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Exam Review

Key Facts: VWO Scheikunde Exam

VWO Scheikunde is a core subject in the Nature & Health (N&G) and Nature & Technology (N&T) upper-secondary profiles in the Netherlands.

The exam is administered under the authority of the College voor Toetsen en Examens (CvTE).

The official syllabus is organized into Domain B (Stoffen en bindingen), Domain C (Processen en kinetiek), Domain D (Zuur-base en redox), Domain E (Synthese), Domain F (Industrie), and Domain G (Groene chemie).

Calculators and approved reference books (Binas 7th ed.) are permitted during the Centraal Examen.

The written CE exam duration is exactly 3 hours (180 minutes).

The final mark is determined 50% by the school-based SE and 50% by the national CE.

Numeric grades range from 1.0 to 10.0, with 5.5 being the passing threshold.

These 100 practice items offer complete coverage of all syllabus domains in English adaptation.

Official VWO Scheikunde 2026 is CE (3-hour written exam) plus school SE. These 100 English MCQs provide practice aligned to CvTE VWO chemistry CE subdomains with numerical stoichiometry and equilibrium calculations.

Sample VWO Scheikunde Practice Questions

Try these sample questions to review concepts for the VWO Scheikunde exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1What is the ground-state valence electron configuration of a neutral sulfur atom (Z = 16)?
A.3s² 3p⁴
B.3s² 3p⁶
C.2s² 2p⁴
D.3s¹ 3p⁵
Explanation: Sulfur has atomic number 16. Its full electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁴. The outer (valence) shell is n = 3, containing 2 electrons in the 3s orbital and 4 electrons in the 3p subshell, giving 6 valence electrons (3s² 3p⁴).
2Which of the following molecules has a permanent dipole moment due to its polar bonds and bent molecular geometry?
A.Carbon dioxide (CO₂)
B.Methane (CH₄)
C.Water (H₂O)
D.Boron trifluoride (BF₃)
Explanation: Water (H₂O) has two polar O-H bonds (ΔEN ≈ 1.2) and two lone pairs on oxygen, giving it a bent VSEPR shape (bond angle ~104.5°). The individual bond dipoles do not cancel out, resulting in a net permanent dipole moment. CO₂ is linear, CH₄ is tetrahedral, and BF₃ is trigonal planar; their symmetric shapes cause bond dipoles to cancel.
3According to VSEPR theory, what is the molecular geometry of sulfur hexafluoride (SF₆)?
A.Tetrahedral
B.Trigonal bipyramidal
C.Octahedral
D.Square planar
Explanation: In SF₆, the central sulfur atom forms 6 single covalent bonds with fluorine atoms and has no lone pairs (steric number = 6). The electron-pair arrangement and molecular geometry are both octahedral, with 90° bond angles between adjacent F-S-F bonds.
4Which requirement must be met for a substance to form intermolecular hydrogen bonds between its molecules?
A.The molecule must contain any hydrogen atom bonded to carbon.
B.The molecule must have a hydrogen atom covalently bonded directly to a highly electronegative atom (N, O, or F).
C.The molecule must contain a metal ion bonded to a nonmetal.
D.The molecule must possess a net negative ionic charge.
Explanation: Hydrogen bonding requires a hydrogen donor (a hydrogen atom bonded to N, O, or F) and a hydrogen acceptor (an electronegative atom N, O, or F with a lone pair). C-H bonds are not sufficiently polar to participate in hydrogen bonding.
5Comparing propane (C₃H₈), propan-1-ol (C₃H₇OH), and propanoic acid (C₂H₅COOH), which list correctly arranges them in order of INCREASING boiling point?
A.Propane < propan-1-ol < propanoic acid
B.Propanoic acid < propan-1-ol < propane
C.Propan-1-ol < propane < propanoic acid
D.Propane < propanoic acid < propan-1-ol
Explanation: Propane experiences only weak London dispersion forces (b.p. -42 °C). Propan-1-ol forms intermolecular hydrogen bonds via its -OH group (b.p. 97 °C). Propanoic acid forms stable hydrogen-bonded dimers via its -COOH group, leading to stronger intermolecular attraction and a higher boiling point (141 °C). Thus: propane < propan-1-ol < propanoic acid.
6How many resonance structures are required to accurately represent the electron delocalization in the nitrate anion (NO₃⁻)?
A.1
B.2
C.3
D.4
Explanation: The nitrate ion (NO₃⁻) has 24 valence electrons (5 from N, 3×6 from O, +1 charge). Its Lewis structure contains one N=O double bond and two N-O single bonds. The double bond can be placed on any of the three oxygen atoms, giving 3 equivalent resonance structures with a N-O bond order of 1.33.
7What are the hybridization states of the carbon atoms in ethane (C₂H₆), ethene (C₂H₄), and ethyne (C₂H₂), respectively?
A.sp³, sp², sp
B.sp², sp³, sp
C.sp, sp², sp³
D.sp³, sp, sp²
Explanation: In ethane (C₂H₆), carbon forms 4 single sigma bonds (sp³). In ethene (C₂H₄), carbon forms 3 sigma bonds and 1 pi bond in a double bond (sp²). In ethyne (C₂H₂), carbon forms 2 sigma bonds and 2 pi bonds in a triple bond (sp).
8Which predominant type of bonding or interaction holds together solid iodine (I₂), solid ice (H₂O), and solid sodium chloride (NaCl), respectively?
A.London dispersion forces; Hydrogen bonding; Ionic bonding
B.Ionic bonding; Covalent bonding; Metallic bonding
C.Dipole-dipole forces; London dispersion forces; Ionic bonding
D.Hydrogen bonding; Covalent bonding; Ionic bonding
Explanation: Solid I₂ consists of nonpolar diatomic molecules held in a crystal lattice by London dispersion forces. Solid ice consists of polar H₂O molecules held by intermolecular hydrogen bonds. Solid NaCl is an ionic crystal held together by strong electrostatic ionic bonds.
9Iron crystallizes in a body-centered cubic (BCC) unit cell with an edge length of 286.6 pm. If the molar mass of iron is 55.85 g/mol, what is the theoretical density of iron? (Avogadro's number = 6.022 × 10²³ mol⁻¹)
A.7.88 g/cm³
B.3.94 g/cm³
C.15.76 g/cm³
D.5.25 g/cm³
Explanation: A BCC unit cell contains Z = 2 atoms. Volume V = a³ = (2.866 × 10⁻⁸ cm)³ = 2.354 × 10⁻²³ cm³. Mass of unit cell m = Z × M / N_A = 2 × 55.85 / (6.022 × 10²³) = 1.855 × 10⁻²² g. Density ρ = m / V = 1.855 × 10⁻²² / 2.354 × 10⁻²³ = 7.88 g/cm³.
10Why does carbon dioxide (CO₂) have a zero dipole moment while sulfur dioxide (SO₂) has a non-zero dipole moment?
A.CO₂ is linear so its C=O bond dipoles cancel out, whereas SO₂ is bent due to a lone pair on sulfur.
B.SO₂ contains ionic bonds while CO₂ contains covalent bonds.
C.C=O bonds are nonpolar while S=O bonds are polar.
D.CO₂ is a gas at room temperature while SO₂ is a solid.
Explanation: CO₂ has 2 electron domains around carbon (AX₂), giving a linear shape (180°) where two equal and opposite C=O bond dipoles cancel (μ = 0). SO₂ has 3 electron domains around sulfur (AX₂E, 1 lone pair), creating a bent shape (~119°), so its S=O bond dipoles do not cancel (μ > 0).

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