UCLA CHEMISTRY 14A MIDTERM
EXAM 2026
149 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
UCLA CHEMISTRY 14A MIDTERM EXAM 2026. It contains 149 carefully selected questions that reflect the most
current exam content and testing strategies. Each question is accompanied by a correct answer and a detailed
rationale that explains the underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 149 Questions
Foundations - Application - UCLA Chemistry 14a 2026 Chemistry 14a Atomic Structure Bonding
Stoichiometry Gases Thermochemistry Solutions Undergraduate YEAR 1 General Chemistry FOR LIFE
Sciences
All answers with rationales
,Table of Contents
Content Area Questions Key Topics
Atomic Structure AND 1-25 Sample, Electron, Polar, Element, Ground-state
Periodicity
Chemical Bonding AND 26-50 Reaction, Electron, Point, Polar, Correctly Describes
Molecular Geometry
Stoichiometry AND Chemical 51-75 Solution, Electron, Energy, Molecules, Sample
Reactions
Gases AND GAS LAWS 76-100 Reaction, Electron, Sample, Buffer, Statements
Thermochemistry AND 101-125 Temperature, Electron, Reaction, Equilibrium, Point
Enthalpy
Solutions AND Concentration 126-149 Reaction, Electron, Polar, Statements, Aqueous
TOTAL 149 All questions include answers and detailed rationales
,Section A - Atomic Structure AND Periodicity
Q1.
An element has the ground-state electron configuration [Ar] 4s² 3d¹ 4p³. Which statement
about this element is correct?
A. It is a transition metal with variable B. It is a main-group metalloid in Group 15
oxidation states. (VA).
C. It is an alkali metal that forms a 1+ cation. D. It is a halogen that forms a 1 anion.
Correct: B - It is a main-group metalloid in Group 15 (VA).
Rationale:The configuration ends in 4p³, placing the element in Group 15 (As, arsenic), a
metalloid. The filled 3d subshell is core-like and does not make it a transition metal; alkali
metals and halogens have s¹ and p valence configurations, respectively.
Why the other answers are wrong:
A. The 3d subshell is full and not valence; the element behaves as a main-group p-block
element.
C. Alkali metals have an ns¹ valence configuration, not np³.
D. Halogens have an np valence configuration, not np³.
Reference: Zumdahl & DeCoste, Chemistry, 10th Ed., Ch. 7
Q2.
For the reaction N(g) + 3H(g) -> 2NH(g), a 2.0 L flask initially contains 1.0 mol N and 3.0
mol H. If 0.60 mol NH is present at equilibrium, what is the equilibrium concentration of H?
A. 1.05 M B. 0.60 M
C. 1.20 M D. 0.90 M
Correct: A - 1.05 M
Rationale:0.60 mol NH ƒ forms from 0.90 mol H ‚ consumed (3:2 ratio), leaving 3.0 " 0.90 =
2.10 mol H in 2.0 L = 1.05 M. Distractors arise from incorrect stoichiometric ratios or
forgetting the volume.
Why the other answers are wrong:
B. Uses 1:1 H:NH stoichiometry instead of 3:2.
C. Fails to subtract consumed H or ignores volume.
D. Uses 0.60 mol H consumed rather than 0.90 mol.
Reference: Brown et al., Chemistry: The Central Science, 15th Ed., Ch. 15
Page 3
, Section A - Atomic Structure AND Periodicity
Q3.
Which of the following best explains why the first ionization energy of nitrogen is greater
than that of oxygen?
A. Nitrogen has a larger atomic radius, B. Oxygen's 2p electrons experience greater
weakening electron attraction. electron-electron repulsion due to pairing.
C. Nitrogen has a half-filled 2p subshell, D. Oxygen has a higher effective nuclear
which is anomalously stable. charge, which should increase its ionization
energy.
Correct: C - Nitrogen has a half-filled 2p subshell, which is anomalously stable.
Rationale:Nitrogen's 2p³ half-filled subshell has exchange-energy stabilization and minimized
pairing repulsion, making electron removal harder than from oxygen's 2p. Oxygen's extra
paired electron actually lowers its IE relative to N despite higher Z_eff.
Why the other answers are wrong:
A. Atomic radius alone does not explain the N > O anomaly; N is actually smaller.
B. Pairing repulsion in O is a contributing factor but the half-filled stability of N is the standard
explanation.
D. Higher Z_eff would predict O > N, opposite to observation.
Reference: Zumdahl & DeCoste, Chemistry, 10th Ed., Ch. 7
Q4.
Which molecule is polar despite having polar bonds?
A. CO B. CCl
C. BF D. CHCl
Correct: D - CHCl
Rationale:CH ‚Cl ‚ is tetrahedral but not symmetric: the C–H and C–Cl bond dipoles do not
cancel, giving a net dipole. CO (linear), CCl (tetrahedral), and BF (trigonal planar) have
symmetric arrangements that cancel bond dipoles.
Why the other answers are wrong:
A. Linear geometry makes the two C=O dipoles cancel.
B. Tetrahedral symmetry cancels the four C-Cl dipoles.
C. Trigonal planar symmetry cancels the three B-F dipoles.
Reference: Brown et al., Chemistry: The Central Science, 15th Ed., Ch. 8
Q5.
A 25.0 mL sample of 0.200 M HCl is titrated with 0.100 M NaOH. What is the pH after 25.0
mL of NaOH has been added?
