Chapter 8 — Bonding and chemical reactions
From particles to substances
Matter may be described at different levels: atoms, molecules, ions and bulk substances. An atom is the smallest unit of an element that retains its identity in ordinary chemical description. A molecule is a bonded group of atoms with a definite composition. Some substances, such as water, are commonly discussed as molecules. An ionic solid such as sodium chloride is better described as an extended array of ions; NaCl is its formula unit, not a little independent NaCl molecule in the crystal. Examinations often use “molecule” loosely. When choosing an answer, follow the explicit physical description and do not confuse a written formula with proof of molecular structure.
An element consists of one kind of atom, even if its particles are molecules such as O₂. A compound combines atoms of different elements in a fixed chemical ratio. A mixture contains substances together without the same kind of chemical combination, and its proportions can vary. Air is a mixture of gases; pure water is a compound with a fixed H:O atom ratio of 2:1. A compound can be separated into simpler substances by appropriate chemical changes; a mixture's components can often be separated by physical methods based on differences in boiling point, solubility, particle size or magnetism. This classification is more reliable than appearance: a clear liquid can still be a mixture.
Ionic and covalent models
In a school-level ionic model, an electron may be transferred from one atom to another, producing oppositely charged ions whose electrostatic attraction holds a solid together. Sodium and chlorine give a familiar example: Na loses one electron to form Na⁺; Cl gains one to form Cl⁻. Charge balance gives NaCl. Magnesium can form Mg²⁺ and oxygen O²⁻, which gives MgO in a one-to-one ion ratio. The model explains formula writing and several broad properties. It does not mean a single physical electron transfer event describes every detail of every ionic solid.
In a covalent model, atoms share electron pairs. A water molecule has two O–H bonds; a methane molecule has four C–H bonds. Sharing is a useful first picture, but “one pair belongs equally to both atoms” is not a general rule. Different atoms can attract shared electrons unequally, giving polar covalent bonds. Some solids are giant covalent networks rather than isolated small molecules. Diamond, made of a carbon network, is hard and has very different properties from methane even though carbon is involved in both. The type of bonding and the structure of the whole substance matter.
Do not infer bond type solely from whether a substance dissolves in water or conducts electricity. Many ionic compounds conduct when molten or dissolved because mobile ions are available; a solid ionic crystal generally does not conduct by freely moving ions. Yet solubility varies, and some molecular substances can create ions in water. A multiple-choice option using “all” or “never” may hide the exception. For a basic exam item, select the property pattern stated with the right conditions, such as conduction in the molten state for a typical ionic substance.
Read a formula as a count
Subscripts in a chemical formula give the relative number of atoms of each element in a molecule or formula unit. H₂O contains two H atoms and one O atom per water molecule. Ca(OH)₂ contains one Ca, two O and two H atoms per formula unit because the 2 multiplies the entire OH group. A coefficient in front of a formula counts whole particles or formula units: 3H₂O represents three water molecules in the molecular picture, or six H atoms and three O atoms in total. Subscripts should not be changed to balance a chemical equation because that changes the substance.
Chemical equations represent reactants on the left and products on the right. Coefficients balance the number of atoms of each element across the arrow. For the simple example H₂ + O₂ → H₂O, the first draft has two oxygen atoms on the left and one on the right. Put 2 before H₂O to make two oxygen atoms on the right, then put 2 before H₂ to restore four hydrogen atoms on the left: 2H₂ + O₂ → 2H₂O. The left and right now each contain four H and two O atoms. These coefficients are ratios, not a requirement to have exactly that many individual molecules in every physical reaction.
The conservation of mass in an ordinary chemical reaction follows from accounting for atoms within a closed system. Products may have a different appearance, volume or state, but the constituent atoms have not vanished. If a reaction in an open container produces a gas that escapes, a balance may show an apparent loss for what remains in the container. That does not overturn the accounting; the gas is part of the products and must be included. If oxygen from the air enters a reaction, the final solid may gain mass relative to the starting solid because an additional reactant was omitted from that limited comparison.
