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100+ Free Galicia PAU Chemistry / Química 2º Bachillerato CIUG Practice Questions
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2026 Statistics
Key Facts: Galicia PAU Chemistry / Química 2º Bachillerato CIUG Exam
90 min
Exam duration (1.5 hours)
CIUG PAU Guidelines
0–10
Grading scale
CIUG Evaluation Criteria
4.0
Minimum Access Phase mark required
Spanish University Access Regulations
63.67 €
Ordinary registration fee
CIUG Official Fees
100
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Galicia PAU Chemistry (CIUG 2026) is a 90-minute examination assessing 2º Bachillerato chemistry competencies across 5 key domain areas under LOMLOE guidelines.
Sample Galicia PAU Chemistry / Química 2º Bachillerato CIUG Practice Questions
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1Which set of quantum numbers (n, l, ml, ms) is allowed for a valence electron in a ground-state sulfur atom (Z = 16)?
A.n = 3, l = 1, ml = -1, ms = -1/2
B.n = 3, l = 2, ml = 0, ms = +1/2
C.n = 2, l = 1, ml = +1, ms = -1/2
D.n = 3, l = 1, ml = +2, ms = +1/2
Explanation: Sulfur (Z = 16) has the ground-state electron configuration 1s² 2s² 2p⁶ 3s² 3p⁴. The valence electrons occupy the n = 3 shell. For 3p electrons, n = 3 and l = 1. Permissible values for ml range from -l to +l (-1, 0, +1), and ms can be +1/2 or -1/2. Therefore, n = 3, l = 1, ml = -1, ms = -1/2 is a valid set.
2What is the ground-state electron configuration of the iron(III) ion, Fe³⁺ (Z = 26)?
A.[Ar] 3d⁵
B.[Ar] 4s² 3d³
C.[Ar] 4s¹ 3d⁴
D.[Ar] 3d⁶
Explanation: Neutral iron (Z = 26) has the electron configuration [Ar] 4s² 3d⁶. When transition metals ionize, electrons are removed first from the outermost 4s subshell, then from the 3d subshell. Removing three electrons yields Fe³⁺ with configuration [Ar] 3d⁵, representing a stable half-filled d subshell.
3Why is the first ionization energy of phosphorus (Z = 15) higher than that of sulfur (Z = 16)?
A.Phosphorus has a stable half-filled 3p³ subshell, whereas sulfur has paired electrons in one 3p orbital experiencing electron-electron repulsion.
B.Sulfur has a smaller nuclear charge than phosphorus, resulting in weaker attraction for valence electrons.
C.Phosphorus has a significantly smaller atomic radius than sulfur.
D.Sulfur's valence electrons are shielded by an extra inner electron shell.
Explanation: Phosphorus ([Ar] 3s² 3p³) has three unpaired electrons in three degenerate 3p orbitals, giving an extra exchange energy stability due to a half-filled subshell. Sulfur ([Ar] 3s² 3p⁴) has two paired electrons in one 3p orbital. The interelectronic repulsion between these paired electrons lowers the energy required to remove one electron from sulfur.
4Which list arranges the isoelectronic species N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺ in order of DECREASING ionic radius?
A.N³⁻ > O²⁻ > F⁻ > Na⁺ > Mg²⁺
B.Mg²⁺ > Na⁺ > F⁻ > O²⁻ > N³⁻
C.F⁻ > O²⁻ > N³⁻ > Mg²⁺ > Na⁺
D.N³⁻ > F⁻ > O²⁻ > Na⁺ > Mg²⁺
Explanation: All five species are isoelectronic with 10 electrons (configuration 1s² 2s² 2p⁶). The ionic radius in an isoelectronic series is determined by nuclear charge (Z). As Z increases from N (Z=7) to Mg (Z=12), the greater nuclear pull contracts the electron cloud. Thus, N³⁻ (Z=7) is largest and Mg²⁺ (Z=12) is smallest.
5Given standard enthalpies for NaCl formation: Sublimation of Na(s) = +107 kJ/mol, 1st Ionization Energy of Na(g) = +496 kJ/mol, Bond dissociation of Cl₂(g) = +244 kJ/mol, Electron affinity of Cl(g) = -349 kJ/mol, and Standard enthalpy of formation ΔH°f(NaCl, s) = -411 kJ/mol. Calculate the lattice enthalpy (ΔH_lattice) for NaCl(s) → Na⁺(g) + Cl⁻(g).
