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Inorganic & Physical Chemistry | 2025/ & Mark Scheme for Exam
Success | Questions & Answers (Verified Answers) With Rationales (2026 /
2027 Update)
This Document Contains:
40 Questions with Correct, Detailed and Verified Answers
2026/2027 Actual Exam Testbank
Questions & Answers (Verified Answers) With Rationales
100% Guaranteed Pass
Complete A+ Guide
Page 1
,Question 1
Consider the first ionization energies (IE1) of elements in period 3. Which of the following
statements best explains why the IE1 of sulfur is lower than that of phosphorus?
A) Sulfur has a larger atomic radius, so its outer electron experiences less nuclear attraction.
B) In sulfur, the electron removed is from a doubly occupied p orbital, resulting in greater
electron-electron repulsion.
C) Phosphorus has a half-filled p subshell, which is more stable than the configuration of sulfur.
D) The effective nuclear charge of sulfur is lower than that of phosphorus due to increased shielding.
Answer: B) In sulfur, the electron removed is from a doubly occupied p orbital, resulting in greater
electron-electron repulsion.
Explanation: The correct answer is B. In sulfur, the electron removed comes from a p orbital that
contains two electrons, leading to greater repulsion and a lower ionization energy
compared to phosphorus, where the electron is removed from a half-filled p orbital
(more stable). Option A is incorrect because atomic radius decreases across a period, so
sulfur is smaller than phosphorus. Option C is a common misconception: while the
half-filled subshell does provide extra stability, the direct reason for the drop is the
pairing energy. Option D is false because effective nuclear charge increases across a
period.
Question 2
For the reaction 2NO(g) + O2(g) !Ì 2NO2(g), the equilibrium constant Kc = 4.5 × 10^2 at 500 K. If
0.50 mol of NO and 0.30 mol of O2 are placed in a 1.0 L container at 500 K and allowed to reach
equilibrium, what is the equilibrium concentration of NO2?
A) 0.20 M
B) 0.40 M
C) 0.60 M
D) 0.80 M
Answer: B) 0.40 M
Explanation: Let x be the change in concentration of NO2. Initial: [NO] = 0.50 M, [O2] = 0.30 M,
[NO2] = 0. At equilibrium: [NO] = 0.50 - x, [O2] = 0.30 - x/2, [NO2] = x. Kc = x^2 /
[(0.50 - x)^2 (0.30 - x/2)] = 450. Solving (using approximation or quadratic) yields x "H
0.40 M. Option A (0.20) underestimates; option C (0.60) exceeds initial NO; option D
(0.80) is impossible.
Page 2
, Question 3
Which of the following pairs of compounds, when mixed in aqueous solution, will result in the
formation of a precipitate?
A) KCl and NaNO3
B) Ba(OH)2 and HNO3
C) AgNO3 and NaBr
D) NH4Cl and NaOH
Answer: C) AgNO3 and NaBr
Explanation: AgNO3 and NaBr produce AgBr, which is insoluble. Option A: all ions remain in
solution. Option B: acid-base neutralization produces soluble salts. Option D: ammonia
gas may form but no precipitate.
Question 4
The enthalpy change for the reaction C(s) + O2(g) !’ CO2(g) is -393.5 kJ/mol. Given that the
standard enthalpy of formation of CO(g) is -110.5 kJ/mol, calculate the enthalpy change for the
reaction 2C(s) + O2(g) !’ 2CO(g).
A) -110.5 kJ/mol
B) -221.0 kJ/mol
C) -283.0 kJ/mol
D) -566.0 kJ/mol
Answer: B) -221.0 kJ/mol
Explanation: From the given data: ”Hf°(CO2) = -393.5 kJ/mol. For the reaction C(s) + 1/2 O2(g) !’
CO(g), ”H = ”Hf°(CO) = -110.5 kJ/mol. Therefore, for 2 moles of CO: 2 × (-110.5) =
-221.0 kJ. Option A is per mole of CO, not for 2 moles. Options C and D are incorrect
because they involve CO2 formation.
Question 5
In the Born-Haber cycle for MgCl2, which of the following quantities is the largest in magnitude
(most exothermic or most endothermic)?
A) First electron affinity of chlorine
B) Second ionization energy of magnesium
C) Lattice enthalpy of MgCl2
D) Enthalpy of atomization of magnesium
Answer: C) Lattice enthalpy of MgCl2
Explanation: Lattice enthalpy is highly exothermic due to strong electrostatic forces between Mg2+
and Cl- ions. The second ionization energy of Mg is endothermic and large, but lattice
enthalpy is typically much larger in magnitude (exothermic). Electron affinity of Cl is
exothermic but small. Atomization of Mg is endothermic and moderate.