Common reaction patterns
A combination reaction gives one product from two or more reactants in the basic classification: 2Mg + O₂ → 2MgO is an introductory example. A decomposition reaction breaks one compound into simpler products: CaCO₃ → CaO + CO₂ is an example when heated under appropriate conditions. A displacement reaction involves one element replacing another in a compound when the chemistry permits it. A double-displacement description involves exchange of ions between compounds, often with a precipitate, gas or weakly ionised product. The name classifies the pattern of a given balanced equation; it does not guarantee that any arbitrarily chosen reactants will react.
Oxidation and reduction are best connected to electron accounting where appropriate. Oxidation means loss of electrons; reduction means gain of electrons. In a redox reaction, one species' loss is paired with another's gain. The school mnemonic can help, but check who actually gained or lost. Oxygen addition and hydrogen removal are older useful patterns in many examples, yet the electron definition is broader. A reaction can be redox without visible oxygen. Conversely, merely seeing oxygen in a formula does not tell you which species is oxidised in a particular equation.
Rusting of iron is a familiar corrosion process involving iron, oxygen and moisture in the broad school account. Rust is not a protective metallic coating. Painting or galvanising can reduce contact with the environment, though different corrosion-control methods work by different mechanisms. Do not equate all surface changes with rust: tarnishing of silver and green patina on copper involve different substances and conditions. For an MCQ, identify the metal and the described change before selecting a product or prevention method.
Combustion is a chemical reaction that releases energy, commonly involving a fuel and oxygen. A candle's flame and a magnesium ribbon both illustrate exothermic reactions, but the products and observations differ. Whether a fuel “burns cleanly” depends on oxygen supply and composition. Incomplete combustion of carbon-containing fuels can yield carbon monoxide and soot; carbon monoxide is hazardous. This is a classification fact, not a demonstration instruction. The book does not ask candidates to perform combustion experiments.
Formula writing and valency checks
To derive a simple ionic formula, start from charges and require net charge zero. For calcium chloride, Ca²⁺ needs two Cl⁻ ions, so CaCl₂. For aluminium oxide, two Al³⁺ contribute +6 and three O²⁻ contribute −6, so Al₂O₃. For ammonium sulfate, NH₄⁺ has charge +1 and sulfate SO₄²⁻ has charge −2; two whole ammonium groups are needed: (NH₄)₂SO₄. The parentheses are essential because the subscript 2 applies to the whole NH₄ ion. If a student writes NH₄₂SO₄, the grouping is unclear and the standard formula is lost.
Balancing a reaction and building a formula are different operations. The formula H₂O is set by the substance's composition; a coefficient may change how many units participate. In 2H₂ + O₂ → 2H₂O, the 2 before water balances the equation while the subscript 2 in water remains fixed. A question may deliberately offer H₂O₂ as a distractor for balancing hydrogen and oxygen. H₂O₂ names hydrogen peroxide, a different compound; it is not a “balanced version” of water.
Count atoms before balancing a reaction
Parentheses and coefficients do different jobs. In 2Al₂(SO₄)₃, one formula unit contains 2 aluminium atoms, 3 sulfur atoms and 12 oxygen atoms. The coefficient 2 doubles the whole count to 4 aluminium, 6 sulfur and 24 oxygen atoms. Multiplying only the atom immediately before a bracket or multiplying a subscript twice gives a wrong total. Work from the innermost group outward: SO₄ supplies one S and four O, the bracket multiplier supplies three such groups, and the coefficient supplies two whole formula units. This method handles long formulas without guessing from their appearance.
An equation coefficient can describe a particle ratio or a mole ratio, but it is not usually a mass ratio. In 2H₂+O₂→2H₂O, hydrogen and oxygen molecules react in the ratio 2:1. Using rounded molar masses, 4 g of hydrogen reacts with 32 g of oxygen to form 36 g of water. Their mass ratio is 1:8, not 2:1. The coefficient tells how many chemical units participate; each unit has its own mass. A distractor that reads coefficients directly as grams ignores this second step. The example is a mass-accounting exercise, not a procedure for carrying out the reaction.
