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2026 Statistics
Key Facts: Madrid PAU Chemistry Exam
90 Minutes
Exam duration
PAU Madrid Commission
EUR 93.02
Base PAU Access Phase registration fee
BOCM Madrid 2026
0–10 Scale
Grading scale (Min 4.0 required in Access Phase)
LOMLOE / Madrid PAU Regulations
5 Core Blocks
Curriculum blocks assessed
2º Bachillerato LOMLOE Chemistry Syllabus
100 Questions
Practice question bank size
OpenExamPrep
Madrid PAU Chemistry 2026 is a 90-minute university entrance exam assessing 2º Bachillerato Chemistry across 5 core curriculum blocks.
Sample Madrid PAU Chemistry Practice Questions
Try these sample questions to test your Madrid PAU Chemistry exam readiness. 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 an electron occupying a 3d orbital?
A.(3, 2, -1, +1/2)
B.(3, 1, -2, +1/2)
C.(3, 3, 0, -1/2)
D.(2, 2, +1, +1/2)
Explanation: For a 3d orbital, the principal quantum number is n = 3 and the angular momentum quantum number is l = 2 (where l = 0 for s, 1 for p, 2 for d). The magnetic quantum number ml can take integer values from -l to +l (i.e., -2, -1, 0, +1, +2), and ms can be +1/2 or -1/2. Therefore, (3, 2, -1, +1/2) is completely valid.
2What is the ground-state electron configuration of a neutral copper atom (Cu, Z = 29)?
A.[Ar] 4s¹ 3d¹⁰
B.[Ar] 4s² 3d⁹
C.[Ar] 4s⁰ 3d¹¹
D.[Ar] 4s² 3d¹⁰
Explanation: Copper is a well-known exception to the Aufbau principle. To achieve the enhanced stability of a fully filled d subshell (3d¹⁰), one electron is promoted from the 4s orbital to the 3d subshell, yielding [Ar] 4s¹ 3d¹⁰.
3Which of the following Period 3 elements possesses the highest first ionization energy (I₁)?
A.Chlorine (Cl)
B.Phosphorus (P)
C.Magnesium (Mg)
D.Sodium (Na)
Explanation: First ionization energy generally increases across a period from left to right due to increasing effective nuclear charge (Zeff) and decreasing atomic radius. Among the given Period 3 elements, chlorine (Cl, Group 17) is furthest to the right and has the highest I₁.
4Calculate the de Broglie wavelength of an electron (me = 9.11 × 10⁻³¹ kg) moving at a velocity of 2.00 × 10⁶ m/s. (Planck 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: According to the de Broglie relation λ = h / (m·v), substituting the given values yields λ = 6.626 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 2.00 × 10⁶) = 6.626 × 10⁻³⁴ / (1.822 × 10⁻²⁴) = 3.636 × 10⁻¹⁰ m = 0.364 nm.
5What is the energy of a single photon of violet light with a wavelength of 400 nm? (c = 3.00 × 10⁸ m/s, h = 6.626 × 10⁻³⁴ J·s)
A.4.97 × 10⁻¹⁹ J
B.2.48 × 10⁻¹⁹ J
C.7.95 × 10⁻¹⁹ J
D.1.66 × 10⁻²⁷ J
Explanation: Photon energy is calculated using E = h·c / λ. Converting λ = 400 nm = 4.00 × 10⁻⁷ m gives E = (6.626 × 10⁻³⁴ × 3.00 × 10⁸) / (4.00 × 10⁻⁷) = 1.9878 × 10⁻²⁵ / (4.00 × 10⁻⁷) = 4.97 × 10⁻¹⁹ J.
6In the photoelectric effect, photoelectrons are ejected from a metal surface only when the incident light frequency is above a threshold frequency ν₀. What does h·ν₀ represent?
A.The work function (Φ) of the metal
B.The maximum kinetic energy of ejected electrons
C.The ionization energy of a free gaseous atom
D.The stopping potential of the photoelectric cell
Explanation: The quantity h·ν₀ represents the minimum energy required to liberate a bound electron from the metal surface, which is defined as the work function (Φ) of the metal.
7Among the isoelectronic species S²⁻, Cl⁻, K⁺, and Ca²⁺, which has the smallest ionic radius?
A.Ca²⁺
B.K⁺
C.Cl⁻
D.S²⁻
Explanation: All four species possess 18 electrons (isoelectronic with argon). As nuclear charge (Z) increases (S=16, Cl=17, K=19, Ca=20), the electrostatic attraction between the nucleus and the electron cloud increases, pulling the electrons closer and reducing the ionic radius. Thus, Ca²⁺ (Z = 20) has the smallest radius.
8Using a Born-Haber cycle for NaCl(s), calculate its lattice energy (U_lat) given: ΔHf°[NaCl(s)] = -411 kJ/mol, ΔHsub[Na(s)] = +108 kJ/mol, 1/2 D[Cl₂(g)] = +122 kJ/mol, IE₁[Na(g)] = +496 kJ/mol, and EA[Cl(g)] = -349 kJ/mol.
A.-788 kJ/mol
B.-654 kJ/mol
C.-411 kJ/mol
D.-936 kJ/mol
Explanation: Applying Hess's law to the Born-Haber cycle: ΔHf° = ΔHsub + 1/2 D + IE₁ + EA + U_lat. Substituting numbers: -411 = 108 + 122 + 496 - 349 + U_lat => -411 = 377 + U_lat => U_lat = -411 - 377 = -788 kJ/mol.
9What is the molecular geometry of sulfur tetrafluoride (SF₄) according to VSEPR theory?
A.See-saw (distorted tetrahedral)
B.Square planar
C.Tetrahedral
D.Trigonal bipyramidal
Explanation: Sulfur in SF₄ has 6 valence electrons, forming 4 single bonds with fluorine and retaining 1 lone pair. This gives 5 electron pairs (AX₄E notation), arranging electron pairs in a trigonal bipyramid. The lone pair occupies an equatorial position to minimize repulsions, producing a see-saw molecular shape.
10Xenon tetrafluoride (XeF₄) possesses 6 electron domains around the central xenon atom. What are its molecular geometry and molecular polarity?
A.Square planar and nonpolar
B.Square pyramidal and polar
C.Octahedral and polar
D.Tetrahedral and nonpolar
Explanation: XeF₄ has 4 Xe-F bonding pairs and 2 lone pairs on Xe (AX₄E₂ notation). The two lone pairs occupy trans positions in an octahedral arrangement to minimize repulsion, yielding a square planar molecular geometry. The four Xe-F bond dipoles cancel out symmetrically, making the molecule nonpolar.
About the Madrid PAU Chemistry Practice Questions
Verified exam format metadata for Madrid PAU Chemistry Examination 2026 (Química 2º Bachillerato) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.