Page 3
Download Download
EXAMS
Inorganic & Physical Chemistry | 2025/ & Mark Scheme for Exam
Success | Questions & Answers (Verified Answers) With Rationales (2026 /
2027 Update)
This Document Contains:
40 Questions with Correct, Detailed and Verified Answers
2026/2027 Actual Exam Testbank
Questions & Answers (Verified Answers) With Rationales
100% Guaranteed Pass
Complete A+ Guide
Page 1
,Question 1
Consider the first ionization energies (IE1) of elements in period 3. Which of the following
statements best explains why the IE1 of sulfur is lower than that of phosphorus?
A) Sulfur has a larger atomic radius, so its outer electron experiences less nuclear attraction.
B) In sulfur, the electron removed is from a doubly occupied p orbital, resulting in greater
electron-electron repulsion.
C) Phosphorus has a half-filled p subshell, which is more stable than the configuration of sulfur.
D) The effective nuclear charge of sulfur is lower than that of phosphorus due to increased shielding.
Answer: B) In sulfur, the electron removed is from a doubly occupied p orbital, resulting in greater
electron-electron repulsion.
Explanation: The correct answer is B. In sulfur, the electron removed comes from a p orbital that
contains two electrons, leading to greater repulsion and a lower ionization energy
compared to phosphorus, where the electron is removed from a half-filled p orbital
(more stable). Option A is incorrect because atomic radius decreases across a period, so
sulfur is smaller than phosphorus. Option C is a common misconception: while the
half-filled subshell does provide extra stability, the direct reason for the drop is the
pairing energy. Option D is false because effective nuclear charge increases across a
period.
Question 2
For the reaction 2NO(g) + O2(g) !Ì 2NO2(g), the equilibrium constant Kc = 4.5 × 10^2 at 500 K. If
0.50 mol of NO and 0.30 mol of O2 are placed in a 1.0 L container at 500 K and allowed to reach
equilibrium, what is the equilibrium concentration of NO2?
A) 0.20 M
B) 0.40 M
C) 0.60 M
D) 0.80 M
Answer: B) 0.40 M
Explanation: Let x be the change in concentration of NO2. Initial: [NO] = 0.50 M, [O2] = 0.30 M,
[NO2] = 0. At equilibrium: [NO] = 0.50 - x, [O2] = 0.30 - x/2, [NO2] = x. Kc = x^2 /
[(0.50 - x)^2 (0.30 - x/2)] = 450. Solving (using approximation or quadratic) yields x "H
0.40 M. Option A (0.20) underestimates; option C (0.60) exceeds initial NO; option D
(0.80) is impossible.
Page 2
, Question 3
Which of the following pairs of compounds, when mixed in aqueous solution, will result in the
formation of a precipitate?
A) KCl and NaNO3
B) Ba(OH)2 and HNO3
C) AgNO3 and NaBr
D) NH4Cl and NaOH
Answer: C) AgNO3 and NaBr
Explanation: AgNO3 and NaBr produce AgBr, which is insoluble. Option A: all ions remain in
solution. Option B: acid-base neutralization produces soluble salts. Option D: ammonia
gas may form but no precipitate.
Question 4
The enthalpy change for the reaction C(s) + O2(g) !’ CO2(g) is -393.5 kJ/mol. Given that the
standard enthalpy of formation of CO(g) is -110.5 kJ/mol, calculate the enthalpy change for the
reaction 2C(s) + O2(g) !’ 2CO(g).
A) -110.5 kJ/mol
B) -221.0 kJ/mol
C) -283.0 kJ/mol
D) -566.0 kJ/mol
Answer: B) -221.0 kJ/mol
Explanation: From the given data: ”Hf°(CO2) = -393.5 kJ/mol. For the reaction C(s) + 1/2 O2(g) !’
CO(g), ”H = ”Hf°(CO) = -110.5 kJ/mol. Therefore, for 2 moles of CO: 2 × (-110.5) =
-221.0 kJ. Option A is per mole of CO, not for 2 moles. Options C and D are incorrect
because they involve CO2 formation.
Question 5
In the Born-Haber cycle for MgCl2, which of the following quantities is the largest in magnitude
(most exothermic or most endothermic)?
A) First electron affinity of chlorine
B) Second ionization energy of magnesium
C) Lattice enthalpy of MgCl2
D) Enthalpy of atomization of magnesium
Answer: C) Lattice enthalpy of MgCl2
Explanation: Lattice enthalpy is highly exothermic due to strong electrostatic forces between Mg2+
and Cl- ions. The second ionization energy of Mg is endothermic and large, but lattice
enthalpy is typically much larger in magnitude (exothermic). Electron affinity of Cl is
exothermic but small. Atomization of Mg is endothermic and moderate.
Page 3