Page 4
EXAM 2026
149 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
UCLA CHEMISTRY 14A MIDTERM EXAM 2026. It contains 149 carefully selected questions that reflect the most
current exam content and testing strategies. Each question is accompanied by a correct answer and a detailed
rationale that explains the underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 149 Questions
Foundations - Application - UCLA Chemistry 14a 2026 Chemistry 14a Atomic Structure Bonding
Stoichiometry Gases Thermochemistry Solutions Undergraduate YEAR 1 General Chemistry FOR LIFE
Sciences
All answers with rationales
,Table of Contents
Content Area Questions Key Topics
Atomic Structure AND 1-25 Sample, Electron, Polar, Element, Ground-state
Periodicity
Chemical Bonding AND 26-50 Reaction, Electron, Point, Polar, Correctly Describes
Molecular Geometry
Stoichiometry AND Chemical 51-75 Solution, Electron, Energy, Molecules, Sample
Reactions
Gases AND GAS LAWS 76-100 Reaction, Electron, Sample, Buffer, Statements
Thermochemistry AND 101-125 Temperature, Electron, Reaction, Equilibrium, Point
Enthalpy
Solutions AND Concentration 126-149 Reaction, Electron, Polar, Statements, Aqueous
TOTAL 149 All questions include answers and detailed rationales
,Section A - Atomic Structure AND Periodicity
Q1.
An element has the ground-state electron configuration [Ar] 4s² 3d¹ 4p³. Which statement
about this element is correct?
A. It is a transition metal with variable B. It is a main-group metalloid in Group 15
oxidation states. (VA).
C. It is an alkali metal that forms a 1+ cation. D. It is a halogen that forms a 1 anion.
Correct: B - It is a main-group metalloid in Group 15 (VA).
Rationale:The configuration ends in 4p³, placing the element in Group 15 (As, arsenic), a
metalloid. The filled 3d subshell is core-like and does not make it a transition metal; alkali
metals and halogens have s¹ and p valence configurations, respectively.
Why the other answers are wrong:
A. The 3d subshell is full and not valence; the element behaves as a main-group p-block
element.
C. Alkali metals have an ns¹ valence configuration, not np³.
D. Halogens have an np valence configuration, not np³.
Reference: Zumdahl & DeCoste, Chemistry, 10th Ed., Ch. 7
Q2.
For the reaction N(g) + 3H(g) -> 2NH(g), a 2.0 L flask initially contains 1.0 mol N and 3.0
mol H. If 0.60 mol NH is present at equilibrium, what is the equilibrium concentration of H?
A. 1.05 M B. 0.60 M
C. 1.20 M D. 0.90 M
Correct: A - 1.05 M
Rationale:0.60 mol NH ƒ forms from 0.90 mol H ‚ consumed (3:2 ratio), leaving 3.0 " 0.90 =
2.10 mol H in 2.0 L = 1.05 M. Distractors arise from incorrect stoichiometric ratios or
forgetting the volume.
Why the other answers are wrong:
B. Uses 1:1 H:NH stoichiometry instead of 3:2.
C. Fails to subtract consumed H or ignores volume.
D. Uses 0.60 mol H consumed rather than 0.90 mol.
Reference: Brown et al., Chemistry: The Central Science, 15th Ed., Ch. 15
Page 3
, Section A - Atomic Structure AND Periodicity
Q3.
Which of the following best explains why the first ionization energy of nitrogen is greater
than that of oxygen?
A. Nitrogen has a larger atomic radius, B. Oxygen's 2p electrons experience greater
weakening electron attraction. electron-electron repulsion due to pairing.
C. Nitrogen has a half-filled 2p subshell, D. Oxygen has a higher effective nuclear
which is anomalously stable. charge, which should increase its ionization
energy.
Correct: C - Nitrogen has a half-filled 2p subshell, which is anomalously stable.
Rationale:Nitrogen's 2p³ half-filled subshell has exchange-energy stabilization and minimized
pairing repulsion, making electron removal harder than from oxygen's 2p. Oxygen's extra
paired electron actually lowers its IE relative to N despite higher Z_eff.
Why the other answers are wrong:
A. Atomic radius alone does not explain the N > O anomaly; N is actually smaller.
B. Pairing repulsion in O is a contributing factor but the half-filled stability of N is the standard
explanation.
D. Higher Z_eff would predict O > N, opposite to observation.
Reference: Zumdahl & DeCoste, Chemistry, 10th Ed., Ch. 7
Q4.
Which molecule is polar despite having polar bonds?
A. CO B. CCl
C. BF D. CHCl
Correct: D - CHCl
Rationale:CH ‚Cl ‚ is tetrahedral but not symmetric: the C–H and C–Cl bond dipoles do not
cancel, giving a net dipole. CO (linear), CCl (tetrahedral), and BF (trigonal planar) have
symmetric arrangements that cancel bond dipoles.
Why the other answers are wrong:
A. Linear geometry makes the two C=O dipoles cancel.
B. Tetrahedral symmetry cancels the four C-Cl dipoles.
C. Trigonal planar symmetry cancels the three B-F dipoles.
Reference: Brown et al., Chemistry: The Central Science, 15th Ed., Ch. 8
Q5.
A 25.0 mL sample of 0.200 M HCl is titrated with 0.100 M NaOH. What is the pH after 25.0
mL of NaOH has been added?
Page 4
