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Sample La Rioja PAU Chemistry Practice Questions

Try these sample questions to review concepts for the La Rioja PAU Chemistry 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, m_l, m_s) correctly describes the highest-energy valence electron in a ground-state sodium atom (Na, Z = 11)?
A.n = 3, l = 0, m_l = 0, m_s = +1/2
B.n = 3, l = 1, m_l = 0, m_s = +1/2
C.n = 2, l = 1, m_l = +1, m_s = -1/2
D.n = 3, l = 2, m_l = 0, m_s = +1/2
Explanation: The ground-state electronic configuration of sodium (Z = 11) is 1s² 2s² 2p⁶ 3s¹. The highest-energy valence electron occupies the 3s orbital, corresponding to principal quantum number n = 3, azimuthal quantum number l = 0 (s-orbital), magnetic quantum number m_l = 0, and spin quantum number m_s = +1/2.
2How does atomic radius change across Period 3 elements from sodium (Na) to chlorine (Cl), and what is the primary cause?
A.It increases because additional electron shells are filled.
B.It decreases because effective nuclear charge increases while core electron shielding remains nearly constant.
C.It remains constant because shielding by inner electrons completely cancels the increased nuclear charge.
D.It decreases because electron-electron repulsions force electrons closer to the nucleus.
Explanation: Across Period 3 (Na to Cl), protons are added to the nucleus while electrons are added to the same principal energy level (n = 3). Because core electron shielding remains approximately constant, the effective nuclear charge (Z_eff) increases, drawing the valence shell closer to the nucleus and decreasing atomic radius.
3What is the ground-state electronic configuration of copper (Cu, Z = 29)?
A.[Ar] 4s² 3d⁹
B.[Ar] 4s¹ 3d¹⁰
C.[Ar] 4s⁰ 3d¹¹
D.[Ar] 4s² 3d¹⁰
Explanation: Copper exhibits an anomalous electronic configuration due to the enhanced thermodynamic stability associated with a completely filled d-subshell. One electron is promoted from the 4s orbital to the 3d subshell, yielding [Ar] 3d¹⁰ 4s¹.
4The first ionization energy of nitrogen (N, Z = 7) is 1402 kJ/mol, whereas that of oxygen (O, Z = 8) is 1314 kJ/mol. Which statement accounts for this anomaly?
A.Nitrogen has a smaller atomic radius than oxygen.
B.Oxygen has a higher effective nuclear charge than nitrogen.
C.Nitrogen possesses a half-filled 2p subshell (2p³), providing extra exchange energy stability, while oxygen has paired electrons in a 2p orbital that experience inter-electron repulsion.
D.Nitrogen's 2s orbital is lower in energy than oxygen's 2s orbital.
Explanation: Nitrogen's valence configuration is 2s² 2p³, possessing a half-filled 2p subshell with parallel spins that maximizes exchange energy. Oxygen (2s² 2p⁴) has one doubly occupied 2p orbital. Electron-electron repulsion between the paired electrons in oxygen lowers the energy required to remove one electron, resulting in a lower first ionization energy for oxygen despite its higher Z.