Balance the substance without changing its identity
To balance Al+O₂→Al₂O₃, first make oxygen counts compatible. Three O₂ units contain six oxygen atoms; two Al₂O₃ units contain six. The product side then has four aluminium atoms, giving 4Al+3O₂→2Al₂O₃. Check aluminium and oxygen separately. The smallest integer coefficients are 4,3,2. Multiplying all by two would still balance the atoms, but the convention is to report the smallest whole-number ratio unless a question asks otherwise. Changing Al₂O₃ to AlO would change the compound, so it cannot serve as a balancing operation.
For a reaction containing an unchanged polyatomic group, keeping that group together can reduce work. In CaCl₂+Na₂CO₃→CaCO₃+NaCl, calcium and carbonate already match as one each. There are two sodium atoms and two chlorine atoms on the left, so place 2 before NaCl. The balanced equation is CaCl₂+Na₂CO₃→CaCO₃+2NaCl. Verify oxygen too: three on each side. The carbonate group is unchanged here, but that shortcut would not work if carbonate decomposed to separate products. Always check whether the group survives the reaction before treating it as one counting unit.
Recognise the driving change in a reaction
Some aqueous ion-exchange reactions produce an insoluble solid, called a precipitate. In the calcium carbonate example, the precipitate forms because calcium and carbonate ions can combine into a sparingly soluble solid under suitable conditions. Sodium and chloride ions remain in solution in the simplified account. Merely exchanging the written partners of two salts does not prove that a visible reaction will occur. A question may supply a precipitate symbol or explicitly name an insoluble product; use that evidence rather than assuming every double-displacement equation yields a solid.
A chemical change can be indicated by a new gas, precipitate, colour change or energy transfer, but a single observation is not conclusive by itself. Boiling water produces gas bubbles through a physical phase change. Dissolving a coloured solid can change a liquid's appearance without creating a new compound. The stronger explanation links the observation to a change in composition. In an assertion–reason question, test whether the reason actually establishes that change rather than merely repeating that something visible happened.
Reaction categories can overlap. Burning magnesium is both combination and oxidation in the usual school description; a reaction need not belong to only one label. Decomposition identifies one reactant forming several products, while endothermic describes energy transfer into the reacting system. A decomposition process can therefore also be endothermic. The stem determines the requested classification. If it asks about number and arrangement of substances, use the reaction pattern; if it asks about energy flow, distinguish exothermic from endothermic.
Electron accounting in a simple redox example
Consider the conceptual equation Zn+Cu²⁺→Zn²⁺+Cu. Zinc loses two electrons and is oxidised; the copper ion gains two electrons and is reduced. The electron loss and gain balance. Zinc is the reducing agent because it supplies electrons that reduce another species. Cu²⁺ is the oxidising agent because it accepts electrons while causing zinc to be oxidised. This naming can feel reversed until the roles are written explicitly: the oxidising agent itself is reduced, and the reducing agent itself is oxidised. No oxygen atom is needed in this example for the process to be redox.
Oxidation numbers are bookkeeping assignments that help recognise electron redistribution. An element in its uncombined form has oxidation number zero in the usual convention. In a simple ion, oxidation number equals ionic charge. Thus zinc changes from 0 to +2 in the example, while copper changes from +2 to 0. An increase indicates oxidation and a decrease reduction. These assigned numbers do not always describe literal charges on individual atoms in a covalent molecule; they are a systematic accounting tool. Keep oxidation number distinct from valency and from the net charge on a whole molecule.
Bond polarity and molecular shape
Unequal sharing of electrons can make a bond polar, but molecular polarity also depends on how the bonds are arranged. Carbon dioxide has polar carbon–oxygen bonds, yet its linear symmetric arrangement allows their bond dipoles to cancel in the simple model. Water has a bent arrangement, so its bond dipoles do not cancel in the same way and the molecule is polar. A question that concludes “every molecule containing a polar bond must be polar” omits geometry. For basic exam reasoning, draw the arrangement or use the shape given in the question before choosing a whole-molecule polarity answer.
