CHEM 104 ACTUAL EXAM 2026/2027 | Modules 1-6
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MODULE 1: Atomic Structure, Periodic Trends, and Chemical Bonding
Section 1: Atomic Structure and Electron Configuration
Q1: Which set of quantum numbers (n, l, ml, ms) is valid for an electron in a 3p orbital?
A. 3, 2, 1, +½
B. 3, 1, -1, +½
C. 3, 1, 2, +½
D. 3, 0, 0, +½
Correct Answer: B
Rationale: For a 3p orbital: n = 3 (principal quantum number), l = 1 (p orbital, where l = 0
for s, 1 for p, 2 for d), ml can be -1, 0, or +1 (values from -l to +l), and ms = ±½. Option A
is incorrect because l = 2 corresponds to a d orbital, not p. Option C is invalid because
ml = 2 exceeds the allowed range for l = 1 (ml must be -1, 0, or +1). Option D represents
a 3s orbital (l = 0), not 3p. Option B correctly specifies n = 3, l = 1 (p), ml = -1 (valid), and
ms = +½.
,Q2: How many unpaired electrons are present in the ground state electron configuration
of vanadium (V, Z = 23)?
A. 2
B. 3
C. 4
D. 5
Correct Answer: B
Rationale: The electron configuration of V is [Ar] 3d³ 4s². The 4s orbital is filled with 2
paired electrons. The 3d subshell contains 3 electrons, which according to Hund's rule
occupy separate orbitals with parallel spins to maximize total spin. Therefore, there are
3 unpaired electrons in the 3d subshell. Option A might result from counting only the 4s
electrons or misapplying pairing. Option C could come from incorrectly including the 4s
electrons as unpaired. Option D represents the maximum possible unpaired electrons in
a d subshell (5), which occurs for elements like Mn, not V.
Q3: Which ion has the same electron configuration as argon (Ar, Z = 18)?
A. Ca²⁺
B. K⁺
C. Cl⁻
D. All of the above
Correct Answer: D
Rationale: Argon has the electron configuration [Ne] 3s² 3p⁶ (18 electrons). Ca²⁺
(calcium loses 2 electrons): [Ar] configuration with 18 electrons. K⁺ (potassium loses 1
,electron): [Ar] configuration with 18 electrons. Cl⁻ (chlorine gains 1 electron): [Ar]
configuration with 18 electrons. All three ions are isoelectronic with argon, having 18
electrons each. This demonstrates the stability of noble gas configurations in ion
formation.
Q4: The wavelength of light emitted when an electron transitions from n = 5 to n = 2 in a
hydrogen atom is:
A. 434 nm
B. 486 nm
C. 656 nm
D. 410 nm
Correct Answer: A
Rationale: Using the Rydberg equation: 1/λ = R(1/n₁² - 1/n₂²) where R = 1.097 × 10⁷ m⁻¹,
n₁ = 2, n₂ = 5. Calculation: 1/λ = 1.097 × 10⁷(1/4 - 1/25) = 1.097 × 10⁷(0.25 - 0.04) =
1.097 × 10⁷(0.21) = 2.304 × 10⁶ m⁻¹. Therefore, λ = 4.34 × 10⁻⁷ m = 434 nm. This
corresponds to the blue-violet line in the Balmer series. Option B (486 nm) is the n=4→2
transition, Option C (656 nm) is the n=3→2 transition (red line), and Option D (410 nm)
is the n=6→2 transition.
Q5: Which orbital has two nodal planes passing through the nucleus?
A. 2s
B. 3p
C. 4d
, D. 5f
Correct Answer: C
Rationale: Nodal planes are regions where the wave function equals zero. The number
of angular nodes equals the azimuthal quantum number l. For s orbitals (l=0): 0 nodal
planes. For p orbitals (l=1): 1 nodal plane. For d orbitals (l=2): 2 nodal planes. For f
orbitals (l=3): 3 nodal planes. Therefore, a 4d orbital has 2 nodal planes. The principal
quantum number (n) determines radial nodes (n-l-1), not angular nodes. Option A (2s)
has 1 radial node but 0 angular nodes. Option B (3p) has 1 angular node. Option D (5f)
has 3 angular nodes.
Section 2: Periodic Trends
Q6: Arrange the following elements in order of increasing first ionization energy: Na, Mg,
Al, Si.
A. Na < Al < Mg < Si
B. Na < Mg < Al < Si
C. Si < Al < Mg < Na
D. Na < Mg < Si < Al
Correct Answer: A
Rationale: Generally, ionization energy increases across a period due to increasing
nuclear charge and decreasing atomic radius. However, there's an exception: Mg ([Ne]
3s²) has a higher IE than Al ([Ne] 3s² 3p¹) because Mg has a filled s-subshell (stable
Comprehensive Bank | Newest Expert Verified Q&A |
Academic Year 2026/2027 | Pass Guaranteed - A+ Graded
MODULE 1: Atomic Structure, Periodic Trends, and Chemical Bonding
Section 1: Atomic Structure and Electron Configuration
Q1: Which set of quantum numbers (n, l, ml, ms) is valid for an electron in a 3p orbital?
