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2026 Statistics
Key Facts: European Baccalaureate Chemistry Exam
180 Minutes
Exam duration
OSGES Syllabus
6.0 / 10
Passing mark (60%)
European Schools Board of Governors
7 Domains
Core curriculum topics
OSGES S6-S7 4P Chemistry Syllabus
100 Questions
Practice question bank size
OpenExamPrep
The European Baccalaureate Chemistry is administered by the Office of the Secretary-General of the European Schools (OSGES). Official assessment involves written/oral components graded on a 0-10 scale (passing score 5.0). Local questions on OpenExamPrep are an English-language MCQ study adaptation designed for syllabus revision.
Sample European Baccalaureate Chemistry Practice Questions
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1Which set of quantum numbers (n, l, m_l, m_s) is valid for an electron occupying a 3d atomic orbital?
A.n = 3, l = 2, m_l = -1, m_s = +1/2
B.n = 3, l = 1, m_l = +2, m_s = -1/2
C.n = 3, l = 3, m_l = 0, m_s = +1/2
D.n = 2, l = 2, m_l = -2, m_s = -1/2
Explanation: For a 3d orbital, the principal quantum number n = 3 and the angular momentum quantum number l = 2 (where l = 0, 1, 2, 3 correspond to s, p, d, f orbitals). The magnetic quantum number m_l can take integer values from -l to +l, so for l = 2, m_l ∈ {-2, -1, 0, +1, +2}, and the spin quantum number m_s is ±1/2. Therefore, n = 3, l = 2, m_l = -1, m_s = +1/2 is completely valid.
2What is the ground-state electron configuration of a neutral chromium atom (Z = 24)?
A.[Ar] 4s² 3d⁴
B.[Ar] 4s¹ 3d⁵
C.[Ar] 4s⁰ 3d⁶
D.[Ar] 4s² 3d⁵
Explanation: Chromium is a classic exception to the standard Aufbau filling rule. Moving one electron from the 4s orbital to the 3d subshell yields a half-filled [Ar] 4s¹ 3d⁵ configuration, which maximizes exchange energy and minimizes electron-electron repulsion.
3Which of the following Period 3 elements possesses the highest first ionization energy?
A.Sodium (Na)
B.Magnesium (Mg)
C.Argon (Ar)
D.Silicon (Si)
Explanation: First ionization energy generally increases across a period from left to right due to increasing effective nuclear charge (Z_eff) with constant core shielding. Argon (Ar) is at the extreme right of Period 3 with a complete octet (3s² 3p⁶), giving it the highest first ionization energy in Period 3.
4According to VSEPR theory, what is the molecular geometry of sulfur tetrafluoride (SF4)?
A.Tetrahedral
B.Square planar
C.Trigonal bipyramidal
D.See-saw
Explanation: Sulfur in SF4 has 6 valence electrons, forming 4 single bonds with fluorine atoms and retaining 1 lone pair (5 electron domains total). The steric number of 5 produces a trigonal bipyramidal electron domain geometry, but the lone pair occupies an equatorial position to minimize 90° repulsions, yielding a see-saw molecular geometry.
5What is the hybridization state of each carbon atom in ethyne (ethyne / acetylene, HC≡CH)?
A.sp
B.sp²
C.sp³
D.sp³d
Explanation: In ethyne (HC≡CH), each carbon forms one σ bond with a hydrogen atom and one σ bond with the neighboring carbon atom (along with two π bonds). Having 2 σ-bonding electron domains requires sp hybridization, producing a linear geometry with a 180° bond angle.
6Given the standard enthalpy of sublimation of Na(s) is +107 kJ/mol, the first ionization energy of Na(g) is +496 kJ/mol, the bond dissociation enthalpy of Cl2(g) is +244 kJ/mol, the electron affinity of Cl(g) is -349 kJ/mol, and the standard enthalpy of formation of NaCl(s) is -411 kJ/mol, what is the lattice enthalpy (ΔH_lattice) of NaCl(s) for the reaction Na+(g) + Cl-(g) → NaCl(s)?
