Matter, Atoms and Chemical Change: Cumulative Review

Bring the four lessons together

Use these 24 original questions to check the distinctions that connect matter, formulae, atomic structure and reactions. They are interleaved so you must choose the right idea before calculating. Each question has one defensible answer under its stated conditions.

This is free, untimed learning practice. Score +1 for a correct answer and 0 for an incorrect or unattempted answer, with no negative marking. There is no imposed pass cutoff. The total 24 follows the distinct learning decisions; it is not a prediction of RRB chemistry weightage or a full CBT simulation. English and Hindi are two versions of the same 24 items, not 48 separate questions.

Try the questions before reading the explanations. Afterwards, explain why your chosen answer works and why the alternatives fail. A correct guess without the reason identifies something to revisit. If you miss a distinction, revisit the matching teaching lesson in this module before trying again.

All masses, ion charges and unfamiliar reactivity/solubility facts needed are supplied. Use the taught shell model for neutral atoms of the first 18 elements only where stated. Average atomic mass must be interpreted, but concentration percentages and weighted-average arithmetic are not tested. The optional electron-accounting bridge is not tested either. No question asks you to taste, smell, mix, heat, generate gases or handle chemical materials.

Figure CFREV-V01. Choose the check that matches the question

What is being asked?	The check to use
State or composition?	Separate physical state from element/compound/mixture
Formula or particle?	Read the complete group, subscript, charge and coefficient
Element, isotope or charge?	Keep proton, neutron and electron counts separate
Atomic mass or mass number?	Distinguish an average/measured mass from integer nucleon count
Reaction or observation?	Use substance identity, fixed formulae, atom account and stated conditions
More than one reaction label?	Ask whether the labels describe structure, energy, precipitation or redox

Questions

1. Two samples are described as pure helium gas and pure liquid water H₂O. Which statement correctly separates composition from physical state?

A. Both are elements because both are pure B. Water is a mixture because H₂O has two different element symbols C. Helium is an element and water is a compound; gas or liquid is a separate classification D. Both are mixtures because both contain very many particles

2. What does the written species CO₃²⁻ represent?

A. One polyatomic ion containing one C atom and three O atoms, with net charge −2 on the group B. A neutral molecule containing one C atom and three O atoms C. Two carbon atoms and three oxygen atoms, each with charge −1 D. Three separate oxide ions, with no carbon atom

3. A species has 12 protons, 13 neutrons and 10 electrons. Which record is correct?

A. Z = 10, A = 23, neutral B. Z = 12, A = 35, charge +2 C. Z = 12, A = 25, charge −2 D. Z = 12, A = 25, charge +2

4. A glass plate breaks into fragments. The description explicitly says no new chemical substance is formed. Which conclusion is justified?

A. It is chemical because joining the original plate back is difficult B. It is physical in this description; difficulty of reversal is not the defining test C. It is chemical because a solid must become a new element when broken D. No change occurred because the chemical substance stayed the same

5. A dry mixture contains metal clips attracted by a magnet and plastic beads that are not attracted. A method card says magnetic separation uses a difference in magnetic response. Which choice uses the supplied property correctly?

A. Magnetic separation can select the clips because the two components respond differently B. Ordinary filtration must separate any two solids even without a liquid C. Evaporation must separate them because plastic is always volatile D. Chromatography must separate them solely because their colours differ

6. Use the supplied ions Zn²⁺ and S²⁻. Which neutral formula gives their smallest whole-number ratio?

A. Zn₂S₂ B. ZnS₂ C. ZnS D. Zn₂S

7. Species 1 is ³⁶₁₈Ar, species 2 is ³⁸₁₈Ar and species 3 is ³⁶₁₆S. Upper-left numbers are A and lower-left numbers are Z. Which pairing is correct?

A. 1 and 2 are isobars; 1 and 3 are isotopes B. 1 and 2 are isotopes; 1 and 3 are isobars C. All three are one element because two have A = 36 D. 2 and 3 are isotopes because their mass numbers are close

8. For the supplied valid reaction N₂ + H₂ → NH₃ under suitable conditions, which coefficients in that order give the smallest positive whole-number balance?