A. 3, 2, 1, +½
B. 3, 1, -1, +½
C. 3, 1, 2, +½
D. 3, 0, 0, +½
Correct Answer: B
Rationale: For a 3p orbital: n = 3 (principal quantum number), l = 1 (p orbital, where l = 0
for s, 1 for p, 2 for d), ml can be -1, 0, or +1 (values from -l to +l), and ms = ±½. Option A
is incorrect because l = 2 corresponds to a d orbital, not p. Option C is invalid because
ml = 2 exceeds the allowed range for l = 1 (ml must be -1, 0, or +1). Option D represents
a 3s orbital (l = 0), not 3p. Option B correctly specifies n = 3, l = 1 (p), ml = -1 (valid), and
ms = +½.
,Q2: How many unpaired electrons are present in the ground state electron configuration
of vanadium (V, Z = 23)?
A. 2
B. 3
C. 4
D. 5
Correct Answer: B
Rationale: The electron configuration of V is [Ar] 3d³ 4s². The 4s orbital is filled with 2
paired electrons. The 3d subshell contains 3 electrons, which according to Hund's rule
occupy separate orbitals with parallel spins to maximize total spin. Therefore, there are
3 unpaired electrons in the 3d subshell. Option A might result from counting only the 4s
electrons or misapplying pairing. Option C could come from incorrectly including the 4s
electrons as unpaired. Option D represents the maximum possible unpaired electrons in
a d subshell (5), which occurs for elements like Mn, not V.
Q3: Which ion has the same electron configuration as argon (Ar, Z = 18)?
A. Ca²⁺
B. K⁺
C. Cl⁻
D. All of the above
Correct Answer: D
Rationale: Argon has the electron configuration [Ne] 3s² 3p⁶ (18 electrons). Ca²⁺
(calcium loses 2 electrons): [Ar] configuration with 18 electrons. K⁺ (potassium loses 1
,electron): [Ar] configuration with 18 electrons. Cl⁻ (chlorine gains 1 electron): [Ar]
configuration with 18 electrons. All three ions are isoelectronic with argon, having 18
electrons each. This demonstrates the stability of noble gas configurations in ion
formation.
Q4: The wavelength of light emitted when an electron transitions from n = 5 to n = 2 in a
hydrogen atom is:
A. 434 nm
B. 486 nm
C. 656 nm
D. 410 nm
Correct Answer: A
Rationale: Using the Rydberg equation: 1/λ = R(1/n₁² - 1/n₂²) where R = 1.097 × 10⁷ m⁻¹,
n₁ = 2, n₂ = 5. Calculation: 1/λ = 1.097 × 10⁷(1/4 - 1/25) = 1.097 × 10⁷(0.25 - 0.04) =
1.097 × 10⁷(0.21) = 2.304 × 10⁶ m⁻¹. Therefore, λ = 4.34 × 10⁻⁷ m = 434 nm. This
corresponds to the blue-violet line in the Balmer series. Option B (486 nm) is the n=4→2
transition, Option C (656 nm) is the n=3→2 transition (red line), and Option D (410 nm)
is the n=6→2 transition.
Q5: Which orbital has two nodal planes passing through the nucleus?
A. 2s
B. 3p
C. 4d
, D. 5f
Correct Answer: C
Rationale: Nodal planes are regions where the wave function equals zero. The number
of angular nodes equals the azimuthal quantum number l. For s orbitals (l=0): 0 nodal
planes. For p orbitals (l=1): 1 nodal plane. For d orbitals (l=2): 2 nodal planes. For f
orbitals (l=3): 3 nodal planes. Therefore, a 4d orbital has 2 nodal planes. The principal
quantum number (n) determines radial nodes (n-l-1), not angular nodes. Option A (2s)
has 1 radial node but 0 angular nodes. Option B (3p) has 1 angular node. Option D (5f)
has 3 angular nodes.
Section 2: Periodic Trends
Q6: Arrange the following elements in order of increasing first ionization energy: Na, Mg,
Al, Si.
A. Na < Al < Mg < Si
B. Na < Mg < Al < Si
C. Si < Al < Mg < Na
D. Na < Mg < Si < Al
Correct Answer: A
Rationale: Generally, ionization energy increases across a period due to increasing
nuclear charge and decreasing atomic radius. However, there's an exception: Mg ([Ne]
3s²) has a higher IE than Al ([Ne] 3s² 3p¹) because Mg has a filled s-subshell (stable
