CHEM 210 Module 2 Exam (Portage Learning
2025/2026) — 30 High-Level Actual Questions with 100%
Correct Answers
1. Which set of quantum numbers (n, l, m , mₛ) is valid for an electron in a
ground-state arsenic atom?
A. 4, 2, −3, +½
B. 3, 1, 0, −½
C. 3, 3, 1, +½
D. 2, 0, 1, −½
Correct Answer: B
Explanation: l must be ≤ n − 1; m ranges from −l to +l. Only B satisfies all rules (n = 3,
l = 1, m = 0, mₛ = ±½).
2. The first ionization energy of magnesium is 738 kJ mol⁻¹. What is the minimum
frequency of light required to ionize a single Mg atom?
A. 1.84 × 10¹⁵ s⁻¹
B. 1.10 × 10¹⁵ s⁻¹
C. 5.51 × 10¹⁴ s⁻¹
D. 3.68 × 10¹⁴ s⁻¹
Correct Answer: C
Calculation: E = 738 kJ mol⁻¹ ÷ 6.022 × 10²³ = 1.225 × 10⁻¹⁸ J atom⁻¹; ν = E/h = 1.225 ×
10⁻¹⁸ J ÷ 6.626 × 10⁻³⁴ J·s = 5.51 × 10¹⁴ s⁻¹.
, 3. How many nodes (total) are present in a 4p orbital?
A. 2
B. 3
C. 4
D. 5
Correct Answer: B
Explanation: Total nodes = n − 1 = 3 (one radial, two angular).
4. Arrange the following species in order of increasing atomic radius: P³⁻, S²⁻, Cl⁻,
Ar.
A. Cl⁻ < S²⁻ < P³⁻ < Ar
B. Ar < Cl⁻ < S²⁻ < P³⁻
C. P³⁻ < S²⁻ < Cl⁻ < Ar
D. Ar < P³⁻ < S²⁻ < Cl⁻
Correct Answer: B
Explanation: All are isoelectronic (18 e⁻); radius increases as nuclear charge decreases
(Ar = 18, Cl⁻ = 17, S²⁻ = 16, P³⁻ = 15).
5. Which element has the largest third ionization energy?
A. Al
B. Si
C. Mg
D. Na
Correct Answer: C
Explanation: Mg loses two 3s electrons first; the third IE removes a 2p core electron,
which is much tighter bound than the 3p electron removed from Al³⁺.
2025/2026) — 30 High-Level Actual Questions with 100%
Correct Answers
1. Which set of quantum numbers (n, l, m , mₛ) is valid for an electron in a
ground-state arsenic atom?
A. 4, 2, −3, +½
B. 3, 1, 0, −½
C. 3, 3, 1, +½
D. 2, 0, 1, −½
Correct Answer: B
Explanation: l must be ≤ n − 1; m ranges from −l to +l. Only B satisfies all rules (n = 3,
l = 1, m = 0, mₛ = ±½).
2. The first ionization energy of magnesium is 738 kJ mol⁻¹. What is the minimum
frequency of light required to ionize a single Mg atom?
A. 1.84 × 10¹⁵ s⁻¹
B. 1.10 × 10¹⁵ s⁻¹
C. 5.51 × 10¹⁴ s⁻¹
D. 3.68 × 10¹⁴ s⁻¹
Correct Answer: C
Calculation: E = 738 kJ mol⁻¹ ÷ 6.022 × 10²³ = 1.225 × 10⁻¹⁸ J atom⁻¹; ν = E/h = 1.225 ×
10⁻¹⁸ J ÷ 6.626 × 10⁻³⁴ J·s = 5.51 × 10¹⁴ s⁻¹.
, 3. How many nodes (total) are present in a 4p orbital?
A. 2
B. 3
C. 4
D. 5
Correct Answer: B
Explanation: Total nodes = n − 1 = 3 (one radial, two angular).
4. Arrange the following species in order of increasing atomic radius: P³⁻, S²⁻, Cl⁻,
Ar.
A. Cl⁻ < S²⁻ < P³⁻ < Ar
B. Ar < Cl⁻ < S²⁻ < P³⁻
C. P³⁻ < S²⁻ < Cl⁻ < Ar
D. Ar < P³⁻ < S²⁻ < Cl⁻
Correct Answer: B
Explanation: All are isoelectronic (18 e⁻); radius increases as nuclear charge decreases
(Ar = 18, Cl⁻ = 17, S²⁻ = 16, P³⁻ = 15).
5. Which element has the largest third ionization energy?
A. Al
B. Si
C. Mg
D. Na
Correct Answer: C
Explanation: Mg loses two 3s electrons first; the third IE removes a 2p core electron,
which is much tighter bound than the 3p electron removed from Al³⁺.