A. 1, 1, 1 B. 2, 3, 1 C. 1, 3, 2 D. 1, 2, 2

9. Pure water vapour condenses into liquid water without a chemical reaction. Which model change is appropriate?

A. H₂O groups become separate H₂ and O₂ molecules B. Each H₂O group grows into a larger kind of particle C. All particles stop moving because the gas becomes liquid D. The H₂O groups keep their identity while their arrangement and movement change

10. How many N, H, C and O atoms are represented by 2(NH₄)₂CO₃?

A. N = 2, H = 8, C = 1, O = 3 B. N = 4, H = 16, C = 2, O = 6 C. N = 4, H = 8, C = 2, O = 6 D. N = 4, H = 16, C = 2, O = 3

11. A neutral atom has Z = 15. In the taught shell model for the first 18 elements, which description correctly gives its arrangement, outer-electron count and simple octet-completion valency?

A. 2, 8, 5; five outer electrons; simple valency 3 B. 2, 8, 5; five outer electrons; simple valency 5 C. 2, 5, 8; eight outer electrons; simple valency 0 D. 2, 8, 3; three outer electrons; simple valency 3

12. A recorded open reaction system increases from 68.5 g to 70.1 g. The supplied account states that only oxygen enters and no matter leaves. Which explanation respects conservation?

A. 1.6 g of matter was created from nothing B. Oxygen has no mass, so the gain cannot be related to it C. 1.6 g of oxygen entered the measured system; the full account includes that incoming material D. Conservation requires every open-system reading to be constant

13. R and T are known mixtures in an idealised school comparison. After 20 minutes R has visible sediment, is retained by ordinary filter paper and scatters light. T does not settle, passes the paper and gives no visible light path under the stated test. Which interpretation is supported?

A. R must be a colloid solely because it scatters light B. T must be pure because no light path is seen C. R and T must both be solutions because they contain liquids D. R fits a suspension and T a solution; R also shows why scattering alone is not a colloid test

14. Ammonia is described here as discrete NH₃ molecules. With supplied rounded masses N = 14 u and H = 1 u, which answer has both the correct value and quantity name?

A. Formula-unit mass 17 u for an ionic crystal B. Molecular mass 17 u C. Molecular mass 15 u D. Molecular mass 17 g

15. Be and N are neutral main-group atoms in period 2. Use the broad across-period trend with comparable atomic-radius definitions. Which statement is appropriate?

A. N must be larger solely because its atomic number is higher B. Be is generally larger than N in this comparison; the across-period attraction trend tends to reduce size C. Their radii must be identical because both occupy two shells D. They cannot share a period because their outer-electron counts differ

16. Two valid reactions are supplied for classification only: I: 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂. II: Mg + ZnSO₄ → MgSO₄ + Zn. For II, magnesium is given as more reactive than zinc under the stated conditions. Which structural pairing is correct?

A. I combination; II decomposition B. I double displacement; II combination C. I decomposition; II displacement D. I precipitation; II always impossible because both are metals

17. A described plan lets water leave a solution as vapour and keeps a stable nonvolatile dissolved solid behind. The goal now changes: the water must also be collected separately. Which conceptual change addresses that new goal?

A. Include vapour condensation and collection as in a suitable distillation design B. Pass the unchanged solution through ordinary filter paper and assume dissolved material is trapped C. Use a magnet regardless of the components’ magnetic properties D. Keep the same open-evaporation plan and call the escaped vapour collected water

18. Given Mg²⁺ and the intact nitrate ion NO₃⁻, which formula is neutral and preserves the nitrate group?

A. Mg₂NO₃ B. MgNO₃ C. Mg₂(NO₃)₂ D. Mg(NO₃)₂

19. A table lists argon Z = 18 and a rounded average atomic mass 39.9 u. A learner subtracts 18 from 39.9 and says a particular argon atom has 21.9 neutrons. Which error is central?

A. The rule neutron count = A − Z is never valid B. Argon cannot contain neutrons because it is a gas C. The learner used a population-average mass as one isotope’s integer mass number A D. An average atomic mass can describe only ions, never atoms

20. The supplied product in Fe + O₂ → Fe₂O₃ must remain iron(III) oxide. A learner changes it to FeO₂ so the atom totals match. Which feedback is correct?

A. Keep Fe₂O₃ fixed and balance coefficients: 4Fe + 3O₂ → 2Fe₂O₃ B. Accept FeO₂ because any formula is allowed if atom counts match C. Keep Fe₂O₃ but use 2Fe + O₂ → Fe₂O₃; all counts then match D. The supplied equation cannot be balanced without changing a subscript

21. The supplied solution reaction is AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq). AgCl is stated to form an insoluble solid. Which description is correct?