5Which of the following orders correctly arranges the isoelectronic species S²⁻, Cl⁻, K⁺, and Ca²⁺ by decreasing ionic radius?
A.S²⁻ > Cl⁻ > K⁺ > Ca²⁺
B.Ca²⁺ > K⁺ > Cl⁻ > S²⁻
C.Cl⁻ > S²⁻ > Ca²⁺ > K⁺
D.K⁺ > Ca²⁺ > S²⁻ > Cl⁻
Explanation: All four species possess 18 electrons (isoelectronic with argon). For isoelectronic ions, ionic radius decreases as nuclear charge (Z) increases. S²⁻ (Z = 16) has the smallest Z and weakest pull on electrons (largest radius), followed by Cl⁻ (Z = 17), K⁺ (Z = 19), and Ca²⁺ (Z = 20, smallest radius).
6Calculate the energy of the photon emitted during an electronic transition from n = 3 to n = 1 in a hydrogen atom. (Rydberg constant R_H = 2.18 × 10⁻¹⁸ J)
A.1.94 × 10⁻¹⁸ J
B.2.42 × 10⁻¹⁹ J
C.1.45 × 10⁻¹⁸ J
D.4.84 × 10⁻¹⁹ J
Explanation: Using the Bohr/Rydberg formula ΔE = R_H (1/n_final² - 1/n_initial²): ΔE = 2.18 × 10⁻¹⁸ J × (1/1² - 1/3²) = 2.18 × 10⁻¹⁸ J × (1 - 1/9) = 2.18 × 10⁻¹⁸ J × (8/9) = 1.938 × 10⁻¹⁸ J ≈ 1.94 × 10⁻¹⁸ J.
7Using Slater's rules, what is the effective nuclear charge (Z_eff) experienced by a 3p valence electron in a chlorine atom (Z = 17)?
A.6.10
B.7.00
C.10.90
D.5.25
Explanation: Chlorine (Z = 17) configuration is (1s²)(2s² 2p⁸)(3s² 3p⁵). The target electron is in the (3s, 3p) group. Shielding constant S = (6 electrons in same group × 0.35) + (8 electrons in n-1 shell × 0.85) + (2 electrons in n-2 shell × 1.00) = (6 × 0.35) + (8 × 0.85) + (2 × 1.00) = 2.10 + 6.80 + 2.00 = 10.90. Thus Z_eff = Z - S = 17 - 10.90 = 6.10.
8What is the de Broglie wavelength of an electron (mass m = 9.11 × 10⁻³¹ kg) traveling at a velocity of 2.00 × 10⁶ m/s? (Planck's constant h = 6.626 × 10⁻³⁴ J·s)
A.0.364 nm
B.3.64 nm
C.0.182 nm
D.1.21 nm
Explanation: Using the de Broglie relation 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 = 3.637 × 10⁻¹⁰ m = 0.364 nm.
9Which set of quantum numbers represents a valid state for an electron in an atom?
A.n = 3, l = 2, m_l = -1, m_s = +1/2
B.n = 2, l = 2, m_l = 0, m_s = -1/2
C.n = 4, l = 1, m_l = +2, m_s = +1/2
D.n = 1, l = 0, m_l = 0, m_s = +1
Explanation: For n = 3, l can take values 0, 1, 2. When l = 2 (d-orbital), m_l can range from -l to +l (-2, -1, 0, +1, +2). The spin m_s can be +1/2 or -1/2. Thus n = 3, l = 2, m_l = -1, m_s = +1/2 is fully valid.
10The first four successive ionization energies (IE_1 to IE_4 in kJ/mol) of a Period 3 element are 738, 1450, 7730, and 10540. To which Group of the periodic table does this element belong?
A.Group 2 (Alkaline earth metals)
B.Group 1 (Alkali metals)
C.Group 13 (Boron group)
D.Group 14 (Carbon group)
Explanation: The ionization energies increase moderately from IE_1 (738) to IE_2 (1450), but there is a massive jump of 6280 kJ/mol between IE_2 and IE_3. This huge energy gap signifies that the third electron is being removed from a stable, filled inner noble-gas core. Therefore, the element has 2 valence electrons and belongs to Group 2 (magnesium).

