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Free Practice Questions for Castilla-La Mancha PAU Chemistry (Química - UCLM 2026)

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Key Facts: Castilla-La Mancha PAU Chemistry (Química - UCLM 2026) Exam

90 min

PAU Chemistry exam duration

Tribunal PAU Universidad de Castilla-La Mancha

0–10

Grading scale (minimum 4.0 in the PAU)

UCLM

5 Blocks

Thematic blocks of the official LOMLOE curriculum

UCLM 2026

100

Questions with answers and detailed solutions

OpenExamPrep

Prepare for the UCLM 2026 PAU Chemistry exam with 100 high-quality practice questions, featuring step-by-step quantitative calculations across all 5 official curriculum areas. This English-language multiple-choice bank is a study adaptation of the official written PAU paper, not a replica of it.

Sample Castilla-La Mancha PAU Chemistry (Química - UCLM 2026) Practice Questions

Try these sample questions to review concepts for the Castilla-La Mancha PAU Chemistry (Química - UCLM 2026) exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Which set of quantum numbers (n, l, ml, ms) is valid for the last electron of a ground-state nitrogen atom (Z = 7)?
A.(2, 1, +1, +1/2)
B.(2, 2, 0, +1/2)
C.(2, 0, +1, -1/2)
D.(1, 1, 0, +1/2)
Explanation: The ground-state electron configuration of nitrogen (Z = 7) is 1s² 2s² 2p³. The valence shell is n = 2. For the 2p subshell, the azimuthal quantum number is l = 1. The magnetic quantum number ml can take values -1, 0, +1, and the spin quantum number ms can be +1/2 or -1/2. Thus, (2, 1, +1, +1/2) is a valid set.
2What is the ground-state electron configuration of the Fe³⁺ ion (Z = 26)?
A.[Ar] 4s² 3d³
B.[Ar] 3d⁵
C.[Ar] 4s¹ 3d⁴
D.[Ar] 3d⁶
Explanation: A neutral iron atom (Z = 26) has the electron configuration [Ar] 4s² 3d⁶. When forming the Fe³⁺ cation, the two electrons in the outermost 4s orbital (n = 4) are lost first, followed by one electron from the 3d subshell, yielding the more stable half-filled d subshell configuration [Ar] 3d⁵.
3Calculate the energy of a photon of electromagnetic radiation with a wavelength λ = 500 nm in a vacuum. (Given: h = 6.626 · 10⁻³⁴ J·s, c = 3 · 10⁸ m/s).
A.1.33 · 10⁻²⁷ J
B.9.94 · 10⁻³¹ J
C.3.98 · 10⁻¹⁹ J
D.5.96 · 10⁻¹⁷ J
Explanation: Applying Planck's equation E = h·ν = h·c/λ with λ = 500 nm = 5.00 · 10⁻⁷ m: E = (6.626 · 10⁻³⁴ J·s · 3.00 · 10⁸ m/s) / (5.00 · 10⁻⁷ m) = 1.9878 · 10⁻²⁵ / 5.00 · 10⁻⁷ = 3.9756 · 10⁻¹⁹ J ≈ 3.98 · 10⁻¹⁹ J.
4Which of the following chemical species has the largest ionic radius?
A.Cl⁻
B.K⁺
C.Ca²⁺
D.S²⁻
Explanation: All four species (S²⁻, Cl⁻, K⁺, Ca²⁺) are isoelectronic with 18 electrons. In an isoelectronic series, ionic radius decreases as nuclear charge Z increases, because greater nuclear attraction pulls the electron cloud tighter. Sulfur (Z = 16) has the lowest nuclear charge, so its anion S²⁻ has the largest radius.
