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Extremadura PAU Chemistry Exam — Química 2º Bachillerato UEx 2026 practice questions are available now; exam metadata is being verified.
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2026 Statistics
Key Facts: Extremadura PAU Chemistry Exam
EUR 78.26
Ordinary registration fee for PAU
Universidad de Extremadura (UEx)
90 Mins
Official examination duration
Comisión Organizadora de la PAU de Extremadura
Min 4.0
Minimum Access Phase score required to combine with Bachillerato GPA
UEx / Junta de Extremadura
7 Topics
Core chemistry syllabus units
2º Bachillerato LOMLOE Chemistry Curriculum
100
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Sample Extremadura PAU Chemistry Practice Questions
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1Which set of quantum numbers (n, l, ml, ms) is valid for the highest-energy valence electron of a ground-state chlorine atom (Z = 17)?
A.(2, 1, -1, +1/2)
B.(3, 2, 0, -1/2)
C.(3, 0, +1, +1/2)
D.(3, 1, 0, +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 n = 3 and l = 1. For l = 1, ml can be -1, 0, or +1, and ms can be +1/2 or -1/2. Thus, (3, 1, 0, +1/2) is a valid set.
2What is the 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] 4s¹ 3d⁶
Explanation: Chromium (Z = 24) is an exception to the standard Aufbau ordering due to the stability associated with a half-filled 3d subshell. Promoting one electron from 4s to 3d gives [Ar] 3d⁵ 4s¹.
3Which of the following general trends correctly describes the first ionization energy across Period 3 elements from sodium to argon?
A.It decreases systematically because atomic radius increases.
B.It remains constant because electrons fill the same principal quantum shell (n = 3).
C.It increases overall due to increasing effective nuclear charge Z*.
D.It reaches a minimum at argon due to electron shielding.
Explanation: Across Period 3 from Na to Ar, nuclear charge Z increases while core shielding remains relatively constant. This increases effective nuclear charge Z*, pulling valence electrons closer and requiring more energy to remove an electron.
4Calculate the de Broglie wavelength of an electron (mass m = 9.11 × 10⁻³¹ kg) moving at a velocity of 2.00 × 10⁶ m/s. (Planck's constant h = 6.63 × 10⁻³⁴ J·s).
A.0.364 nm
B.1.21 nm
C.3.64 nm
D.0.036 nm
Explanation: According to the de Broglie relation, λ = h / (m · v). Substituting values: 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.
5A metal surface has a work function (threshold energy) W₀ = 4.41 × 10⁻¹⁹ J. What is the threshold frequency f₀ required to emit photoelectrons? (h = 6.63 × 10⁻³⁴ J·s).
A.4.41 × 10¹⁴ Hz
B.6.65 × 10¹⁴ Hz
C.1.50 × 10¹⁵ Hz
D.2.92 × 10⁻¹⁵ Hz
Explanation: The photoelectric effect equation gives W₀ = h · f₀. Rearranging for frequency: f₀ = W₀ / h = (4.41 × 10⁻¹⁹ J) / (6.63 × 10⁻³⁴ J·s) = 6.65 × 10¹⁴ Hz.
6Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from the n = 4 level to the n = 2 level. (Rydberg constant / energy level formula Eₙ = -2.18 × 10⁻¹⁸ J / n²).
A.1.36 × 10⁻¹⁹ J
B.5.45 × 10⁻¹⁹ J
C.4.09 × 10⁻¹⁹ J
D.2.18 × 10⁻¹⁸ J
Explanation: E₄ = -2.18 × 10⁻¹⁸ / 4² = -1.3625 × 10⁻¹⁹ J. E₂ = -2.18 × 10⁻¹⁸ / 2² = -5.45 × 10⁻¹⁹ J. ΔE = E₄ - E₂ = (-1.3625 × 10⁻¹⁹) - (-5.45 × 10⁻¹⁹) = +4.0875 × 10⁻¹⁹ J ≈ 4.09 × 10⁻¹⁹ J.
7Which set of isoelectronic species is correctly arranged in order of INCREASING ionic radius?
A.Ca²⁺ < K⁺ < Cl⁻ < S²⁻
B.S²⁻ < Cl⁻ < K⁺ < Ca²⁺
C.K⁺ < Ca²⁺ < S²⁻ < Cl⁻
D.Cl⁻ < S²⁻ < Ca²⁺ < K⁺
Explanation: All four species (Ca²⁺, K⁺, Cl⁻, S²⁻) have 18 electrons (isoelectronic with Ar). As nuclear charge Z increases (Ca=20, K=19, Cl=17, S=16), the electrostatic pull on the 18 electrons increases, making ionic radius smaller. Thus Ca²⁺ (smallest) < K⁺ < Cl⁻ < S²⁻ (largest).
8Why is the second ionization energy of sodium (IE₂ = 4562 kJ/mol) drastically higher than its first ionization energy (IE₁ = 496 kJ/mol)?
A.Sodium forms a covalent network after losing its first electron.
B.The second ionization requires breaking a noble gas core octet of neon by adding an electron.
C.Sodium has a higher electron affinity for cationic states.
D.The second electron is removed from a stable, filled 2p core subshell closer to the nucleus.
Explanation: Neutral Na ([Ne] 3s¹) loses its 3s electron easily to form Na⁺ ([Ne]). Removing a second electron requires pulling an electron from the stable 2p⁶ inner core subshell (n = 2), which experiences a much higher effective nuclear charge Z* and is closer to the nucleus.
9What is the wavelength of the photon emitted in the Lyman series transition from n = 3 to n = 1 in a hydrogen atom? (ΔE = 1.938 × 10⁻¹⁸ J, h = 6.63 × 10⁻³⁴ J·s, c = 3.00 × 10⁸ m/s).
A.103 nm
B.656 nm
C.486 nm
D.121 nm
Explanation: λ = h · c / ΔE = (6.63 × 10⁻³⁴ J·s × 3.00 × 10⁸ m/s) / (1.938 × 10⁻¹⁸ J) = (1.989 × 10⁻²⁵) / (1.938 × 10⁻¹⁸) = 1.026 × 10⁻⁷ m = 103 nm (ultraviolet Lyman line).
10What is the ground-state electron configuration of the iron(III) ion, Fe³⁺ (Z = 26)?
A.[Ar] 4s² 3d³
B.[Ar] 3d⁵
C.[Ar] 4s¹ 3d⁴
D.[Ar] 3d⁶
Explanation: Neutral Fe (Z = 26) has configuration [Ar] 4s² 3d⁶. Transition metals lose valence 4s electrons first during ionization. Removing 3 electrons (two from 4s and one from 3d) yields [Ar] 3d⁵.
About the Extremadura PAU Chemistry Practice Questions
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