About the La Rioja PAU Chemistry Exam

The La Rioja PAU Chemistry examination (Química, Universidad de La Rioja 2026) is the official standardized assessment for 2º Bachillerato science and technology students seeking university entrance in La Rioja. Grounded in the official LOMLOE curriculum, the exam evaluates proficiency across 8 key areas: Atomic Structure & Periodic Properties, Chemical Bonding & Intermolecular Forces, Thermodynamics & Thermochemistry, Chemical Kinetics, Chemical Equilibrium & Le Chatelier, Acid-Base Equilibrium & pH Calculations, Redox & Electrochemistry, and Organic Chemistry.

Exam sponsor: Universidad de La Rioja (UR) / Comisión Organizadora de la PAU de La Rioja. The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

Written 90-minute standardized exam. Practice items are a multiple-choice analytical adaptation of the official open/semi-constructed La Rioja PAU paper, providing comprehensive syllabus coverage.

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 >= 5.0 to pass).

Exam / Certification Fees

EUR 55.43 base registration fee for PAU Access Phase (set by Universidad de La Rioja / Gobierno de La Rioja).

Exam sponsor website

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

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.

12%

Atomic Structure & Periodic Properties

Quantum numbers, electron configurations, periodic trends (atomic/ionic radii, ionization energy, electron affinity, electronegativity), and atomic spectra.

13%

Chemical Bonding & Intermolecular Forces

Ionic lattice energy, Lewis structures, VSEPR geometry, orbital hybridization, polarity, metallic bonding, and intermolecular forces.

13%

Thermodynamics & Thermochemistry

Enthalpy of formation/combustion, Hess's Law, bond enthalpies, entropy, Gibbs free energy, and reaction spontaneity.

12%

Chemical Kinetics

Reaction rates, rate laws, reaction orders, Arrhenius equation, activation energy, collision theory, and catalysis.

13%

Chemical Equilibrium & Le Chatelier

Equilibrium constants (Kc, Kp), degree of dissociation, reaction quotient, Le Chatelier's principle, and solubility products (Ksp).

13%

Acid-Base Equilibrium & pH Calculations

Brønsted-Lowry theory, pH/pOH calculations, weak acids/bases, salt hydrolysis, buffer solutions, and volumetric titrations.

12%

Redox & Electrochemistry

Oxidation states, ion-electron balancing, standard reduction potentials, galvanic cells, Nernst equation, and Faraday's laws of electrolysis.

12%

Organic Chemistry: Functional Groups, Isomerism & Reactions

IUPAC nomenclature, functional groups, structural/stereoisomerism, organic reaction mechanisms (addition, substitution, elimination, oxidation, esterification).

Preparing for the La Rioja PAU Chemistry 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 >= 5.0 to pass).
  • Assessment: Written 90-minute standardized exam. Practice items are a multiple-choice analytical adaptation of the official open/semi-constructed La Rioja PAU paper, providing comprehensive syllabus coverage.
  • Time limit: 90 minutes (1.5 hours)
  • Exam / certification fees: EUR 55.43 base registration fee for PAU Access Phase (set by Universidad de La Rioja / Gobierno de La Rioja). 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

Frequently Asked Questions

Is this practice bank in the same format as the real La Rioja PAU exam?

No, and it is important to know the difference. The official La Rioja PAU paper is a 90-minute examination sat in Spanish featuring open-ended quantitative problems, written mechanisms, and conceptual explanations. Under the 2026 PAU design rules agreed by the Comisión Organizadora de la PAU de La Rioja and CRUE, open and semi-constructed responses must account for at least 70% of every paper, so the real exam contains no multiple-choice section. This bank is an English-language multiple-choice study adaptation of the same official 2º Bachillerato syllabus — not an official translation, not a past paper, and not a simulation of the exam format. Use it to drill the underlying knowledge and calculations quickly, then practise solving full open numerical problems in Spanish separately.

What is the format of the La Rioja PAU Chemistry exam?

It is a 90-minute written examination administered at Universidad de La Rioja designated centers, featuring stoichiometry problems, thermochemical calculations, equilibrium problems, acid-base titrations, redox balancing, and organic synthesis.

How is the exam scored for university entrance in Spain?

The exam is graded on a 0–10 scale. In the compulsory Access Phase, a minimum mark of 4.0 is required to combine with Bachillerato GPA (60% Bachillerato + 40% PAU Access Phase >= 5.0). In the Admission Phase, it serves as a high-weighting subject (0.2 factor) for health, chemistry, engineering, and science degrees.

Who should take this exam?

Students in the Science and Technology pathway of 2º Bachillerato in La Rioja who are applying to medicine, pharmacy, chemistry, biotechnology, chemical engineering, or related STEM degree programs.

Which body manages the PAU in La Rioja?

The exam is organized by the Universidad de La Rioja (UR) and the Comisión Organizadora de la PAU de La Rioja.