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Exam Review

Key Facts: ЦТ Chemistry Exam

38

total tasks (16 Part A + 22 Part B)

РИКЗ Specification 2026

150 min

exam duration (2.5 hours)

РИКЗ

0–100

score scale

Ministry of Education of Belarus

4.50 BYN

registration fee per subject (0.1 base amount)

РИКЗ 2026

20 / 10

admission threshold points (1st / 2nd profile)

Ministry of Education Rules

The Belarus ЦТ/ЦЭ Chemistry exam contains 38 tasks (16 Part A + 22 Part B) completed in 150 minutes. Scored on a 0–100 scale with admission thresholds of 20 points (1st profile) and 10 points (2nd profile), registration costs 0.1 base amount (4.50 BYN in 2026). Practice with 94 questions covering general, inorganic, and organic chemistry.

Sample ЦТ Chemistry Practice Questions

Try these sample questions to review concepts for the ЦТ Chemistry exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 94+ question experience with AI tutoring.

1Naturally occurring chlorine consists of two stable isotopes: ³⁵Cl and ³⁷Cl. Which statement correctly compares an atom of ³⁵Cl with an atom of ³⁷Cl?
A.Both isotopes have 17 protons and 17 electrons, but ³⁷Cl has 20 neutrons while ³⁵Cl has 18 neutrons.
B.³⁷Cl has 19 protons and 18 neutrons, whereas ³⁵Cl has 17 protons and 18 neutrons.
C.Both isotopes have the same mass number, but ³⁷Cl has a higher atomic number.
D.³⁵Cl has 17 neutrons and ³⁷Cl has 19 neutrons, but both have 18 electrons.
Explanation: Isotopes are nuclides of the same chemical element having identical atomic numbers Z (number of protons and electrons in a neutral atom = 17 for chlorine) but different mass numbers A due to differing numbers of neutrons N (N = A - Z). For ³⁵Cl, N = 35 - 17 = 18 neutrons; for ³⁷Cl, N = 37 - 17 = 20 neutrons.
2What is the ground-state valence electron configuration and the number of unpaired electrons in an isolated phosphorus atom (Z = 15)?
A.3s² 3p³ with 3 unpaired electrons
B.3s² 3p⁴ with 2 unpaired electrons
C.3s¹ 3p⁴ with 3 unpaired electrons
D.3s² 3p¹ with 1 unpaired electron
Explanation: Phosphorus (atomic number 15) has the full ground-state electron configuration 1s² 2s² 2p⁶ 3s² 3p³. The valence shell (n = 3) contains 3s² 3p³. According to Hund's rule, the three 3p electrons occupy three separate p-orbitals (3px¹, 3py¹, 3pz¹) with parallel spins, giving exactly 3 unpaired electrons.
3Which of the following represents the anomalous ground-state electron configuration of a neutral copper atom (₂₉Cu)?
A.[Ar] 3d⁹ 4s²
B.[Ar] 3d⁸ 4s² 4p¹
C.[Ar] 3d¹⁰ 4s¹
D.[Ar] 3d¹⁰ 4p¹
Explanation: Copper (Z = 29) exhibits an electron configuration anomaly due to the enhanced quantum-mechanical exchange stability of a completely filled d-subshell (3d¹⁰). An electron shifts from the 4s subshell to the 3d subshell, yielding [Ar] 3d¹⁰ 4s¹ rather than the expected [Ar] 3d⁹ 4s².
4How does atomic radius change across Period 3 of the Periodic Table from sodium (Na) to chlorine (Cl), and what is the primary reason?
A.It increases because additional electron shells are added.
B.It remains constant because shielding by core electrons offsets nuclear charge completely.
C.It increases because electron-electron repulsion in the valence shell expands the electron cloud.
D.It decreases because the effective nuclear charge increases while the number of electron shells remains constant.
Explanation: Across Period 3 from Na (Z = 11) to Cl (Z = 17), electrons are added to the same principal energy level (n = 3) while the nuclear charge increases from +11 to +17. The core electron shielding remains essentially constant, resulting in a higher effective nuclear charge (Z_eff) that pulls the valence electron cloud closer to the nucleus, decreasing atomic radius.
