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100+ Free PAU Chemistry (Cantabria) Practice Questions
Cantabria PAU Chemistry 2026 (Pruebas de Acceso a la Universidad, Universidad de Cantabria) practice questions are available now; exam metadata is being verified.
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2026 Statistics
Key Facts: PAU Chemistry (Cantabria) Exam
90 min
Exam time duration for the PAU Chemistry paper
University of Cantabria (UNICAN) PAU Organising Commission
4 Apartados
Compulsory written exam sections, each worth 2.5 of 10 points
UNICAN Chemistry Orientations
0–10 scale
Scoring system (minimum 4.0 required in Access Phase)
Cantabria PAU Regulations
5 Blocks
Core Bachillerato curriculum blocks evaluated in the exam
Real Decreto 243/2022 (LOMLOE)
100
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Master the 2026 Cantabria PAU Chemistry exam with 100 practice questions covering atomic structure and bonding, thermochemistry, chemical equilibrium, acid-base/redox, and organic chemistry. This English-language MCQ bank is a study adaptation of the theory and calculation knowledge behind the official written exam.
Sample PAU Chemistry (Cantabria) Practice Questions
Try these sample questions to test your PAU Chemistry (Cantabria) 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, ml, ms) is permissible for an electron occupying a 3d orbital in an atom's ground state?
A.(3, 2, -1, +1/2)
B.(3, 1, +2, -1/2)
C.(3, 3, 0, +1/2)
D.(2, 2, -1, -1/2)
Explanation: For a 3d orbital, the principal quantum number is n = 3 and the azimuthal quantum number is l = 2. The magnetic quantum number ml can range from -l to +l, so ml = -1 is valid, and ms can be +1/2 or -1/2. Therefore, (3, 2, -1, +1/2) satisfies all quantum rules.
2What is the ground-state electron configuration of the neutral copper atom (Z = 29)?
A.[Ar] 4s² 3d⁹
B.[Ar] 4s¹ 3d¹⁰
C.[Ar] 4s⁰ 3d¹⁰
D.[Ar] 4s² 3d¹⁰
Explanation: Copper exhibits an exception to the standard Aufbau ordering by promoting one electron from 4s to 3d. This complete 3d¹⁰ subshell provides enhanced exchange energy stability. Thus, its ground-state configuration is [Ar] 4s¹ 3d¹⁰.
3How does the atomic radius generally change across Period 3 of the periodic table from sodium (Na) to chlorine (Cl)?
A.It increases because electrons are added to higher energy shells with larger principal quantum numbers.
B.It remains constant because the added protons and electrons cancel out each other's effects completely.
C.It decreases because the effective nuclear charge increases while inner shielding remains nearly constant.
D.It decreases because atomic mass decreases across the period from left to right.
Explanation: Moving from Na to Cl across Period 3, protons are added to the nucleus while valence electrons fill the same n = 3 shell. Inner-core shielding stays constant, so the effective nuclear charge Z* increases. This stronger attraction pulls valence electrons closer, decreasing atomic radius.
4Which element has the highest first ionization energy among oxygen (O), fluorine (F), nitrogen (N), and neon (Ne)?
A.Fluorine (F)
B.Nitrogen (N)
C.Oxygen (O)
D.Neon (Ne)
Explanation: First ionization energy generally increases from left to right across a period due to increasing effective nuclear charge. Noble gases have completed electron octets (2s² 2p⁶), making neon the most stable and difficult element to ionize in Period 2. Therefore, neon has the highest first ionization energy.
5Which molecule serves as a classic example of an incomplete octet on the central atom in its ground-state Lewis structure?
A.BF₃
B.NH₃
C.CH₄
D.H₂O
Explanation: In boron trifluoride (BF₃), the central boron atom has only three valence electrons. Boron forms three single covalent bonds with fluorine atoms, surrounded by six valence electrons. This leaves boron with an electron-deficient incomplete octet.
6Why is carbon dioxide (CO₂) nonpolar despite possessing polar carbon-oxygen double bonds?
A.CO₂ has a bent molecular geometry that creates asymmetric charge distribution across the oxygen atoms.
B.CO₂ has a linear geometry (sp hybridization), so its two equal C=O bond dipole vectors point in opposite directions and cancel completely.
C.The electronegativity difference between carbon and oxygen is zero, making the individual bonds nonpolar.
D.CO₂ exists as an ionic crystal lattice at room temperature where dipoles cannot exist.
Explanation: Carbon dioxide has a central sp-hybridized carbon atom forming a linear O=C=O structure (180° bond angle). Although each C=O bond is polar due to oxygen's higher electronegativity, the two equal bond dipole vectors are oriented anti-parallel. Thus, the net molecular dipole moment cancels to zero.
7What 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? (Planck's constant h = 6.626 × 10⁻³⁴ J·s)
A.1.82 × 10⁻¹⁰ m
B.7.27 × 10⁻¹⁰ m
C.3.64 × 10⁻¹⁰ m
D.5.45 × 10⁻⁹ m
Explanation: According to de Broglie's relation λ = h / (m · v), momentum determines wavelength. Substituting the given values yields λ = (6.626 × 10⁻³⁴) / ((9.11 × 10⁻³¹) × (2.00 × 10⁶)) = 3.637 × 10⁻¹⁰ m. This corresponds to 0.364 nm, which falls in the X-ray wavelength range.
8Which transition metal ion is paramagnetic with exactly 5 unpaired electrons in its ground state?
A.Cu²⁺ (Z = 29)
B.Zn²⁺ (Z = 30)
C.Ti³⁺ (Z = 22)
D.Fe³⁺ (Z = 26)
Explanation: Neutral iron (Z = 26) has configuration [Ar] 4s² 3d⁶. When forming Fe³⁺, two 4s electrons and one 3d electron are lost, yielding [Ar] 3d⁵. By Hund's rule, all 5 d-orbitals are singly occupied with parallel spins, producing 5 unpaired electrons and strong paramagnetism.
9According to VSEPR theory, what is the molecular geometry and approximate bond angle of ammonia (NH₃)?
A.Trigonal pyramidal geometry with bond angles of approximately 107°
B.Tetrahedral geometry with bond angles of exactly 109.5°
C.Trigonal planar geometry with bond angles of exactly 120°
D.T-shaped geometry with bond angles of approximately 90°
Explanation: Ammonia has four electron domains around central nitrogen (three N-H single bonds and one lone pair), adopting a tetrahedral electron domain geometry. Because lone pair-bonding pair repulsions are stronger than bonding pair-bonding pair repulsions, the H-N-H bond angles compress from 109.5° to ~107°. This produces a trigonal pyramidal molecular geometry.
10Comparing magnesium oxide (MgO) and sodium chloride (NaCl), why is the lattice energy of MgO significantly larger than that of NaCl?
A.MgO contains covalent double bonds whereas NaCl contains single ionic bonds.
B.Mg²⁺ and O²⁻ have higher ionic charges (+2/-2) and smaller interionic distances than Na⁺ and Cl⁻ (+1/-1).
C.Sodium has a larger electronegativity than magnesium, reducing its ionic bond strength.
D.NaCl forms a face-centered cubic structure whereas MgO forms a liquid crystal structure.
Explanation: Lattice energy by Coulomb's law is proportional to (q₁ · q₂) / r. MgO consists of divalent ions (+2 and -2, product = 4) with smaller ionic radii, while NaCl consists of monovalent ions (+1 and -1, product = 1). The fourfold charge product and smaller radius give MgO a vastly higher lattice energy (~3790 kJ/mol vs ~786 kJ/mol).
About the PAU Chemistry (Cantabria) Practice Questions
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