Intermolecular attractions also differ from the bonds within a molecule. When molecular water boils, molecules separate from neighbouring molecules; boiling does not normally split each water molecule into hydrogen and oxygen. The energy overcomes enough intermolecular attraction for vaporisation. Chemical decomposition would be a different process involving changed chemical composition. This distinction helps explain why a substance can undergo a physical phase change while keeping its molecular formula. It also prevents calling every energy-requiring change a chemical reaction.
Mole ratios connect equations to measurable amounts
The mole counts specified entities: one mole contains exactly 6.02214076×10²³ of them. Name the entity. One mole of O₂ molecules contains two moles of oxygen atoms, because each molecule has two oxygen atoms. It does not contain two moles of O₂ molecules. The Avogadro constant has unit mol⁻¹. If a question supplies the rounded value 6.022×10²³ mol⁻¹, use that value in the arithmetic rather than adding unnecessary digits. The number of entities is N=nNₐ, where n is amount in moles. For 0.25 mol of O₂ molecules, the count is about 1.506×10²³ molecules and twice as many oxygen atoms.
Molar mass is mass per mole, commonly expressed in g/mol for these questions. With the supplied rounded atomic masses Na=23 and Cl=35.5, NaCl has molar mass 58.5 g/mol. A 29.25 g sample therefore contains n=m/M=29.25/58.5=0.50 mol of NaCl formula units. The answer is an amount, not a mass or concentration. For CaCO₃, rounded masses Ca=40, C=12 and O=16 give 40+12+3×16=100 g/mol. The three oxygen atoms must all be included. A formula-mass calculation is a count followed by addition; it does not involve balancing a reaction.
A balanced equation then supplies a mole ratio. In CH₄+2O₂→CO₂+2H₂O, one mole of methane produces one mole of carbon dioxide if combustion is complete and enough oxygen is available. With rounded molar masses 16 g/mol for CH₄ and 44 g/mol for CO₂, 8 g of methane represents 0.5 mol and produces 0.5×44=22 g of carbon dioxide. The carbon dioxide can weigh more than the starting methane because oxygen also enters the product. Applying the coefficient ratio directly to grams would omit that mass contribution. Use the sequence mass → moles → equation ratio → requested mass.
Limiting reactants and percentage yield
When two reactant amounts are supplied, first ask which runs out. In N₂+3H₂→2NH₃, suppose 2 mol of nitrogen and 3 mol of hydrogen are available. Reacting all 2 mol of nitrogen would require 6 mol of hydrogen, which is not available. Hydrogen is limiting: 3 mol of it consumes 1 mol of nitrogen and produces 2 mol of ammonia. One mole of nitrogen remains. Do not add the reactant mole amounts and label the sum as product moles. Atom counts are conserved in the equation; total molecule count need not be.
A quick general check divides each available reactant amount by its equation coefficient. Here nitrogen gives 2/1=2 and hydrogen gives 3/3=1. The smaller value fixes how many complete reaction ratios can occur. Multiply that smaller value by a product coefficient to find the theoretical product amount. This method assumes the stated reaction proceeds as represented; an actual reaction can be affected by equilibrium, side reactions or incomplete conversion. A question that says “reaction proceeds completely” and gives one equation is specifying the simpler model.
Percentage yield compares product actually obtained with the theoretical maximum under the stated stoichiometry: actual/theoretical ×100. If a calculation predicts 50 g of product but only 42 g is collected, percentage yield is 84%. The shortfall is 8 g, or 16% of theoretical yield. The denominator is the theoretical product amount, not the total mass of all starting materials. An apparent yield above 100% can indicate wet or impure product, a measurement error or an incorrect theoretical calculation; it does not establish that mass has been created. Yield and purity are different ratios and need separate denominators.