5Why is the first ionization energy of nitrogen (N, Z = 7) higher than that of oxygen (O, Z = 8), despite oxygen having a higher nuclear charge?
A.Nitrogen has a stable half-filled 2p³ subshell, whereas oxygen has paired electrons in a 2p orbital that experience inter-electronic repulsion.
B.Oxygen has its valence electrons in a higher principal quantum shell than nitrogen.
C.Nitrogen has greater electronegativity and higher nuclear charge than oxygen.
D.Oxygen has a completely filled 2p⁶ subshell that shields valence electrons more effectively.
Explanation: Nitrogen has a 2s² 2p³ configuration with three singly occupied 2p orbitals (half-filled subshell, extra exchange stability). Oxygen has a 2s² 2p⁴ configuration where one 2p orbital contains two paired electrons. The electrostatic repulsion between the two electrons sharing the same 2p orbital in oxygen makes it easier to remove one electron, resulting in a lower first ionization energy than nitrogen.
6Which of the following lists the elements in order of strictly INCREASING electronegativity according to the Pauling scale?
A.F < O < N < C < Li
B.Si < Al < Mg < Na < K
C.Cs < Ca < Si < P < F
D.Cl < Br < I < At < F
Explanation: Electronegativity increases across a period from left to right and decreases down a group. Cesium (Cs, ~0.79) is an alkali metal with very low electronegativity, followed by calcium (Ca, 1.00), silicon (Si, 1.90), phosphorus (P, 2.19), and fluorine (F, 3.98, the most electronegative element).
7Which pair of substances contains ONLY compounds with purely ionic bonding in their crystalline state?
A.HCl and CH₄
B.CO₂ and SiO₂
C.NH₃ and H₂O
D.NaCl and BaCl₂
Explanation: Ionic bonding occurs between elements with a large difference in electronegativity (typically an active metal and a non-metal). In NaCl and BaCl₂, electrons are transferred to form cations (Na⁺, Ba²⁺) and anions (Cl⁻), which are held together by electrostatic forces in an ionic crystal lattice.
8Which of the following molecules contains polar covalent bonds but has a net dipole moment of zero (is non-polar) due to its symmetrical molecular geometry?
A.H₂O
B.SO₂
C.NH₃
D.CCl₄
Explanation: Carbon tetrachloride (CCl₄) contains four polar C-Cl covalent bonds. Because carbon is sp³ hybridized with a symmetrical tetrahedral geometry, the four individual bond dipole vectors point toward the vertices of a regular tetrahedron and vectorially sum to zero, giving a non-polar molecule with μ = 0.
9In the formation of the ammonium cation (NH₄⁺) from ammonia (NH₃) and a hydrogen ion (H⁺), what type of covalent bonding mechanism occurs?
A.Donor-acceptor (coordinate covalent) mechanism
B.Free-radical exchange mechanism
C.Metallic delocalization mechanism
D.Intermolecular hydrogen bonding
Explanation: In NH₃, the nitrogen atom has an unshared electron pair (lone pair). The hydrogen ion (H⁺) has an empty 1s orbital. Nitrogen acts as an electron pair donor and H⁺ acts as an electron pair acceptor, forming a coordinate (donor-acceptor) covalent bond: NH₃ + H⁺ → NH₄⁺. Once formed, all four N-H bonds in NH₄⁺ are completely equivalent.
10Which of the following compounds exhibits strong intermolecular hydrogen bonding in the liquid state, resulting in an anomalously high boiling point compared to its heavier group congeners?
A.HCl
B.HBr
C.HF
D.HI
Explanation: Hydrogen fluoride (HF) exhibits strong intermolecular hydrogen bonds (H···F) because fluorine is the most electronegative element with a very small atomic radius. This creates a large dipole and strong electrostatic attraction between molecules. In contrast, HCl, HBr, and HI have lower electronegativity and larger radii, so their boiling points are dominated by weaker London dispersion and dipole forces.

