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2026 Statistics
Key Facts: PAU Chemistry (Aragón) Exam
90 min
Exam Duration
UNIZAR PAU Commission
0–10
Grading Scale
Gobierno de Aragón
4.0 / 10
Min. Access Phase Mark
PAU Regulations
EUR 75.00
Ordinary Registration Fee
UNIZAR PAU inscription page 2026
The Aragón PAU Chemistry exam (Química) is administered by the PAU Organising Commission of Aragón / UNIZAR for 2nd Bachillerato students entering university. Lasting 90 minutes and graded on a 0–10 scale (minimum 4.0 required in the Access Phase), it tests mastery of physical, analytical, inorganic, and organic chemistry. Questions here serve as an English MCQ study adaptation for the Aragón PAU 2026 exam.
Sample PAU Chemistry (Aragón) Practice Questions
Try these sample questions to test your PAU Chemistry (Aragón) 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 allowed for an electron occupying a 3p orbital?
A.n = 3, l = 0, m_l = 1, m_s = +1/2
B.n = 2, l = 1, m_l = -1, m_s = +1/2
C.n = 3, l = 1, m_l = -1, m_s = +1/2
D.n = 3, l = 2, m_l = 0, 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 integer values from -l to +l (-1, 0, +1), and the spin quantum number m_s must be +1/2 or -1/2. The set (n = 3, l = 1, m_l = -1, m_s = +1/2) satisfies all these quantum mechanical rules.
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 electron configuration [Ar] 4s² 3d⁶. When transition metal atoms lose electrons to form cations, the outer 4s electrons are removed before 3d electrons. Thus, Fe²⁺ loses its two 4s electrons, resulting in [Ar] 3d⁶.
3Which of the following ground-state neutral atoms has the highest first ionization energy?
A.Fluorine (F)
B.Neon (Ne)
C.Chlorine (Cl)
D.Oxygen (O)
Explanation: First ionization energy generally increases across a period from left to right and decreases down a group. Neon (Ne) is a noble gas in Period 2 with a complete octet valence shell (2s² 2p⁶), giving it an exceptionally high effective nuclear charge and high ionization energy compared to F, Cl, and O.
4According to Pauling's scale, which element is the most electronegative?
A.Fluorine (F)
B.Oxygen (O)
C.Chlorine (Cl)
D.Helium (He)
Explanation: Fluorine has the highest electronegativity value on the Pauling scale (3.98), reflecting its strong tendency to attract shared electron pairs in covalent bonds due to its small radius and high effective nuclear charge.
5Which principle states that no two electrons in an atom can have the exact same four quantum numbers?
A.Aufbau Principle
B.Pauli Exclusion Principle
C.Heisenberg Uncertainty Principle
D.Hund's Rule of Maximum Multiplicity
Explanation: The Pauli Exclusion Principle dictates that no two electrons in a single atom can possess an identical set of four quantum numbers (n, l, m_l, m_s). Consequently, an atomic orbital can hold at most two electrons with opposite spins (+1/2 and -1/2).
6How many unpaired electrons are present in a ground-state neutral Nitrogen atom (Z = 7)?
A.2
B.3
C.0
D.1
Explanation: Nitrogen (Z = 7) has the ground-state electron configuration 1s² 2s² 2p³. According to Hund's rule, the three electrons in the 2p subshell occupy three separate degenerate orbitals (2px, 2py, 2pz) with parallel spins, giving 3 unpaired electrons.
7Which of the following species is isoelectronic with the Argon atom (Z = 18)?
A.S²⁻
B.Na⁺
C.F⁻
D.Mg²⁺
Explanation: Argon has 18 electrons (1s² 2s² 2p⁶ 3s² 3p⁶). Neutral Sulfur (Z = 16) gains 2 electrons to form S²⁻, giving it 16 + 2 = 18 electrons. Thus, S²⁻ is isoelectronic with Ar.
8Which list correctly orders the ionic radii of the isoelectronic species N³⁻, O²⁻, F⁻, and Na⁺ from largest to smallest?
A.F⁻ > O²⁻ > N³⁻ > Na⁺
B.N³⁻ > O²⁻ > F⁻ > Na⁺
C.N³⁻ > F⁻ > O²⁻ > Na⁺
D.Na⁺ > F⁻ > O²⁻ > N³⁻
Explanation: For an isoelectronic series (all having 10 electrons), ionic radius decreases as nuclear charge (Z) increases because more protons pull the electron cloud inward. Here Z(N)=7, Z(O)=8, Z(F)=9, Z(Na)=11. Thus, N³⁻ is largest and Na⁺ is smallest: N³⁻ > O²⁻ > F⁻ > Na⁺.
9What is the maximum number of electrons that can occupy a subshell with azimuthal quantum number l = 2?
A.10
B.6
C.14
D.2
Explanation: An azimuthal quantum number l = 2 corresponds to a d subshell. The number of orbitals in a subshell is 2l + 1 = 2(2) + 1 = 5. Since each orbital holds up to 2 electrons of opposite spin, the maximum electron capacity is 2(2l + 1) = 2(5) = 10 electrons.
10Why is the second ionization energy of Sodium (Na, Z = 11) significantly larger than its first ionization energy?
A.The 3s subshell is doubly degenerate and stabilizes the Na⁺ ion.
B.Sodium forms covalent bonds after losing one electron.
C.The nuclear charge decreases after losing the first valence electron.
D.The second electron must be removed from a stable, lower-shell noble gas core ([Ne]), which experiences a higher effective nuclear charge.
Explanation: Na has the electron configuration [Ne] 3s¹. Removing the first electron (IE1 = 496 kJ/mol) leaves the Na⁺ ion with the stable noble-gas configuration [Ne] (2s² 2p⁶). Removing a second electron (IE2 = 4562 kJ/mol) requires breaking into this fully-filled n=2 inner shell, which is closer to the nucleus and experiences a dramatically higher effective nuclear charge.
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Verified exam format metadata for Aragón PAU Chemistry / Química 2º Bachillerato Exam is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.