Bond directions and molecular shapes
In the VSEPR model, regions of electron density around a central atom arrange themselves to reduce repulsion. A single, double or triple bond to one neighbouring atom counts as one direction or electron domain for the basic shape prediction; a lone pair also counts as a domain. Molecular shape describes positions of atoms, while electron-domain geometry includes lone pairs. Confusing those two descriptions is the main trap in ammonia and water questions.
Carbon dioxide has two bonding directions around carbon and a linear shape, with an O–C–O angle of 180°. Boron trifluoride has three bonding directions around boron and a trigonal planar shape, with ideal angles of 120°. Methane has four bonding directions and a tetrahedral shape, with angles about 109.5°. Ammonia has three bonds and one lone pair around nitrogen; its molecular shape is trigonal pyramidal, with an H–N–H angle about 107°. Water has two bonds and two lone pairs around oxygen; its molecular shape is bent, with an H–O–H angle about 104.5°. These familiar values describe the ordinary molecules, not every compound with a superficially similar formula.
The molecular PCl₅ model is trigonal bipyramidal: three equatorial bonds lie in one plane, and two axial bonds point above and below it. Equatorial angles are 120°, axial-to-equatorial angles 90°, and the two axial directions are 180° apart. SF₆ is octahedral, with adjacent bond directions at 90° and opposite directions at 180°. These are molecular descriptions; phase and chemical environment can matter. For basic carbon-centred examples, sp, sp² and sp³ labels correspond to linear, trigonal planar and tetrahedral orbital arrangements. Avoid treating older sp³d labels as literal proof that substantial d-orbital participation is required to explain all hypervalent bonding.
Sigma bonds, pi bonds and resonance
A single covalent bond is normally described as one sigma bond. A double bond consists of one sigma and one pi bond, and a triple bond of one sigma and two pi bonds. Count every bond in the whole molecule before choosing a total. Ethene, H₂C=CH₂, contains four C–H sigma bonds plus the sigma component of the C=C bond: five sigma bonds and one pi bond altogether. Ethyne, HC≡CH, contains two C–H sigma bonds plus one C–C sigma bond and two pi bonds: three sigma and two pi bonds. A triple bond is not three sigma bonds. Rotation about the ethene double bond is restricted because unrestricted rotation would disrupt its pi-bond overlap.
Resonance is used when one ordinary Lewis drawing does not adequately represent the electron distribution. Equivalent carbonate contributors each place one formal C=O double bond and two C–O single bonds around the same set of nuclei. The actual carbonate ion has equivalent C–O bonds; it does not alternate among three distinct molecules. The simple average bond order is (2+1+1)/3=4/3. For ozone, two equivalent contributors give a simplified mean O–O bond order of 1.5. These averages are bookkeeping models that help explain equivalent bonds, not claims that a fraction of a physical stick joins two atoms. Resonance contributors retain the nuclear arrangement; isomers have different structures.
Hydrogen bonding and empirical formulas
An introductory hydrogen-bond example involves hydrogen covalently attached to a strongly electronegative atom such as nitrogen, oxygen or fluorine and attracted to an appropriate electron-rich site. The hydrogen bond between water molecules is different from the covalent O–H bond within one molecule. Methane does not form the usual strong hydrogen-bond network between its own molecules. Hydrogen bonding can also occur within one molecule when suitable groups are positioned close enough; the nearby hydroxyl and nitro groups in o-nitrophenol provide a common textbook example. Merely counting hydrogen atoms does not establish hydrogen bonding.
A molecular formula records actual atom counts in a molecule; an empirical formula gives the simplest whole-number ratio. Hydrogen peroxide has molecular formula H₂O₂ and empirical formula HO, obtained by dividing both counts by two. Glucose has molecular formula C₆H₁₂O₆ and empirical formula CH₂O. Different compounds can share an empirical formula, so the simplest ratio alone does not identify the molecule. Never reduce the subscripts of a molecular formula when a question asks for its actual molecular composition; reduction is appropriate only when an empirical ratio is requested.