A. Combination, because there is only one product substance B. Decomposition, because one substance alone breaks apart C. An element displaces another element, with no exchange between compounds D. Double displacement and precipitation both apply

22. A supplied reaction description explicitly says the overall process absorbs heat from its surroundings. What follows from that energy statement alone?

A. It is exothermic and must be combination B. It is endothermic; its structural reaction type is not fixed by this statement alone C. It must be decomposition because every heat-absorbing reaction is decomposition D. It cannot be a chemical reaction because energy is required

23. For the supplied reaction ZnO + C → Zn + CO, identify the paired changes by oxygen transfer. Which account is correct?

A. ZnO loses oxygen and is reduced; C gains oxygen and is oxidised B. ZnO loses oxygen and is oxidised; C gains oxygen and is reduced C. Both are reduced because oxygen is conserved D. Neither changes chemically because each element has equal atom counts on both sides

24. A report identifies a newly formed green corrosion layer on copper and rust on an iron fitting. Which statement uses the everyday terms correctly?

A. Every metal corrosion product must be called iron rust B. The copper change must be physical merely because a colour is mentioned C. The two layers must be the same compound because both are corrosion D. Rusting is a specific iron-corrosion case; copper corrosion need not be iron rust

Answers and every-option diagnosis

1. C. Helium is an element and water is a compound; gas or liquid is a separate classification

C is correct. Pure substances include both elements and compounds. Helium contains one element; water has hydrogen and oxygen chemically combined in a fixed composition. A incorrectly restricts purity to elements. B treats chemically combined constituents as a mixture. D confuses the number of particles with the number of chemical substances. The state labels do not decide any of these composition categories.

2. A. One polyatomic ion containing one C atom and three O atoms, with net charge −2 on the group

A reads the positions correctly: C has an implied subscript 1, O has subscript 3, and the superscript 2− is the charge of the whole carbonate ion. B ignores the charge. C reads the charge magnitude as a carbon subscript and invents individual charges. D breaks up the written group and deletes carbon. Atom count and net charge are separate information.

3. D. Z = 12, A = 25, charge +2

D follows Z=p = 12, A=p+N = 12 + 13 = 25 and charge number p−e = 12 − 10=+2. A wrongly uses electron count to change the element identity and ignores the imbalance of charges. B includes electrons in A. C reverses the sign: fewer electrons than protons leaves positive net charge. The isotope’s mass number stays 25 when its electron count changes.

4. B. It is physical in this description; difficulty of reversal is not the defining test

B uses the supplied substance-identity evidence. Shape and size can change physically even when restoring the object is difficult. A substitutes reversibility for the definition. C invents an element change. D ignores a real physical change merely because it is not chemical. The question classifies the described breaking, not every possible microscopic event in every material.

5. A. Magnetic separation can select the clips because the two components respond differently

A matches the supplied method principle to the stated difference. B invents a universal filtering rule without the needed setup or size behaviour. C supplies a volatility claim that was not given and is not universally true. D treats colour as the separation mechanism; chromatography needs the relevant differing interactions and a suitable system. No unknown chemical reaction is needed to justify A.

6. C. ZnS

C gives one ion of each kind: +2 − 2 = 0 and ratio 1: 1. A is charge-neutral in its written counts but leaves the reducible ratio 2: 2. B gives +2 − 4=−2; D gives +4 − 2=+2. The question asks a simplest ionic ratio, not reduction of an independently supplied molecular formula.

7. B. 1 and 2 are isotopes; 1 and 3 are isobars

B is correct. Species 1 and 2 have equal Z = 18 and different A, making them isotopes of argon. Species 1 and 3 have equal A = 36 and different Z, making them isobars. A reverses the tests. C ignores the sulfur proton number 16. D replaces equality of Z with closeness of A, which is not an isotope criterion.

8. C. 1, 3, 2

C gives N₂ + 3H₂ → 2NH₃, with N 2/2 and H 6/6. A has N 2/1 and H 2/3. B has N 4/1 and H 6/3. D balances nitrogen but leaves H 4/6. The formulas N₂, H₂ and NH₃ stay fixed; only coefficients change. This numerical interpretation is not an instruction for preparing ammonia.

9. D. The H₂O groups keep their identity while their arrangement and movement change

D models a physical state change. A instead breaks and reforms substances, which the description excludes. B confuses bulk-state behaviour with growth into a new particle identity. C falsely removes particle motion in a liquid. A representative drawing should keep the water groups intact and should not be read as a photograph to scale.