5Calculate the de Broglie wavelength associated with an electron (m = 9.11 · 10⁻³¹ kg) moving at a velocity v = 2.00 · 10⁶ m/s. (Given: h = 6.626 · 10⁻³⁴ J·s).
A.0.364 nm
B.1.21 nm
C.3.64 nm
D.0.036 nm
Explanation: The de Broglie wavelength is given by λ = h / (m·v). λ = (6.626 · 10⁻³⁴ J·s) / (9.11 · 10⁻³¹ kg · 2.00 · 10⁶ m/s) = (6.626 · 10⁻³⁴) / (1.822 · 10⁻²⁴) = 3.6366 · 10⁻¹⁰ m = 0.364 nm.
6According to the VSEPR model, what is the molecular geometry of the ammonia molecule (NH₃)?
A.Tetrahedral
B.Trigonal pyramidal
C.Trigonal planar
D.Linear
Explanation: The central N atom in NH₃ has 4 valence electron pairs (3 N-H bonding pairs and 1 non-bonding lone pair). The electron geometry is tetrahedral (AX₃E), but the molecular geometry (arrangement of the atoms) is trigonal pyramidal.
7Determine the formal charge of the central nitrogen atom in the nitrate ion (NO₃⁻), given that it is bonded to one oxygen via a double bond and to two oxygens via single bonds.
A.0
B.-1
C.+1
D.+2
Explanation: Formal charge is calculated as FC = V - N - B/2, where V is valence electrons (N has 5), N is non-bonding electrons (0 on central N of NO₃⁻), and B is shared bonding electrons (1 double bond + 2 single bonds = 4 bonds = 8 shared electrons). FC(N) = 5 - 0 - (8/2) = 5 - 4 = +1.
8Using the Born-Haber cycle, calculate the lattice energy (U) of NaCl(s) in kJ/mol given the following data: ΔH°f(NaCl,s) = -411 kJ/mol, ΔH°sub(Na,s) = +108 kJ/mol, E_ioniz(Na,g) = +496 kJ/mol, ΔH°dis(Cl₂,g) = +244 kJ/mol, A_electr(Cl,g) = -349 kJ/mol.
A.-910 kJ/mol
B.-666 kJ/mol
C.-544 kJ/mol
D.-788 kJ/mol
Explanation: The Born-Haber cycle states: ΔH°f = ΔH°sub(Na) + E_ioniz(Na) + (1/2)ΔH°dis(Cl₂) + A_electr(Cl) + U Substituting the values: -411 = 108 + 496 + (1/2)(244) + (-349) + U -411 = 108 + 496 + 122 - 349 + U -411 = 377 + U ⇒ U = -411 - 377 = -788 kJ/mol.
9What hybridization of the central atom and what net dipole moment are exhibited by the boron trifluoride molecule (BF₃)?
A.sp² hybridization, dipole moment μ = 0 (nonpolar)
B.sp³ hybridization, dipole moment μ ≠ 0 (polar)
C.sp hybridization, dipole moment μ = 0 (nonpolar)
D.sp²d hybridization, dipole moment μ ≠ 0 (polar)
Explanation: Boron in BF₃ forms three covalent bonds with no lone pairs (AX₃), adopting sp² hybridization with trigonal planar geometry and 120° bond angles. Due to molecular symmetry, the vector sum of the three individual C-F dipole moments yields a net zero dipole moment (μ = 0), making the molecule nonpolar.
10Arrange the following pure substances in order of increasing boiling point: CH₄, CH₃OH, CH₃CH₂OH, H₂O.
A.CH₄ < H₂O < CH₃OH < CH₃CH₂OH
B.CH₄ < CH₃OH < CH₃CH₂OH < H₂O
C.CH₃OH < CH₄ < CH₃CH₂OH < H₂O
D.H₂O < CH₃CH₂OH < CH₃OH < CH₄
Explanation: CH₄ exhibits only very weak London dispersion forces (bp -161°C). Alcohols form hydrogen bonds; CH₃CH₂OH has a larger molar mass and surface area than CH₃OH, increasing additional London dispersion forces (bp CH₃OH = 65°C, CH₃CH₂OH = 78°C). H₂O forms a 3D network of up to 4 hydrogen bonds per molecule (bp 100°C).

