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100+ Free Castilla y León PAU Chemistry Practice Questions
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2026 Statistics
Key Facts: Castilla y León PAU Chemistry Exam
90 min
Exam Time Limit
Junta de Castilla y León
0–10
Grading Scale
USAL / UVA / UBU / ULE
4.0
Min. Access Phase Score
Comisión PAU Castilla y León
EUR 76.82
Base Registration Fee
Junta de Castilla y León
5 Core Units
Curriculum Content Areas
2º Bachillerato Chemistry Syllabus
The Castilla y León PAU Chemistry exam (Química) is administered by public universities in Castilla y León (USAL, UVA, UBU, ULE) 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 Castilla y León PAU Chemistry Practice Questions
Try these sample questions to test your Castilla y León 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 specific atomic orbital is characterized by the set of quantum numbers n = 3, l = 1, and ml = 0?
A.3s orbital
B.3p orbital
C.3d orbital
D.4p orbital
Explanation: The principal quantum number n = 3 specifies the third electron shell. The angular momentum quantum number l = 1 corresponds specifically to a p subshell. Therefore, n = 3 and l = 1 designate a 3p orbital.
2What is the ground-state electron configuration of a neutral iron atom (Fe, atomic number Z = 26)?
A.[Ar] 3d⁶ 4s²
B.[Ar] 3d⁸
C.[Ar] 3d⁵ 4s³
D.[Ar] 4s² 4p⁶
Explanation: Argon has 18 electrons. Iron has 26 electrons, meaning 8 valence electrons beyond the noble gas core. Following the Aufbau principle, the 4s orbital fills first with 2 electrons, and the remaining 6 electrons enter the 3d subshell, yielding [Ar] 3d⁶ 4s².
3How does the first ionization energy generally change across Period 3 of the periodic table from Sodium (Na) to Argon (Ar)?
A.It decreases continuously because atomic mass increases.
B.It increases generally due to increasing effective nuclear charge.
C.It remains constant because electrons are added to the same principal shell.
D.It fluctuates unpredictably with no observable trend.
Explanation: Across Period 3, protons are added to the nucleus, increasing the nuclear charge while core shielding remains relatively constant. This increases the effective nuclear charge (Zeff), pulling valence electrons more tightly and requiring more energy to remove an electron.
4Which of the following statements correctly compares the atomic/ionic radii of Na vs Na⁺ and Cl vs Cl⁻?
A.Na⁺ is larger than Na, and Cl⁻ is smaller than Cl.
B.Na⁺ is smaller than Na, and Cl⁻ is larger than Cl.
C.Both Na⁺ and Cl⁻ are larger than their neutral parent atoms.
D.Both Na⁺ and Cl⁻ are smaller than their neutral parent atoms.
Explanation: Cations are smaller than their parent neutral atoms because removing electrons decreases electron-electron repulsion and often removes an entire outer shell (Na → Na⁺). Anions are larger than their parent atoms because adding electrons increases electron-electron repulsion, expanding the electron cloud (Cl → Cl⁻).
5According to Hund's rule of maximum multiplicity, how do electrons populate degenerate 2p orbitals in a ground-state carbon atom (Z = 6)?
A.They pair up in a single 2p orbital with antiparallel spins.
B.They occupy two separate 2p orbitals with parallel spins.
C.They occupy two separate 2p orbitals with antiparallel spins.
D.They enter the 3s orbital before filling the 2p subshell.
Explanation: Hund's rule states that when electrons enter degenerate orbitals (orbitals of equal energy), one electron enters each orbital until all degenerate orbitals are singly occupied, and all singly occupied electrons have parallel spins to minimize electron repulsion.
6Which of the following sets of quantum numbers (n, l, ml, ms) is physically impossible for an electron in an atom?
A.n = 3, l = 2, ml = -1, ms = +1/2
B.n = 2, l = 2, ml = 0, ms = -1/2
C.n = 4, l = 0, ml = 0, ms = +1/2
D.n = 1, l = 0, ml = 0, ms = -1/2
Explanation: The angular momentum quantum number l can only take integer values from 0 up to n - 1. For n = 2, the maximum allowed value of l is 1 (2s or 2p). A value of l = 2 would require n ≥ 3 (d subshell).
7What is the ground-state electron configuration of the ferric cation Fe³⁺ (atomic number Z = 26)?
A.[Ar] 3d⁵
B.[Ar] 3d³ 4s²
C.[Ar] 3d⁶
D.[Ar] 3d⁴ 4s¹
Explanation: Neutral Fe is [Ar] 3d⁶ 4s². When transition metals ionize, electrons are removed first from the outermost principal shell (4s) before removing electrons from 3d. Removing 3 electrons takes 2 from 4s and 1 from 3d, leaving [Ar] 3d⁵.
8Which element on the Pauling scale has the highest electronegativity value (3.98)?
A.Oxygen (O)
B.Chlorine (Cl)
C.Fluorine (F)
D.Helium (He)
Explanation: Fluorine is the most electronegative element in the periodic table (assigned ~4.0 on the Pauling scale) due to its small atomic radius and high effective nuclear charge.
9Arranging the isoelectronic species O²⁻, F⁻, Na⁺, and Mg²⁺ in order of decreasing ionic radius yields which sequence?
A.O²⁻ > F⁻ > Na⁺ > Mg²⁺
B.Mg²⁺ > Na⁺ > F⁻ > O²⁻
C.F⁻ > O²⁻ > Mg²⁺ > Na⁺
D.Na⁺ > Mg²⁺ > O²⁻ > F⁻
Explanation: All four species are isoelectronic with 10 electrons (configuration 1s² 2s² 2p⁶). In an isoelectronic series, ionic radius decreases as nuclear charge (atomic number Z) increases. O²⁻ (Z=8) has the smallest Z and largest radius, while Mg²⁺ (Z=12) has the largest Z and smallest radius.
10What is effective nuclear charge (Zeff), and how does core electron shielding affect it?
A.Zeff is the total charge of all neutrons; shielding increases it.
B.Zeff is the net positive charge felt by a valence electron, reduced by core electron shielding (Zeff = Z - S).
C.Zeff is equal to the atomic mass number minus the number of valence electrons.
D.Zeff is the negative charge exerted by outer shell electrons on the nucleus.
Explanation: Effective nuclear charge (Zeff) represents the net positive attractive force experienced by valence electrons from the nucleus. Core electrons shield valence electrons from the full nuclear charge Z, giving Zeff = Z - S, where S is the shielding constant.
About the Castilla y León PAU Chemistry Practice Questions
Verified exam format metadata for Castilla y León PAU Chemistry (Química 2º Bachillerato) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.