10. B. N = 4, H = 16, C = 2, O = 6

B is correct. One (NH₄)₂CO₃ unit represents N = 2, H = 8, C = 1, O = 3. The outside coefficient 2 doubles every count, giving N = 4, H = 16, C = 2, O = 6. A ignores that outside coefficient. C does not apply the NH₄ group multiplier to all four H atoms. D doubles the other totals but not oxygen. These are represented atom counts, not isolated molecular packets in an ionic solid.

11. A. 2, 8, 5; five outer electrons; simple valency 3

A preserves 15 electrons as 2 + 8 + 5. The outer count is 5; three more complete eight in this introductory model. B equates outer-electron count with valency. C uses the wrong distribution for this taught range. D totals 13 electrons, not 15. The simple model is not a claim that this element has only one possible valency in every compound.

12. C. 1.6 g of oxygen entered the measured system; the full account includes that incoming material

C is correct: 70.1 − 68.5 = 1.6 g. Oxygen was initially outside the measured boundary and is included afterwards. A ignores this supplied source of matter. B denies a gas’s mass. D applies a closed-account conclusion to an open subsystem. No conclusion about mass creation follows from the increased reading.

13. D. R fits a suspension and T a solution; R also shows why scattering alone is not a colloid test

D combines settling, ordinary filtration and the beam observation. A discards the evidence for a coarse suspension and treats scattering as unique. B contradicts the known-mixture premise and overclaims from one observation. C ignores R’s sediment and retained particles. These classifications use the supplied comparison conditions, not an unlimited claim about every real sample or filter.

14. B. Molecular mass 17 u

B is correct: 14 + 3 × 1 = 17 u for the given molecule. A uses the wrong structural description and quantity name. C counts hydrogen only once. D copies the number in u into grams without a conversion. Properly converted single-particle masses can be written in grams, but 17 u is not 17 g.

15. B. Be is generally larger than N in this comparison; the across-period attraction trend tends to reduce size

B is the taught bounded main-group trend: across period 2, increasing proton number with electrons added in the same principal outer shell generally reduces comparable atomic size. A uses higher atomic number to infer the opposite size change. C treats equal shell count as equal radius. D confuses the period criterion with equal outer-electron count. No universal rule for every radius definition, transition element or reactivity is claimed.

16. C. I decomposition; II displacement

C is correct. I has one reactant substance, even though its coefficient is 2, and it forms several product substances: decomposition. In II, magnesium replaces zinc using the supplied reactivity fact: displacement. A reverses inappropriate pattern labels. B invents an exchange in I and overlooks the two products in II. D has no supplied precipitation evidence and wrongly forbids metal displacement. A coefficient counts amount, not the number of different reactant substances.

17. A. Include vapour condensation and collection as in a suitable distillation design

A adds the missing recovery step for the solvent. B falsely assumes ordinary paper removes dissolved material. C supplies no relevant property difference. D changes a label, not the physical recovery: escaped vapour has not been collected. The goal determines why evaporation alone is insufficient here. The answer makes no drinking-safety claim about any recovered water.

18. D. Mg(NO₃)₂

D gives one Mg²⁺ and two NO₃⁻ ions: +2 + 2 × (−1)=0. The brackets repeat all of nitrate. A gives +4 − 1=+3; B gives +2 − 1=+1; C gives +4 − 2=+2. None is neutral with the supplied ions. Keep the 3 within NO₃ fixed rather than changing nitrate to a different group.

19. C. The learner used a population-average mass as one isotope’s integer mass number A

C identifies the quantity mismatch. Neutron count requires the particular isotope’s A, not the average in u. A rejects a valid rule used with the right quantities. B confuses bulk physical state with nuclear composition. D invents an ions-only restriction. Rounding 39.9 would not repair the missing isotope information. The measured mass of an individual isotope is also distinct from A and can be noninteger in u.

20. A. Keep Fe₂O₃ fixed and balance coefficients: 4Fe + 3O₂ → 2Fe₂O₃

A preserves the stated product and gives Fe 4/4 and O 6/6. B changes the chemical description instead of balancing it; matching counts for a different product is not a repair of this equation. C leaves oxygen 2/3. D is disproved by A’s valid coefficient set. The issue is the specified product identity, not a claim about whether every conceivable formula could exist under some other conditions.

