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100+ Free Castilla-La Mancha PAU Chemistry (Química - UCLM 2026) Practice Questions
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2026 Statistics
Key Facts: Castilla-La Mancha PAU Chemistry (Química - UCLM 2026) Exam
90 min
PAU Chemistry exam duration
Tribunal PAU Universidad de Castilla-La Mancha
0–10
Grading scale (minimum 4.0 in the PAU)
UCLM
5 Blocks
Thematic blocks of the official LOMLOE curriculum
UCLM 2026
100
Questions with answers and detailed solutions
OpenExamPrep
Prepare for the UCLM 2026 PAU Chemistry exam with 100 high-quality practice questions, featuring step-by-step quantitative calculations across all 5 official curriculum areas. This English-language multiple-choice bank is a study adaptation of the official written PAU paper, not a replica of it.
Sample Castilla-La Mancha PAU Chemistry (Química - UCLM 2026) Practice Questions
Try these sample questions to test your Castilla-La Mancha PAU Chemistry (Química - UCLM 2026) 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 valid for the last electron of a ground-state nitrogen atom (Z = 7)?
A.(2, 1, +1, +1/2)
B.(2, 2, 0, +1/2)
C.(2, 0, +1, -1/2)
D.(1, 1, 0, +1/2)
Explanation: The ground-state electron configuration of nitrogen (Z = 7) is 1s² 2s² 2p³. The valence shell is n = 2. For the 2p subshell, the azimuthal quantum number is l = 1. The magnetic quantum number ml can take values -1, 0, +1, and the spin quantum number ms can be +1/2 or -1/2. Thus, (2, 1, +1, +1/2) is a valid set.
2What is the ground-state electron configuration of the Fe³⁺ ion (Z = 26)?
A.[Ar] 4s² 3d³
B.[Ar] 3d⁵
C.[Ar] 4s¹ 3d⁴
D.[Ar] 3d⁶
Explanation: A neutral iron atom (Z = 26) has the electron configuration [Ar] 4s² 3d⁶. When forming the Fe³⁺ cation, the two electrons in the outermost 4s orbital (n = 4) are lost first, followed by one electron from the 3d subshell, yielding the more stable half-filled d subshell configuration [Ar] 3d⁵.
3Calculate the energy of a photon of electromagnetic radiation with a wavelength λ = 500 nm in a vacuum. (Given: h = 6.626 · 10⁻³⁴ J·s, c = 3 · 10⁸ m/s).
A.1.33 · 10⁻²⁷ J
B.9.94 · 10⁻³¹ J
C.3.98 · 10⁻¹⁹ J
D.5.96 · 10⁻¹⁷ J
Explanation: Applying Planck's equation E = h·ν = h·c/λ with λ = 500 nm = 5.00 · 10⁻⁷ m:
E = (6.626 · 10⁻³⁴ J·s · 3.00 · 10⁸ m/s) / (5.00 · 10⁻⁷ m) = 1.9878 · 10⁻²⁵ / 5.00 · 10⁻⁷ = 3.9756 · 10⁻¹⁹ J ≈ 3.98 · 10⁻¹⁹ J.
4Which of the following chemical species has the largest ionic radius?
A.Cl⁻
B.K⁺
C.Ca²⁺
D.S²⁻
Explanation: All four species (S²⁻, Cl⁻, K⁺, Ca²⁺) are isoelectronic with 18 electrons. In an isoelectronic series, ionic radius decreases as nuclear charge Z increases, because greater nuclear attraction pulls the electron cloud tighter. Sulfur (Z = 16) has the lowest nuclear charge, so its anion S²⁻ has the largest radius.
5Calculate the de Broglie wavelength associated with an electron (m = 9.11 · 10⁻³¹ kg) moving at a velocity v = 2.00 · 10⁶ m/s. (Given: h = 6.626 · 10⁻³⁴ J·s).
A.0.364 nm
B.1.21 nm
C.3.64 nm
D.0.036 nm
Explanation: The de Broglie wavelength is given by λ = h / (m·v).
