100+ Free Scio NSZ Chemistry Practice Questions

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Key Facts: Scio NSZ Chemistry Exam

The Scio NSZ Chemie is a 30-task, 40-minute standardized university admissions chemistry exam with 4 choices, a candidate-supplied calculator allowed, and a -1/3 penalty; scored in percentiles.

Sample Scio NSZ Chemistry Practice Questions

Try these sample questions to test your Scio NSZ Chemistry exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Which of the following sets of quantum numbers (n, l, ml, ms) is physically permissible for an electron occupying a 3d atomic orbital?

A.n = 3, l = 2, ml = -1, ms = +1/2

B.n = 3, l = 1, ml = 0, ms = +1/2

C.n = 3, l = 3, ml = +1, ms = -1/2

D.n = 4, l = 2, ml = -2, ms = +1/2

Explanation: For a 3d orbital, the principal quantum number is n = 3 and the azimuthal (orbital angular momentum) quantum number is l = 2 (where l = 0 is s, 1 is p, 2 is d, 3 is f). The magnetic quantum number ml can take integer values from -l to +l (i.e., -2, -1, 0, +1, +2), and the spin quantum number ms must be either +1/2 or -1/2. Therefore, n = 3, l = 2, ml = -1, ms = +1/2 is a valid set.

2What is the ground-state electron configuration of the iron(III) cation, Fe3+ (atomic number Z = 26)?

A.[Ar] 4s2 3d3

B.[Ar] 3d5

C.[Ar] 4s1 3d4

D.[Ar] 3d6

Explanation: Neutral iron has 26 electrons with the ground-state electron configuration [Ar] 4s2 3d6. When first-row transition metals undergo ionization, electrons are preferentially removed first from the outermost 4s subshell before any 3d electrons are removed. Ionization to Fe2+ removes the two 4s electrons yielding [Ar] 3d6. Further ionization to Fe3+ removes one 3d electron, resulting in the stable half-filled d-subshell configuration [Ar] 3d5.

3Based on Pauling electronegativity values, which of the following diatomic chemical bonds possesses the greatest percentage of ionic character?

A.C–H (electronegativities: C = 2.55, H = 2.20)

B.H–F (electronegativities: H = 2.20, F = 3.98)

C.Cs–F (electronegativities: Cs = 0.79, F = 3.98)

D.Na–Cl (electronegativities: Na = 0.93, Cl = 3.16)

Explanation: The percentage of ionic character in a chemical bond increases monotonically with the magnitude of the electronegativity difference (ΔEN) between the two bonded atoms. For Cs–F, ΔEN = 3.98 - 0.79 = 3.19, which is the largest electronegativity difference possible between stable elements in the periodic table, giving Cs–F over 90% ionic character.

4According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, what is the molecular geometry of sulfur tetrafluoride, SF4?

A.Regular tetrahedral

B.Square planar

C.Trigonal pyramidal

D.Seesaw (sawhorse)

Explanation: In SF4, the central sulfur atom has 6 valence electrons and forms 4 single covalent bonds with fluorine atoms, leaving one non-bonding electron pair (lone pair). This results in a steric number of 5 (AX4E notation). The electron pair geometry is trigonal bipyramidal. To minimize equatorial-axial repulsions, the lone pair occupies an equatorial position, resulting in a seesaw (sawhorse) molecular shape with bond angles of slightly less than 90° and 120°.

5What is the hybridization state of the central carbon atom and the approximate O–C–O bond angle in the carbonate anion, CO3^2-?

A.sp2 hybridization, 120°

B.sp3 hybridization, 109.5°

C.sp hybridization, 180°

D.sp3d hybridization, 90° and 120°

Explanation: In the carbonate ion (CO3^2-), the central carbon atom forms three equivalent sigma bonds to three oxygen atoms through resonance and has zero non-bonding electron pairs. The steric number is 3, corresponding to sp2 hybridization. The three sp2 hybrid orbitals arrange in a symmetrical trigonal planar geometry with identical O–C–O bond angles of exactly 120°.

6Which of the following series arranges the Group 16 binary hydrides in order of increasing normal boiling point?