21. D. Double displacement and precipitation both apply

D is correct: the reactants exchange ionic partners, and the supplied insoluble AgCl is a precipitate. A ignores the second product. B ignores the two-reactant exchange. C introduces a free-element displacement that the equation does not show. The two correct labels describe different features of the same reaction; they are not mutually exclusive alternatives.

22. B. It is endothermic; its structural reaction type is not fixed by this statement alone

B uses the stated net heat direction without adding a structural conclusion. A reverses the energy definition and invents combination. C treats energy and reactant/product pattern as the same classification. D wrongly excludes reactions that absorb energy. To decide combination, decomposition or another structural type, inspect the supplied chemical equation and conditions.

23. A. ZnO loses oxygen and is reduced; C gains oxygen and is oxidised

A is correct. Oxygen goes from zinc oxide into carbon monoxide: zinc oxide is reduced and carbon is oxidised in the oxygen-transfer model. B reverses both terms. C confuses conservation of oxygen with the direction of oxygen transfer and misses the gaining partner. D mistakes conserved atoms for unchanged substances. The equation is supplied for interpretation only; no heating or gas-handling procedure is implied.

24. D. Rusting is a specific iron-corrosion case; copper corrosion need not be iron rust

D preserves the distinction between the broad process and the specific iron example. A extends the iron term to every metal. B ignores the supplied identification of a newly formed corrosion product and uses colour alone to reverse that evidence. C assumes a shared process label means identical chemical products. The question supplies the report; a surface colour by itself would not establish all those facts.

Use the explanations to choose what to revisit

• Lesson 1, Matter, Mixtures and Separation: items 1, 5, 9, 13, 17
• Lesson 2, Atoms, Molecules and Formulae: items 2, 6, 10, 14, 18
• Lesson 3, Atomic Structure and Periodic Patterns: items 3, 7, 11, 15, 19
• Lesson 4, Reactions and Chemical Equations: items 4, 8, 12, 16, 20–24

For a repeated error, return to the relevant worked case, cover its resolved answer, and explain the next step yourself. Keep the stated assumptions with the question: a supplied reactivity ordering, a known-mixture premise or a specified system boundary is evidence, not optional decoration.

Sources and scope

The official RRB CEN 09/2025 notice, §14.1, printed p.28, gives a Class 10 CBSE-level General Science boundary and an illustrative, non-exhaustive topic list. It does not prescribe these individual lesson titles or review counts.

Scientific references are NCERT Exploration Grade 9 Chapter 5, §§5.1–5.5, pp.73–88; Chapter 8, §§8.4–8.9, pp.148–157; Chapter 9, §§9.1–9.5 and 9.7–9.8, pp.165–180, first edition April 2026; and NCERT Class X Science Chapter 1, §§1.1–1.3, pp.2–13, Reprint 2026–27. CBSE's periodic-classification reading material, Unit 1, §§1.3–1.3.2, PDF pp.6–9, supports bounded periodic comparisons. Its annual formative-assessment status is not an RRB inclusion or exclusion rule. The teaching lessons carry further source and model limitations.

All review wording, options, explanations and the decision panel are original. Common formulae and equations are scientific facts, not copied source exercises. This is one foundation module within a developing course, not the full General Science syllabus.

Analogy

Use a checklist before choosing a tool: identify what the label describes, what must remain fixed and what evidence is supplied. A formula-count problem, an isotope comparison and an open-system mass change need different checks. The analogy helps organise your decision; it cannot replace scientific definitions or make an unsupported reaction occur.

Quick reference

• Pure substance: element or compound; mixture contains more than one chemical substance
• Molecule is neutral; an ion has net charge; an ionic formula unit is a ratio in a crystal
• Coefficients repeat whole formulae; brackets repeat whole groups; subscripts stay fixed when balancing
• Neutral ionic formula: choose the smallest whole-number ratio giving total charge zero
• Molecular/formula-unit mass: multiply atom counts by supplied masses and preserve the unit
• Z counts protons; A counts protons plus neutrons; charge number = protons − electrons
• Same Z/different A: isotopes; same A/different Z: isobars
• Average atomic mass is not one isotope’s A; actual isotope mass is also a distinct quantity
• The shell model stays within the neutral atoms of the first 18 elements as taught; main-group trend limits remain in force
• Separation depends on the differing property and the recovery goal
• A closed mass account includes gas; open measurements need transfer information
• Atom balance does not prove chemical feasibility
• Structure, heat direction, precipitation and redox are different classification questions