
B. Boron (B)
C. Carbon (C)
D. Nitrogen (N)
Correct Answer: D
Rationale: First ionization energy generally increases left to right across a period, but
there is an exception between Be (2s², fully filled subshell) and B (2p¹, less stable).
Nitrogen has a half-filled 2p³ subshell, which provides extra stability, giving it the
highest first ionization energy among these elements. [CORRECT]
,3
Q5. Arrange the following species in order of increasing ionic radius: Mg²⁺, Na⁺, F⁻,
O²⁻.
A. Mg²⁺ < Na⁺ < F⁻ < O²⁻
B. O²⁻ < F⁻ < Na⁺ < Mg²⁺
C. Mg²⁺ < O²⁻ < Na⁺ < F⁻
D. Na⁺ < Mg²⁺ < F⁻ < O²⁻
Correct Answer: A
Rationale: All four species are isoelectronic with neon (10 electrons). Ionic radius
decreases as nuclear charge increases because the same number of electrons is
pulled closer to a larger nucleus. The nuclear charges are: Mg²⁺ (12), Na⁺ (11), F⁻ (9),
O²⁻ (8). Thus Mg²⁺ is smallest and O²⁻ is largest. [CORRECT]
Q6. Which statement about effective nuclear charge (Z_eff) is correct?
A. Z_eff increases down a group because the principal quantum number increases
B. Z_eff decreases across a period because more electrons are added
C. Z_eff increases across a period due to poor shielding by electrons in the same shell
D. Z_eff is constant for all elements in the same period
Correct Answer: C
Rationale: Effective nuclear charge increases across a period because electrons
added to the same principal shell shield each other poorly, while proton number
increases. This pulls valence electrons closer to the nucleus, explaining decreasing
atomic radius across a period. Down a group, Z_eff remains relatively constant.
[CORRECT]
Q7. Which of the following is the correct electron configuration for the Fe³⁺ ion?
A. [Ar] 4s² 3d³
B. [Ar] 4s² 3d⁶
, 4
C. [Ar] 3d⁵
D. [Ar] 3d⁶
Correct Answer: C
Rationale: Neutral Fe is [Ar] 4s² 3d⁶. Transition metals lose electrons from the s
orbital before the d orbital. Fe³⁺ loses all three valence electrons (two 4s and one 3d),
leaving [Ar] 3d⁵. Choice A incorrectly removes two 3d and one 4s electron; choice B is
Fe²⁺. [CORRECT]
Q8. The Heisenberg uncertainty principle states that:
A. It is impossible to know both the exact position and exact momentum of an
electron simultaneously
B. Electrons occupy discrete energy levels in the atom
C. No two electrons in an atom can have the same set of four quantum numbers
D. Electrons fill degenerate orbitals singly before pairing
Correct Answer: A
Rationale: The Heisenberg uncertainty principle mathematically states that Δx · Δp ≥
h/4π, meaning the more precisely position is known, the less precisely momentum
can be known, and vice versa. Choice B describes Bohr's postulate; choice C is the
Pauli exclusion principle; choice D is Hund's rule. [CORRECT]
Q9. Which element has the most negative electron affinity?
A. Sodium (Na)
B. Magnesium (Mg)
C. Chlorine (Cl)
D. Argon (Ar)
Correct Answer: C
Rationale: Electron affinity is most negative (most energy released) for halogens
because adding one electron gives them a stable noble gas configuration. Chlorine
has a very high electron affinity because the added electron enters the 3p subshell
