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100+ Free Catalonia PAU Chemistry Practice Questions
Catalonia PAU Chemistry Examination — CIC / Generalitat de Catalunya (Química 2n Batxillerat 2026) practice questions are available now; exam metadata is being verified.
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2026 Statistics
Key Facts: Catalonia PAU Chemistry Exam
EUR 110.00+ access phase (EUR 41.30 exam right + EUR 68.70 access phase; plus EUR 13.80 per admission exercise, Generalitat de Catalunya 2026)
Base registration fee set by Generalitat de Catalunya
Generalitat de Catalunya PAU Rules
90 Mins
Official examination duration
Consell Interuniversitari de Catalunya (CIC)
Min 4.0
Minimum Access Phase score required to combine with Bachillerato GPA
CIC Catalonia PAU Guidelines
7 Blocks
Core chemistry syllabus units
2n Batxillerat LOMLOE Chemistry Curriculum
100
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Sample Catalonia PAU Chemistry Practice Questions
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1Which set of quantum numbers (n, l, ml, ms) represents a valid valence electron in the highest-energy populated orbital of a ground-state chlorine atom (Z = 17)?
A.(3, 1, 0, +1/2)
B.(3, 2, 0, -1/2)
C.(2, 1, -1, +1/2)
D.(3, 0, +1, -1/2)
Explanation: Chlorine (Z = 17) has the ground-state electron configuration 1s² 2s² 2p⁶ 3s² 3p⁵. The highest-energy valence electrons reside in the 3p subshell, corresponding to principal quantum number n = 3 and azimuthal quantum number l = 1. For l = 1, magnetic quantum number ml can be -1, 0, or +1, and spin quantum number ms can be +1/2 or -1/2, making (3, 1, 0, +1/2) valid.
2What is the correct ground-state electron configuration of a neutral chromium atom (Z = 24)?
A.[Ar] 4s² 3d⁴
B.[Ar] 3d⁵ 4s¹
C.[Ar] 4s² 3d⁵
D.[Ar] 3d⁶
Explanation: Chromium (Z = 24) exhibits an anomaly to the standard Aufbau ordering because transferring one electron from the 4s orbital to the 3d subshell creates a half-filled 3d subshell ([Ar] 3d⁵ 4s¹), which confers extra exchange stability.
3How does atomic radius vary across Period 3 of the periodic table from Sodium (Na, Z = 11) to Chlorine (Cl, Z = 17)?
A.Atomic radius increases because additional electrons increase electron-electron repulsion.
B.Atomic radius remains constant because electrons are added to the same principal shell n = 3.
C.Atomic radius decreases because increasing nuclear charge increases effective nuclear charge (Zeff), pulling valence shell electrons closer.
D.Atomic radius decreases initially then increases sharp at Phosphorus due to subshell half-filling.
Explanation: Across Period 3, atomic radius decreases from Sodium to Chlorine because nuclear charge Z increases while inner-shell shielding remains nearly constant. The resulting higher effective nuclear charge (Zeff) exerts a stronger electrostatic pull on the n = 3 valence electrons, drawing them closer to the nucleus.
4Why is the first ionization energy of Magnesium (Z = 12, 738 kJ/mol) higher than that of Aluminum (Z = 13, 578 kJ/mol)?
A.Magnesium has a larger nuclear charge than Aluminum.
B.Aluminum's valence electron is removed from a higher-energy 3p subshell that is shielded by filled 3s orbitals.
C.Magnesium has a smaller atomic radius than Aluminum.
D.Aluminum forms a stable noble gas core upon losing one electron.
Explanation: Magnesium has a ground-state configuration of [Ne] 3s², whereas Aluminum is [Ne] 3s² 3p¹. The 3p electron removed during Aluminum's first ionization is higher in energy and shielded by the fully filled 3s subshell, requiring less energy to remove than a 3s electron from Magnesium.
5Why is the electron affinity of Chlorine (-349 kJ/mol) more negative (more exothermic) than that of Fluorine (-328 kJ/mol)?
A.Fluorine has a lower electronegativity than Chlorine.
B.Chlorine has a smaller nuclear charge than Fluorine.
C.Fluorine's compact 2p subshell experiences strong electron-electron repulsion, weakening the net attraction for an incoming electron.