λ = (6.626 · 10⁻³⁴ J·s) / (9.11 · 10⁻³¹ kg · 2.00 · 10⁶ m/s) = (6.626 · 10⁻³⁴) / (1.822 · 10⁻²⁴) = 3.6366 · 10⁻¹⁰ m = 0.364 nm.
6According to the VSEPR model, what is the molecular geometry of the ammonia molecule (NH₃)?
A.Tetrahedral
B.Trigonal pyramidal
C.Trigonal planar
D.Linear
Explanation: The central N atom in NH₃ has 4 valence electron pairs (3 N-H bonding pairs and 1 non-bonding lone pair). The electron geometry is tetrahedral (AX₃E), but the molecular geometry (arrangement of the atoms) is trigonal pyramidal.
7Determine the formal charge of the central nitrogen atom in the nitrate ion (NO₃⁻), given that it is bonded to one oxygen via a double bond and to two oxygens via single bonds.
A.0
B.-1
C.+1
D.+2
Explanation: Formal charge is calculated as FC = V - N - B/2, where V is valence electrons (N has 5), N is non-bonding electrons (0 on central N of NO₃⁻), and B is shared bonding electrons (1 double bond + 2 single bonds = 4 bonds = 8 shared electrons). FC(N) = 5 - 0 - (8/2) = 5 - 4 = +1.
8Using the Born-Haber cycle, calculate the lattice energy (U) of NaCl(s) in kJ/mol given the following data: ΔH°f(NaCl,s) = -411 kJ/mol, ΔH°sub(Na,s) = +108 kJ/mol, E_ioniz(Na,g) = +496 kJ/mol, ΔH°dis(Cl₂,g) = +244 kJ/mol, A_electr(Cl,g) = -349 kJ/mol.
A.-910 kJ/mol
B.-666 kJ/mol
C.-544 kJ/mol
D.-788 kJ/mol
Explanation: The Born-Haber cycle states:
ΔH°f = ΔH°sub(Na) + E_ioniz(Na) + (1/2)ΔH°dis(Cl₂) + A_electr(Cl) + U
Substituting the values:
-411 = 108 + 496 + (1/2)(244) + (-349) + U
-411 = 108 + 496 + 122 - 349 + U
-411 = 377 + U ⇒ U = -411 - 377 = -788 kJ/mol.
9What hybridization of the central atom and what net dipole moment are exhibited by the boron trifluoride molecule (BF₃)?
A.sp² hybridization, dipole moment μ = 0 (nonpolar)
B.sp³ hybridization, dipole moment μ ≠ 0 (polar)
C.sp hybridization, dipole moment μ = 0 (nonpolar)
D.sp²d hybridization, dipole moment μ ≠ 0 (polar)
Explanation: Boron in BF₃ forms three covalent bonds with no lone pairs (AX₃), adopting sp² hybridization with trigonal planar geometry and 120° bond angles. Due to molecular symmetry, the vector sum of the three individual C-F dipole moments yields a net zero dipole moment (μ = 0), making the molecule nonpolar.
10Arrange the following pure substances in order of increasing boiling point: CH₄, CH₃OH, CH₃CH₂OH, H₂O.
A.CH₄ < H₂O < CH₃OH < CH₃CH₂OH
B.CH₄ < CH₃OH < CH₃CH₂OH < H₂O
C.CH₃OH < CH₄ < CH₃CH₂OH < H₂O
D.H₂O < CH₃CH₂OH < CH₃OH < CH₄
Explanation: CH₄ exhibits only very weak London dispersion forces (bp -161°C). Alcohols form hydrogen bonds; CH₃CH₂OH has a larger molar mass and surface area than CH₃OH, increasing additional London dispersion forces (bp CH₃OH = 65°C, CH₃CH₂OH = 78°C). H₂O forms a 3D network of up to 4 hydrogen bonds per molecule (bp 100°C).
About the Castilla-La Mancha PAU Chemistry (Química - UCLM 2026) Practice Questions
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