About the ЦТ Chemistry Exam

The Chemistry Centralized Testing (ЦТ) and Centralized Exam (ЦЭ) is the standardized admissions examination administered by RIKZ for medical, chemical, and pharmaceutical university programs in Belarus. The exam consists of 38 tasks (16 closed-ended Part A items and 22 open-ended Part B items) completed over 150 minutes (2.5 hours). A simple single-line calculator (arithmetic operations, square root, percentages, one memory cell) is permitted. This practice bank provides 94 comprehensive questions with detailed explanations and step-by-step chemical calculations.

Exam sponsor: Republican Institute for Knowledge Control (РИКЗ). The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Questions

38 questions

Time Limit

150 minutes

Passing Score

20/100 (1st profile) or 10/100 (2nd profile)

Exam / Certification Fees

4.50 BYN

Exam sponsor website

Fees, eligibility, and exam policies can change. Confirm them with the exam sponsor before applying or paying.

Our practice resources: topics covered

We aim to reflect publicly available exam outlines and topic information in our study resources. Coverage, format, and difficulty may differ from the actual exam, and we cannot guarantee that every detail is accurate or current. Confirm exam requirements, fees, and policies with the official exam sponsor.

~35%

General Chemistry

Electronic configurations, electronegativity, types of bonds, reaction rate, Le Chatelier's principle, hydrolysis, and redox balancing.

~30%

Inorganic Chemistry

Chemical properties and reactions of main group non-metals, s/p/d metals, amphoteric oxides/hydroxides, and complex salts.

~25%

Organic Chemistry

Homologous series, isomerism, reaction mechanisms, functional groups, genetic links between classes, and biopolymers.

~10%

Calculations & Lab Methods

Mole concepts, mass fractions in solutions, excess/limiting reagents, gas laws, and yield calculations.

Preparing for the ЦТ Chemistry Exam

What You Need to Know

  • Passing score: 20/100 (1st profile) or 10/100 (2nd profile)
  • Exam length: 38 questions
  • Time limit: 150 minutes
  • Exam / certification fees: 4.50 BYN Official sources

Using Our Practice Resources

  • Work through all 94 available questions
  • Review every answer and explanation
  • Track weak areas and revisit them
  • Use our AI tutor for tough concepts

ЦТ Chemistry: Suggested Study Strategy

1Master electronic configurations and orbital diagrams of elements in both ground and excited states.
2Practice balancing redox reactions using the electron balance method and ion-electron method.
3Memorize characteristic qualitative reactions for cations (Fe2+, Fe3+, Cu2+, NH4+, Ba2+, Ag+) and anions (SO42-, Cl-, Br-, I-, CO32-, PO43-).
4Work through the 100 practice questions to master multi-step stoichiometry and genetic link reaction chains.

Frequently Asked Questions

What is the structure of the ЦТ Chemistry exam in Belarus?

The official RIKZ exam consists of 38 tasks (16 in Part A and 22 in Part B) to be solved within 150 minutes.

Are calculators and reference tables allowed in ЦТ Chemistry?

Yes, simple non-programmable calculators are permitted. The exam booklet includes standard reference tables: the Periodic Table, Solubility Table, and Electrochemical Series of Metals.

What is the passing cutoff for medical university admission?

When Chemistry is the 1st profile subject (such as for medical and pharmaceutical faculties), the minimum threshold is 20 points out of 100. As a 2nd profile subject, it is 10 points.

How much is the ЦТ Chemistry exam fee?

The registration fee is 0.1 base amount (4.50 BYN in 2026) for Centralized Testing (ЦТ). Graduating high school students taking the Centralized Exam (ЦЭ) sit it for free.

Is this bank the same language and format as the real ЦТ Chemistry paper?

No. The official RIKZ paper is sat in Belarusian or Russian and combines Part A items with five answer options and 22 open-response Part B tasks. This bank is an English-language MCQ study adaptation with four options per question — a study aid for the official Программа вступительных испытаний content, not an official translation or a format simulation.