About the Castilla-La Mancha PAU Chemistry (Química - UCLM 2026) Exam

Comprehensive question bank for the Castilla-La Mancha PAU Chemistry exam (Química - UCLM 2026). Features 100 rigorous multiple-choice questions including extensive quantitative calculations (pH, stoichiometry, enthalpy, equilibrium constants, cell potential) and detailed explanations in Spanish.

Exam sponsor: University of Castilla-La Mancha PAU Tribunal. The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

The PAU Chemistry exam for Castilla-La Mancha (UCLM) tests 2º Bachillerato Chemistry competencies across 5 core areas: Atomic Structure & Bonding (20%), Thermodynamics & Kinetics (20%), Chemical Equilibrium (20%), Acid-Base Equilibria (20%), and Redox/Electrochemistry/Organic Chemistry (20%).

Time Limit

90 minutes (1.5 hours)

Passing Score

Marked on a 0-10 scale. Minimum 4.0 required in Access Phase to combine with Bachillerato GPA (60% Bachillerato + 40% PAU Access Phase >= 5.0 to pass).

Exam / Certification Fees

EUR 52.99 base registration fee for PAU Access Phase (ordinary sitting, set by Universidad de Castilla-La Mancha).

Exam sponsor website

Fees, eligibility, and exam policies can change. Confirm them with the exam sponsor before applying or paying.

Official sources

Our practice resources: topics covered

We aim to reflect publicly available exam outlines and topic information in our study resources. Coverage, format, and difficulty may differ from the actual exam, and we cannot guarantee that every detail is accurate or current. Confirm exam requirements, fees, and policies with the official exam sponsor.

20%

Atomic Structure, Periodic Table & Chemical Bonding

Quantum numbers, electron configurations, periodic trends, ionic/covalent/metallic bonding, VSEPR geometry, polarity, and intermolecular forces.

20%

Chemical Thermodynamics & Kinetics

Enthalpy of reaction, Hess's law, entropy, Gibbs free energy spontaneity, reaction rates, rate laws, collision theory, and catalysis.

20%

Chemical Equilibrium (Kc, Kp, Le Chatelier)

Homogeneous and heterogeneous equilibria, Kc and Kp relationships, degree of dissociation, Le Chatelier's principle, and solubility products (Ksp).

20%

Acid-Base Equilibria & pH Calculations

Brønsted-Lowry theory, pH/pOH calculations of strong and weak acids/bases, salt hydrolysis, buffer solutions, and acid-base titrations.

20%

Redox Reactions, Electrochemistry & Organic Chemistry

Balancing redox reactions, standard cell potential (E°cell), Nernst equation, electrolysis, IUPAC organic nomenclature, isomerism, and organic reaction mechanisms.

Preparing for the Castilla-La Mancha PAU Chemistry (Química - UCLM 2026) Exam

What You Need to Know

  • Passing score: Marked on a 0-10 scale. Minimum 4.0 required in Access Phase to combine with Bachillerato GPA (60% Bachillerato + 40% PAU Access Phase >= 5.0 to pass).
  • Assessment: The PAU Chemistry exam for Castilla-La Mancha (UCLM) tests 2º Bachillerato Chemistry competencies across 5 core areas: Atomic Structure & Bonding (20%), Thermodynamics & Kinetics (20%), Chemical Equilibrium (20%), Acid-Base Equilibria (20%), and Redox/Electrochemistry/Organic Chemistry (20%).
  • Time limit: 90 minutes (1.5 hours)
  • Exam / certification fees: EUR 52.99 base registration fee for PAU Access Phase (ordinary sitting, set by Universidad de Castilla-La Mancha). Official sources

Using Our Practice Resources

  • Work through all 100 available questions
  • Review every answer and explanation
  • Track weak areas and revisit them
  • Use our AI tutor for tough concepts

Castilla-La Mancha PAU Chemistry (Química - UCLM 2026): Suggested Study Strategy

1Review carefully the units and algebraic signs in thermochemistry calculations (ΔH, ΔS, and ΔG = ΔH - TΔS).
2Practice calculating pH for weak acids and bases using the acid dissociation constant Ka and the initial concentration approximation.
3Make sure you master the relationship between Kc and Kp (Kp = Kc · (RT)^Δn) and the law of mass action.
4Review balancing redox reactions by the ion-electron method in both acidic and basic media.
5Memorize the IUPAC nomenclature of organic functional groups and the identification of chain, position, functional isomers, and stereoisomers.

Frequently Asked Questions

What is the structure of the PAU Chemistry exam in Castilla-La Mancha (UCLM 2026)?

The exam consists of a 90-minute written test, weighted across 5 fundamental thematic blocks of the 2º Bachillerato curriculum.

How is the access mark calculated with the PAU Chemistry exam at UCLM?

The final Access Phase mark is calculated by combining 60% of the Bachillerato GPA and 40% of the PAU score, requiring at least 4.0 in the PAU to be averaged.

How many numerical calculation questions does this question bank include?

It includes at least 40 quantitative calculation problems explained step by step (pH, equilibrium constants Kc/Kp, Gibbs free energy change ΔG, cell potentials E°, and stoichiometry).

In which language are the question bank explanations presented?

All explanations and distractor analyses are rigorously written in Castilian (Spanish).

Is this practice bank in the same format as the real Castilla-La Mancha PAU Chemistry exam?

No. The official Castilla-La Mancha PAU Chemistry paper is a 90-minute written examination in Spanish consisting of open-ended, semi-constructed, and short-answer questions — not multiple choice. This bank is an English-language multiple-choice adaptation designed to test and reinforce the core concepts and skills of the 2º Bachillerato curriculum.