D.Chlorine's 3p subshell can accommodate an extra electron without requiring spin pairing.
Explanation: Fluorine is extremely small, resulting in high electron density within its 2p subshell. The strong inter-electronic repulsions in this compact region partially offset the attractive force of the nucleus, making Fluorine's electron affinity slightly less exothermic than Chlorine's 3p subshell addition.
6Which of the following orders represents the correct trend of decreasing ionic radius for the isoelectronic series S²⁻, Cl⁻, K⁺, Ca²⁺?
A.Ca²⁺ > K⁺ > Cl⁻ > S²⁻
B.S²⁻ > Cl⁻ > K⁺ > Ca²⁺
C.Cl⁻ > S²⁻ > Ca²⁺ > K⁺
D.K⁺ > Ca²⁺ > S²⁻ > Cl⁻
Explanation: For an isoelectronic series (all species have 18 electrons, [Ar] configuration), ionic radius decreases as nuclear charge Z increases. S²⁻ (Z=16) has the smallest Z and largest radius, followed by Cl⁻ (Z=17), K⁺ (Z=19), and Ca²⁺ (Z=20) which has the highest nuclear charge pulling electrons tightest.
7What is the energy of a single photon of ultraviolet radiation with a wavelength of λ = 250 nm? (Planck's constant h = 6.63 × 10⁻³⁴ J·s, speed of light c = 3.00 × 10⁸ m/s)
A.7.96 × 10⁻¹⁹ J
B.1.66 × 10⁻³¹ J
C.4.97 × 10⁻¹⁹ J
D.2.65 × 10⁻²⁷ J
Explanation: Using E = hc/λ: λ = 250 × 10⁻⁹ m. E = (6.63 × 10⁻³⁴ J·s × 3.00 × 10⁸ m/s) / (250 × 10⁻⁹ m) = 1.989 × 10⁻²⁵ / 2.50 × 10⁻⁷ = 7.956 × 10⁻¹⁹ J ≈ 7.96 × 10⁻¹⁹ J.
8In the hydrogen atom emission spectrum, what is the wavelength of the light emitted during the Balmer series transition from n = 3 to n = 2? (Rydberg constant R_H = 1.097 × 10⁷ m⁻¹)
A.434 nm
B.656 nm
C.486 nm
D.122 nm
Explanation: Using the Rydberg equation 1/λ = R_H (1/n1² - 1/n2²): 1/λ = 1.097 × 10⁷ (1/4 - 1/9) = 1.097 × 10⁷ (5/36) = 1.5236 × 10⁶ m⁻¹. Inverting gives λ = 6.563 × 10⁻⁷ m = 656 nm (H-alpha red visible line).
9What is the de Broglie wavelength of an electron (mass m = 9.11 × 10⁻³¹ kg) moving at a velocity of 2.00 × 10⁶ m/s? (h = 6.63 × 10⁻³⁴ J·s)
A.3.64 × 10⁻¹⁰ m
B.1.21 × 10⁻⁹ m
C.7.28 × 10⁻¹¹ m
D.5.47 × 10⁻⁷ m
Explanation: According to de Broglie's equation λ = h / (m·v): λ = (6.63 × 10⁻³⁴ J·s) / (9.11 × 10⁻³¹ kg × 2.00 × 10⁶ m/s) = 6.63 × 10⁻³⁴ / 1.822 × 10⁻²⁴ = 3.64 × 10⁻¹⁰ m (0.364 nm).
10An element X in Period 3 has the following successive ionization energies (in kJ/mol): IE1 = 578, IE2 = 1817, IE3 = 2745, IE4 = 11577, IE5 = 14842. To which group of the periodic table does element X belong?
A.Group 1 (Alkali metals)
B.Group 2 (Alkaline earth metals)
C.Group 13 (Boron group)
D.Group 14 (Carbon group)
Explanation: A dramatic spike occurs between IE3 (2745 kJ/mol) and IE4 (11577 kJ/mol), over a 4-fold increase. This indicates that element X has 3 valence electrons. Removing the 4th electron requires breaking into a stable noble gas core, identifying element X as Aluminum in Group 13.
About the Catalonia PAU Chemistry Practice Questions
Verified exam format metadata for Catalonia PAU Chemistry Examination — CIC / Generalitat de Catalunya (Química 2n Batxillerat 2026) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.