A.H2O < H2S < H2Se < H2Te

B.H2S < H2Se < H2Te < H2O

C.H2Te < H2Se < H2S < H2O

D.H2S < H2O < H2Se < H2Te

Explanation: Water (H2O) has an anomalously high boiling point (+100.0 °C) due to strong intermolecular hydrogen bonding formed between the highly electronegative oxygen atom and hydrogen atoms. Among the heavier hydrides (H2S, H2Se, H2Te), hydrogen bonding is absent because S, Se, and Te are insufficiently electronegative. In these molecules, dispersion forces dominate, which increase with increasing molecular mass and polarizability: H2S (-60.3 °C) < H2Se (-41.2 °C) < H2Te (-2.2 °C). Therefore, the correct order is H2S < H2Se < H2Te < H2O.

7Complete combustion of 0.440 g of an organic compound containing only C, H, and O yields 0.880 g of CO2 (M = 44.01 g/mol) and 0.360 g of H2O (M = 18.02 g/mol). What is the empirical formula of this compound?

A.CH2O

B.C2H6O

C.C2H4O

D.C3H6O2

Explanation: Moles of C = n(CO2) = 0.880 g / 44.01 g/mol = 0.0200 mol; mass of C = 0.0200 mol * 12.011 g/mol = 0.240 g. Moles of H = 2 * n(H2O) = 2 * (0.360 g / 18.02 g/mol) = 0.0400 mol; mass of H = 0.0400 mol * 1.008 g/mol = 0.0403 g. Mass of O = total mass - m(C) - m(H) = 0.440 g - (0.240 g + 0.0403 g) = 0.160 g; moles of O = 0.160 g / 16.00 g/mol = 0.0100 mol. Molar ratio C : H : O = 0.0200 : 0.0400 : 0.0100 = 2 : 4 : 1. The empirical formula is C2H4O.

8Aluminum reacts with iron(III) oxide via the thermite reaction: 2 Al(s) + Fe2O3(s) -> Al2O3(s) + 2 Fe(l) If 54.0 g of Al (M = 27.0 g/mol) is mixed with 200.0 g of Fe2O3 (M = 160.0 g/mol) and ignited, what is the maximum theoretical yield of metallic iron (M = 55.85 g/mol)?

A.55.9 g

B.139.6 g

C.223.4 g

D.111.7 g

Explanation: Calculate available moles of reactants: n(Al) = 54.0 g / 27.0 g/mol = 2.00 mol. n(Fe2O3) = 200.0 g / 160.0 g/mol = 1.25 mol. According to stoichiometry, 2.00 mol Al requires 1.00 mol Fe2O3. Since 1.25 mol Fe2O3 is present, Al is the limiting reactant and Fe2O3 is in excess. From 2 mol Al, the reaction produces 2 mol Fe. Theoretical yield of Fe = 2.00 mol * 55.85 g/mol = 111.7 g.

9How many grams of distilled water must be added to 300 g of a 40.0% (w/w) sodium hydroxide (NaOH) solution to dilute it to a 15.0% (w/w) solution?

A.500 g

B.400 g

C.600 g

D.800 g

Explanation: Mass of solute NaOH in the initial solution: m(NaOH) = 300 g * 0.400 = 120 g. Let m(water) be the mass of water added. The target mass fraction is w = 0.150: w = m(NaOH) / [m(initial solution) + m(water)] = 0.150 120 g / (300 g + m(water)) = 0.150 300 g + m(water) = 120 g / 0.150 = 800 g m(water) = 800 g - 300 g = 500 g.

10A chemist needs to prepare 500.0 mL of a 0.200 M sulfuric acid (H2SO4, M = 98.08 g/mol) aqueous solution. The stock solution is concentrated sulfuric acid with a mass fraction of 96.0% (w/w) and a density of 1.84 g/cm3. What volume of concentrated stock solution is required?

A.2.78 mL

B.5.55 mL

C.10.2 mL

D.18.0 mL

Explanation: Step 1: Calculate moles of H2SO4 required: n = c * V = 0.200 mol/L * 0.5000 L = 0.1000 mol. Step 2: Calculate mass of pure H2SO4 needed: m(pure) = 0.1000 mol * 98.08 g/mol = 9.808 g. Step 3: Calculate mass of 96.0% stock solution needed: m(stock) = 9.808 g / 0.960 = 10.217 g. Step 4: Convert stock solution mass to volume using density: V(stock) = m(stock) / rho = 10.217 g / (1.84 g/mL) = 5.55 mL.

