CHEM 210 Module 2 Exam (Portage Learning
2025/2026) — 30 High-Level Practice Questions
with 100% Correct Answers
Which quantum-number set (n, l, m , mₛ) is permissible for an electron in a
ground-state potassium atom?
1. A. 4, 2, −1, +½
B. 3, 3, 0, −½
C. 4, 1, +2, +½
D. 3, 0, +1, −½
Correct Answer: A
Explanation: l must be ≤ n−1; m ranges from −l to +l; mₛ = ±½. Only set A (4d
electron) satisfies all rules.
The first ionization energy of aluminum is lower than that of magnesium because:
2. A. Al has a larger nuclear charge
B. the 3p electron in Al is farther from the nucleus on average
C. Mg contains paired 3s electrons
D. Al has a filled 3s subshell
Correct Answer: B
, Explanation: The 3p electron in Al is higher in energy (less penetration) than Mg’s 3s
electron, so less energy is needed to remove it.
How many nodes (total) are present in a 4s orbital?
3. A. 2
B. 3
C. 4
D. 5
Correct Answer: B
Explanation: Total nodes = n − 1 = 4 − 1 = 3 (one radial, two angular).
Arrange the following species in order of increasing atomic radius: P³⁻, S²⁻, Cl⁻, Ar.
4. A. Cl⁻ < Ar < S²⁻ < P³⁻
B. Ar < Cl⁻ < S²⁻ < P³⁻
C. P³⁻ < S²⁻ < Cl⁻ < Ar
D. Ar < P³⁻ < S²⁻ < Cl⁻
Correct Answer: A
Explanation: All are isoelectronic (18 e⁻); radius increases as nuclear charge decreases
(P³⁻ has smallest Zeff).
Which pair of elements would form the most polar single bond?
2025/2026) — 30 High-Level Practice Questions
with 100% Correct Answers
Which quantum-number set (n, l, m , mₛ) is permissible for an electron in a
ground-state potassium atom?
1. A. 4, 2, −1, +½
B. 3, 3, 0, −½
C. 4, 1, +2, +½
D. 3, 0, +1, −½
Correct Answer: A
Explanation: l must be ≤ n−1; m ranges from −l to +l; mₛ = ±½. Only set A (4d
electron) satisfies all rules.
The first ionization energy of aluminum is lower than that of magnesium because:
2. A. Al has a larger nuclear charge
B. the 3p electron in Al is farther from the nucleus on average
C. Mg contains paired 3s electrons
D. Al has a filled 3s subshell
Correct Answer: B
, Explanation: The 3p electron in Al is higher in energy (less penetration) than Mg’s 3s
electron, so less energy is needed to remove it.
How many nodes (total) are present in a 4s orbital?
3. A. 2
B. 3
C. 4
D. 5
Correct Answer: B
Explanation: Total nodes = n − 1 = 4 − 1 = 3 (one radial, two angular).
Arrange the following species in order of increasing atomic radius: P³⁻, S²⁻, Cl⁻, Ar.
4. A. Cl⁻ < Ar < S²⁻ < P³⁻
B. Ar < Cl⁻ < S²⁻ < P³⁻
C. P³⁻ < S²⁻ < Cl⁻ < Ar
D. Ar < P³⁻ < S²⁻ < Cl⁻
Correct Answer: A
Explanation: All are isoelectronic (18 e⁻); radius increases as nuclear charge decreases
(P³⁻ has smallest Zeff).
Which pair of elements would form the most polar single bond?
