CHEM 210 Module 2 Exam (Portage Learning
2025/2026) — 30 High-Level Actual Questions with
100% Correct Answers
Which quantum-numbers set (n, l, m , mₛ) is permissible for an electron in a
ground-state arsenic atom?
1. A. 3, 3, 0, +½
B. 4, 1, –2, –½
C. 3, 2, –1, +½
D. 2, 2, 1, –½
Correct Answer: C
Explanation: l must be ≤ n – 1; for n = 3, l = 2 (d orbital) is allowed, |m | ≤ l, and mₛ =
±½.
The first ionization energy of aluminum is 578 kJ mol⁻¹. How much energy (kJ) is
required to ionize 5.00 g of Al atoms?
2. A. 107
B. 214
C. 53.5
D. 1.07 × 10³
Correct Answer: A
Calculation: (5.00 g / 26.98 g mol⁻¹) × 578 kJ mol⁻¹ ≈ 107 kJ.
Arrange the following species in order of increasing atomic radius: Mg²⁺, Na⁺, Ne, F⁻.
3. A. Mg²⁺ < Na⁺ < Ne < F⁻
B. F⁻ < Ne < Na⁺ < Mg²⁺
C. Ne < Mg²⁺ < Na⁺ < F⁻
D. Mg²⁺ < Na⁺ < F⁻ < Ne
Correct Answer: A
, Explanation: Iso-electronic with 10 e⁻; radius increases as nuclear charge decreases (Mg²⁺
12+ → F⁻ 9+).
How many nodes (total) does a 4d orbital possess?
4. A. 3
B. 4
C. 2
D. 1
Correct Answer: B
Explanation: Total nodes = n – 1 = 4 – 1 = 3; of these, 2 are angular (l = 2) and 1 is radial
→ 3 total (question asks total, not radial).
Which element has the largest third ionization energy?
5. A. Al
B. Si
C. Mg
D. Na
Correct Answer: C
Explanation: Removing the third electron from Mg breaks the stable [Ne]3s²
configuration → large jump.
The electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ represents a neutral
atom of:
6. A. Xe
B. Cs
C. Ba
D. Rn
Correct Answer: A
Explanation: 54 electrons → Xe (Z = 54).
2025/2026) — 30 High-Level Actual Questions with
100% Correct Answers
Which quantum-numbers set (n, l, m , mₛ) is permissible for an electron in a
ground-state arsenic atom?
1. A. 3, 3, 0, +½
B. 4, 1, –2, –½
C. 3, 2, –1, +½
D. 2, 2, 1, –½
Correct Answer: C
Explanation: l must be ≤ n – 1; for n = 3, l = 2 (d orbital) is allowed, |m | ≤ l, and mₛ =
±½.
The first ionization energy of aluminum is 578 kJ mol⁻¹. How much energy (kJ) is
required to ionize 5.00 g of Al atoms?
2. A. 107
B. 214
C. 53.5
D. 1.07 × 10³
Correct Answer: A
Calculation: (5.00 g / 26.98 g mol⁻¹) × 578 kJ mol⁻¹ ≈ 107 kJ.
Arrange the following species in order of increasing atomic radius: Mg²⁺, Na⁺, Ne, F⁻.
3. A. Mg²⁺ < Na⁺ < Ne < F⁻
B. F⁻ < Ne < Na⁺ < Mg²⁺
C. Ne < Mg²⁺ < Na⁺ < F⁻
D. Mg²⁺ < Na⁺ < F⁻ < Ne
Correct Answer: A
, Explanation: Iso-electronic with 10 e⁻; radius increases as nuclear charge decreases (Mg²⁺
12+ → F⁻ 9+).
How many nodes (total) does a 4d orbital possess?
4. A. 3
B. 4
C. 2
D. 1
Correct Answer: B
Explanation: Total nodes = n – 1 = 4 – 1 = 3; of these, 2 are angular (l = 2) and 1 is radial
→ 3 total (question asks total, not radial).
Which element has the largest third ionization energy?
5. A. Al
B. Si
C. Mg
D. Na
Correct Answer: C
Explanation: Removing the third electron from Mg breaks the stable [Ne]3s²
configuration → large jump.
The electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ represents a neutral
atom of:
6. A. Xe
B. Cs
C. Ba
D. Rn
Correct Answer: A
Explanation: 54 electrons → Xe (Z = 54).