About the Scio NSZ Chemistry Exam

The Scio NSZ Chemie is the standardized national comparative examination in chemistry utilized for university admissions across the Czech Republic and Slovakia by chemical engineering, pharmaceutical, medical, biochemical, and agricultural faculties (including University of Chemistry and Technology Prague — VŠCHT Praha, Charles University Faculty of Pharmacy, and faculties of science). The test consists of 30 multiple-choice tasks completed in 40 minutes. Candidates may bring their own calculator (not a phone, tablet or laptop); the online sitting supplies one in the test application. Each question offers 4 choices, with a -1/3 point penalty for incorrect answers. Scio states the syllabus as chemistry 'in the scope taught at gymnázia', publishing topic areas for inorganic chemistry, organic chemistry and biochemistry but no percentage weightings. This practice bank is an English-language study adaptation of that scope, emphasizing stoichiometry, equilibrium problem-solving, organic synthesis pathways, and biochemical mechanisms; the official test is administered in Czech.

Questions

30 scored questions

Time Limit

40 minutes.

Passing Score

Harmonized percentile score (0–100); admissions requirements set by participating faculties (such as VŠCHT Praha, Faculty of Pharmacy UK in Hradec Králové, and Science faculties).

Exam Fee

920 CZK (Mini), 1,290 CZK (Standard) or 1,490 CZK (Komplet) per exam date; free with Scio's social discount or a faculty voucher. (Scio s.r.o. (Národní srovnávací zkoušky — NSZ))

Scio NSZ Chemistry Exam Content Outline

30%

Obecna a fyzikalni chemie (General & Physical Chemistry)

Atomic orbitals, electronegativity, stoichiometry calculations, thermochemistry, reaction orders, chemical equilibrium, pH calculations, and redox potentials.

25%

Anorganicka chemie (Inorganic Chemistry)

Properties and reactions of main-group and transition elements, acid-base behavior of oxides, coordination chemistry, and inorganic nomenclature.

30%

Organicka chemie (Organic Chemistry)

IUPAC systematic nomenclature, structural/stereoisomerism, electrophilic addition/substitution, nucleophilic substitution/elimination, and functional group conversions.

15%

Biochemie (Biochemistry)

Biomolecules (carbohydrates, lipids, proteins, vitamins), enzyme catalysis, Michaelis-Menten concepts, ATP energetics, and central metabolic pathways.

How to Pass the Scio NSZ Chemistry Exam

What You Need to Know

Passing score: Harmonized percentile score (0–100); admissions requirements set by participating faculties (such as VŠCHT Praha, Faculty of Pharmacy UK in Hradec Králové, and Science faculties).
Exam length: 30 questions
Time limit: 40 minutes.
Exam fee: 920 CZK (Mini), 1,290 CZK (Standard) or 1,490 CZK (Komplet) per exam date; free with Scio's social discount or a faculty voucher.

Keys to Passing

Work through all 100 available questions
Review every answer and explanation
Track weak areas and revisit them
Use our AI tutor for tough concepts

Scio NSZ Chemistry Study Tips from Top Performers

1Master fast stoichiometric calculations: molar mass calculations, mass fraction (w = m_solute / m_solution), and molar concentration (c = n / V).

2Review the pH formulas for strong acids/bases (pH = -log[H+]) and weak acids/bases (pH = 1/2(pKa - log c)), and know how buffer solutions resist pH shifts.

3Understand organic reaction mechanisms: Markovnikov addition to alkenes, electrophilic aromatic substitution directing effects (ortho/para vs. meta), and esterification equilibria.

4Memorize the key biochemical tests and structures: Fehling's/Tollens' reagent for reducing sugars, biuret test for peptide bonds, and the structure of alpha-amino acids.

5Practice past Scio chemistry papers with your calculator to get accustomed to performing multi-step stoichiometric conversions within the 80-second-per-question pace.

Frequently Asked Questions

Can I use a calculator during the Scio NSZ Chemie test?

Yes. A basic or scientific non-programmable calculator without memory storage, text-display capability, or graphical functions is permitted during the exam.

Is a periodic table provided in the examination?

No. Scio's only stated aid for the chemistry test is a calculator that you bring yourself (or the one built into the online test application). The published NSZ chemistry papers contain the 30 tasks and nothing else — no periodic table and no constants sheet — so learn the common relative atomic masses before the sitting.

How many questions are on the Scio Chemie test and what is the time limit?

The examination contains exactly 30 multiple-choice questions and lasts 40 minutes (an average of 80 seconds per question).

What is the penalty for guessing incorrectly on Scio Chemie?

Each correct answer awards 1 point, unanswered questions award 0 points, and each wrong answer subtracts 1/3 of a point (-0.33).

100+ Free Scio NSZ Chemistry Practice Questions

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