A.-787 kJ/mol
B.-788 kJ/mol
C.-641 kJ/mol
D.-909 kJ/mol
Explanation: According to Hess's Law for a Born-Haber cycle: ΔH_f°[NaCl(s)] = ΔH_sub[Na] + IE1[Na] + 1/2 ΔH_diss[Cl2] + EA1[Cl] + ΔH_lattice. Substituting the values: -411 = +107 + 496 + 1/2(244) + (-349) + ΔH_lattice ⇒ -411 = +107 + 496 + 122 - 349 + ΔH_lattice ⇒ -411 = +376 + ΔH_lattice ⇒ ΔH_lattice = -411 - 376 = -787 kJ/mol (or -788 kJ/mol depending on rounding). Thus -788 kJ/mol is the correct value.
7Ethanol (CH3CH2OH) and dimethyl ether (CH3OCH3) are constitutional isomers with the molecular formula C2H6O. Why is the boiling point of ethanol (78.4 °C) significantly higher than that of dimethyl ether (-24 °C)?
A.Ethanol is a non-polar molecule with strong London dispersion forces.
B.Dimethyl ether forms stronger hydrogen bonds than ethanol.
C.Ethanol can form intermolecular hydrogen bonds due to its O-H group, whereas dimethyl ether cannot.
D.Ethanol has a much larger molar mass than dimethyl ether.
Explanation: Ethanol contains a hydrogen atom directly bonded to an electronegative oxygen atom (O-H group), allowing it to form extensive intermolecular hydrogen networks. Dimethyl ether lacks an O-H bond (it has C-O-C), so its primary intermolecular interactions are weaker dipole-dipole forces and London dispersion forces.
8What are the formal charges on the central oxygen atom and one of the terminal oxygen atoms in the resonance structure of ozone (O3) represented as O=O⁺-O⁻?
A.Central = 0, Double-bonded terminal = +1
B.Central = -1, Single-bonded terminal = +1
C.Central = +1, Double-bonded terminal = -1
D.Central = +1, Single-bonded terminal = -1
Explanation: Formal charge = (Valence e⁻) - (Non-bonding e⁻) - 1/2(Bonding e⁻). Oxygen has 6 valence electrons. For the central oxygen: 6 - 2 (1 lone pair) - 1/2(6 bonding e⁻ in 3 bonds) = +1. For the single-bonded terminal oxygen: 6 - 6 (3 lone pairs) - 1/2(2 bonding e⁻) = -1. For the double-bonded terminal oxygen: 6 - 4 - 2 = 0.
9According to Molecular Orbital (MO) theory, why is diatomic oxygen (O2) paramagnetic?
A.O2 contains two unpaired electrons in two degenerate π*2p antibonding orbitals.
B.O2 contains one unpaired electron in a σ2p bonding orbital.
C.O2 has a lone pair of electrons localized on each oxygen atom.
D.O2 rapidly dissociates into oxygen radicals in the gas phase.
Explanation: In the MO energy diagram for O2, the 12 valence electrons fill orbitals up to the antibonding level: (σ2s)² (σ*2s)² (σ2p)² (π2p)⁴ (π*2p)². By Hund's rule, the two electrons in the degenerate π*2px and π*2py antibonding orbitals remain unpaired with parallel spins, causing paramagnetism.
10Which sequence correctly arranges CH4, NH3, and H2O in order of DECREASING bond angle (largest to smallest)?
A.H2O > NH3 > CH4
B.CH4 > NH3 > H2O
C.NH3 > CH4 > H2O
D.CH4 > H2O > NH3
Explanation: All three molecules have 4 electron domains (tetrahedral domain geometry). CH4 has 0 lone pairs (109.5°), NH3 has 1 lone pair (107°), and H2O has 2 lone pairs (104.5°). Lone-pair–bonding-pair repulsion is stronger than bonding-pair–bonding-pair repulsion, compressing the bond angles as lone pairs increase.
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