A.+787 kJ/mol
B.+665 kJ/mol
C.+388 kJ/mol
D.+909 kJ/mol
Explanation: According to Hess's Law applied to the Born-Haber cycle:
ΔH°f = ΔH_sub(Na) + IE₁(Na) + 1/2 ΔH_diss(Cl₂) + EA₁(Cl) - ΔH_lattice
-411 = +107 + 496 + 1/2(244) + (-349) - ΔH_lattice
-411 = 107 + 496 + 122 - 349 - ΔH_lattice = +376 - ΔH_lattice
ΔH_lattice = +376 - (-411) = +787 kJ/mol.
6In the preferred Lewis structure of the sulfate ion, SO₄²⁻, with expanded octet to minimize formal charges, what are the formal charges on the sulfur atom and the double-bonded oxygen atoms?
A.Formal charge on S = 0; formal charge on double-bonded O = 0
B.Formal charge on S = +2; formal charge on double-bonded O = -1
C.Formal charge on S = +1; formal charge on double-bonded O = 0
D.Formal charge on S = 0; formal charge on double-bonded O = -1
Explanation: In SO₄²⁻, sulfur has 6 valence electrons. In the structure with two S=O double bonds and two S-O⁻ single bonds, sulfur forms 6 covalent bonds (12 shared electrons) and has no lone pairs: FC(S) = 6 - 0 - 6 = 0. The double-bonded oxygens have 2 lone pairs and 2 bonds: FC(O=) = 6 - 4 - 2 = 0. The single-bonded oxygens have 3 lone pairs and 1 bond: FC(O⁻) = 6 - 6 - 1 = -1.
7According to VSEPR theory, what are the electron domain geometry and molecular geometry of sulfur tetrafluoride, SF₄?
A.Trigonal bipyramidal electron domain geometry; seesaw molecular geometry
B.Tetrahedral electron domain geometry; tetrahedral molecular geometry
C.Square planar electron domain geometry; square planar molecular geometry
D.Octahedral electron domain geometry; square pyramidal molecular geometry
Explanation: Sulfur in SF₄ has 6 valence electrons + 4 single bonds = 10 electrons around S (5 electron pairs: 4 bonding pairs and 1 lone pair). 5 electron domains adopt a trigonal bipyramidal domain geometry. To minimize lone pair-bonding pair repulsions, the lone pair occupies an equatorial position, giving a seesaw molecular shape.
8What are the hybridizations of the carbon atoms in ethyne (acetylene, C₂H₂), ethene (ethylene, C₂H₄), and ethane (C₂H₆), respectively?
A.C₂H₂: sp; C₂H₄: sp²; C₂H₆: sp³
B.C₂H₂: sp²; C₂H₄: sp³; C₂H₆: sp
C.C₂H₂: sp³; C₂H₄: sp²; C₂H₆: sp
D.C₂H₂: sp; C₂H₄: sp³; C₂H₆: sp²
Explanation: In ethyne (H-C≡C-H), each carbon forms 2 σ-bonds (linear shape, 180° bond angle) requiring sp hybridization. In ethene (H₂C=CH₂), each carbon forms 3 σ-bonds (trigonal planar shape, 120° bond angle) requiring sp² hybridization. In ethane (H₃C-CH₃), each carbon forms 4 σ-bonds (tetrahedral shape, 109.5° bond angle) requiring sp³ hybridization.
9Which of the following molecules has a non-zero net dipole moment (is polar)?
A.Sulfur dioxide, SO₂
B.Carbon dioxide, CO₂
C.Boron trifluoride, BF₃
D.Carbon tetrachloride, CCl₄
Explanation: SO₂ has a bent molecular shape due to 1 lone pair and 2 bonding domains on sulfur (VSEPR angle ~119°). The S-O bond dipoles do not cancel out, resulting in a net dipole moment μ > 0. CO₂ (linear), BF₃ (trigonal planar), and CCl₄ (tetrahedral) are highly symmetric, causing their bond dipoles to cancel out completely (μ = 0).
10Which of the following compounds has the highest boiling point, and what is the primary intermolecular force responsible?
A.Ethanol (CH₃CH₂OH); hydrogen bonding
B.Dimethyl ether (CH₃OCH₃); dipole-dipole interactions
C.Ethane (CH₃CH₃); London dispersion forces
D.Fluoromethane (CH₃F); ion-dipole interactions
Explanation: Ethanol contains an -OH group where hydrogen is covalently bonded to highly electronegative oxygen, allowing strong intermolecular hydrogen bonding. Dimethyl ether (polar, dipole-dipole) and ethane (nonpolar, London dispersion) lack hydrogen bond donors, giving ethanol a significantly higher boiling point (78 °C vs -24 °C for dimethyl ether).
About the Galicia PAU Chemistry / Química 2º Bachillerato CIUG Practice Questions
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