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2026 Statistics
Key Facts: Andalusia PAU Chemistry Exam
90 min
Exam Time Limit
Distrito Único Andaluz
0–10
Grading Scale
Junta de Andalucía
4.0
Min. Access Phase Score
Comisión Interuniversitaria
EUR 58.70
Base Registration Fee
Junta de Andalucía
8 Core Units
Curriculum Content Areas
2º Bachillerato Chemistry Syllabus
The Andalusia PAU Chemistry exam (Química) is administered by the Distrito Único Andaluz and Comisión Interuniversitaria de Andalucía for students completing 2nd Bachillerato. The exam lasts 90 minutes and is graded on a 0–10 scale (minimum 4.0 required in the Access Phase). Note that local questions on this platform are an English-language MCQ study adaptation created to help students master the underlying 2nd Bachillerato curriculum.
Sample Andalusia PAU Chemistry Practice Questions
Try these sample questions to test your Andalusia 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, m_l, m_s) is permissible for an electron in a 3p atomic orbital?
A.n = 3, l = 1, m_l = -1, m_s = +1/2
B.n = 3, l = 2, m_l = 0, m_s = -1/2
C.n = 3, l = 0, m_l = 1, m_s = +1/2
D.n = 2, l = 1, m_l = -1, m_s = +1/2
Explanation: For a 3p orbital, the principal quantum number is n = 3 and the azimuthal quantum number is l = 1. The magnetic quantum number m_l can take values from -l to +l (-1, 0, +1), and the spin quantum number m_s can be +1/2 or -1/2. The set (n=3, l=1, m_l=-1, m_s=+1/2) satisfies all these quantum mechanical restrictions.
2What is the ground-state electron configuration of the Fe²⁺ ion (atomic number Z = 26)?
A.[Ar] 3d⁶
B.[Ar] 4s² 3d⁴
C.[Ar] 4s¹ 3d⁵
D.[Ar] 3d⁵ 4s¹
Explanation: Neutral iron (Z = 26) has the ground-state configuration [Ar] 4s² 3d⁶. When transition metals form cations, electrons are removed first from the outermost s orbital (4s) before the 3d orbitals. Removing 2 electrons from neutral iron yields [Ar] 3d⁶.
3Which of the following elements has the highest first ionization energy?
A.Fluorine (F)
B.Oxygen (O)
C.Chlorine (Cl)
D.Nitrogen (N)
Explanation: First ionization energy increases across a period from left to right (due to increasing effective nuclear charge) and decreases down a group (due to increased atomic radius and shielding). Fluorine is at the top right of the main-group elements (excluding noble gases) and has the highest first ionization energy among the options.
4Why does nitrogen (Z = 7) have a higher first ionization energy than oxygen (Z = 8), despite oxygen being to the right of nitrogen in Period 2?
A.Nitrogen has a half-filled 2p subshell (2p³), which confers extra exchange stability.
B.Oxygen has a smaller nuclear charge than nitrogen.
C.Nitrogen has a larger atomic radius than oxygen, making electron removal easier.
D.Oxygen experiences no electron-electron repulsion in its 2p orbitals.
Explanation: Nitrogen's valence electron configuration is 2s² 2p³, possessing a half-filled 2p subshell with maximum spin multiplicity (Hund's rule), which is extra stable. In oxygen (2s² 2p⁴), the fourth 2p electron is paired in one 2p orbital, experiencing inter-electronic repulsion that makes its removal easier than expected.
5Arrange the following species in order of INCREASING ionic/atomic radius: K⁺, Ar, Cl⁻, S²⁻.
A.K⁺ < Ar < Cl⁻ < S²⁻
B.S²⁻ < Cl⁻ < Ar < K⁺
C.Ar < K⁺ < Cl⁻ < S²⁻
D.K⁺ < Cl⁻ < Ar < S²⁻
Explanation: These four species are isoelectronic, each possessing 18 electrons ([Ar] configuration). For isoelectronic species, radius decreases as nuclear charge (atomic number Z) increases. K⁺ (Z=19) has the most protons and smallest radius, followed by neutral Ar (Z=18), Cl⁻ (Z=17), and S²⁻ (Z=16) with the fewest protons and largest radius.
6According to the de Broglie hypothesis, what is the wavelength of an electron (mass = 9.11 × 10⁻³¹ kg) moving at a velocity of 2.0 × 10⁶ m/s? (h = 6.626 × 10⁻³⁴ J·s)
A.3.64 × 10⁻¹⁰ m
B.3.64 × 10⁻⁷ m
C.1.82 × 10⁻¹⁰ m
D.5.46 × 10⁻⁹ m
Explanation: The de Broglie wavelength is calculated using λ = h / (m · v). Substituting values: λ = (6.626 × 10⁻³⁴ J·s) / [(9.11 × 10⁻³¹ kg) × (2.0 × 10⁶ m/s)] = 6.626 × 10⁻³⁴ / (1.822 × 10⁻²⁴) = 3.64 × 10⁻¹⁰ m (or 0.364 nm).
7Which rule or principle states that no two electrons in the same atom can have identical values for all four quantum numbers?
A.Pauli Exclusion Principle
B.Hund's Rule of Maximum Multiplicity
C.Aufbau Principle
D.Heisenberg Uncertainty Principle
Explanation: The Pauli Exclusion Principle dictates that an orbital can hold a maximum of two electrons, and those two electrons must have opposite spins (m_s = +1/2 and -1/2), ensuring their set of four quantum numbers is unique.
8The work function of potassium metal is 2.30 eV (1 eV = 1.602 × 10⁻¹⁹ J). What is the threshold frequency of light required to cause the photoelectric effect in potassium? (h = 6.626 × 10⁻³⁴ J·s)
A.5.56 × 10¹⁴ Hz
B.3.47 × 10¹⁴ Hz
C.8.84 × 10¹⁴ Hz
D.1.23 × 10¹⁵ Hz
Explanation: First convert the work function Φ to Joules: Φ = 2.30 × 1.602 × 10⁻¹⁹ J = 3.6846 × 10⁻¹⁹ J. The threshold frequency ν₀ is calculated by Φ = h · ν₀, so ν₀ = (3.6846 × 10⁻¹⁹ J) / (6.626 × 10⁻³⁴ J·s) = 5.56 × 10¹⁴ Hz.
9Which electronic transition in a hydrogen atom emits a photon with the shortest wavelength?
A.n = 3 → n = 1
B.n = 2 → n = 1
C.n = 4 → n = 2
D.n = 5 → n = 3
Explanation: According to the Rydberg formula ΔE = E_final - E_initial = h·c/λ, photon energy is inversely proportional to wavelength. Emission to n = 1 (Lyman series) involves the largest energy gaps. Among transitions to n = 1, n = 3 → n = 1 involves a larger energy difference (ΔE = 13.6 × (1 - 1/9) = 12.09 eV) than n = 2 → n = 1 (10.2 eV), producing the shortest wavelength photon.
10What maximum number of electrons can occupy the subshell specified by quantum numbers n = 4, l = 2?
A.10
B.6
C.14
D.2
Explanation: Quantum number l = 2 designates a d subshell. The number of orbitals in any subshell is given by (2l + 1) = 2(2) + 1 = 5 orbitals. Since each orbital holds up to 2 electrons with opposite spins, the maximum capacity is 5 × 2 = 10 electrons.
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