Retrieval exercise with explained answers
Count oxygen atoms in 3Ca(NO₃)₂, give the smallest coefficients for N₂+H₂→NH₃, and name the species reduced in Fe²⁺→Fe³⁺+e⁻. The first answer is 18 oxygen atoms because 3×2×3=18. The second is 1,3,2: one nitrogen molecule and three hydrogen molecules supply two nitrogen and six hydrogen atoms, matching two ammonia molecules. The third prompt is a trap: the written half-reaction shows oxidation of Fe²⁺, not a reduction. An electron is lost and the oxidation number increases. A complete redox reaction would also contain a matching reduction process.
Worked equation audit
Consider the written equation Fe + O₂ → Fe₂O₃. To balance oxygen, use 3O₂ on the left and 2Fe₂O₃ on the right, giving six O atoms on each side. The right then has four Fe atoms, so place 4Fe on the left: 4Fe + 3O₂ → 2Fe₂O₃. Verify both elements independently: Fe 4=4 and O 6=6. The result is a stoichiometric equation for the stated product; it does not claim that every real iron oxidation exposure produces only that one oxide.
In a separate closed-system arithmetic problem, 12 g of carbon reacts completely with 32 g of oxygen to form 44 g of carbon dioxide. The mass sum 12+32=44 illustrates conservation for this specified balanced reaction. If a candidate subtracts reactant mass from product mass and calls the difference “mass of energy,” the chemical accounting is being misread. For ordinary school-scale chemical reactions, the measured mass balance is taught through inclusion of all materials in the system. The numerical example is a conceptual exercise, not a prescription for laboratory work.
Recall and review
Classify O₂, pure H₂O, air and an ionic salt as element, compound or mixture with attention to particle type. Explain why molten sodium chloride can conduct while a solid crystal generally cannot through mobile ions. Count each element in Ca(OH)₂ and (NH₄)₂SO₄. Balance hydrogen combustion and iron-oxide formation without changing a subscript. Name the electron donor and recipient in a simple stated redox example. Finally, explain why a gas leaving an open vessel can create an apparent mass loss without violating conservation. Attempt the chapter practice set, then use the explanations to revisit any uncertain concepts.
Figure 8. Original chapter diagram.
Practice 43 questions
Choose one answer for each item. The answer key and explanations follow this chapter set.
1. How many pairs of electrons are shared between the two nitrogen atoms in a nitrogen molecule (N₂)?
A. 1 B. 3 C. 2 D. 4
2. The bonding in metals, in which positive metal ions are held together by a 'sea' of delocalised electrons, is called:
A. Metallic bonding B. Ionic bonding C. Covalent bonding D. Hydrogen bonding
3. The principle that matter is neither created nor destroyed in a chemical reaction, so that the total mass of reactants equals the total mass of products, is the law of:
A. Definite proportions B. Multiple proportions C. Conservation of mass D. Gaseous volumes
4. In the balanced equation for the combustion of methane, CH₄ + __O₂ → CO₂ + 2H₂O, what is the coefficient of O₂?
A. 1 B. 3 C. 4 D. 2
5. When sodium forms Na⁺ in sodium chloride, the ion has the same electronic configuration as which noble gas?
A. Argon B. Neon C. Helium D. Krypton
6. Sulphur hexafluoride (SF₆) has an octahedral shape. What is the F–S–F bond angle between adjacent fluorine atoms?
A. 120° B. 180° C. 90° D. 109.5°
7. What is the approximate H–O–H bond angle in a water molecule?
A. 109.5° B. 104.5° C. 120° D. 107°
8. What is the geometry of the phosphorus pentachloride (PCl₅) molecule?
A. Tetrahedral B. Octahedral C. Square planar D. Trigonal bipyramidal
9. What is the hybridisation of the carbon atom in carbon dioxide (CO₂)?
A. sp B. sp² C. sp³ D. sp³d
10. Which of the following molecules has zero dipole moment even though its individual bonds are slightly polar?
A. Hydrogen chloride (HCl) B. Ammonia (NH₃) C. Water (H₂O) D. Methane (CH₄)
11. Match the molecule (List-I) with its shape (List-II):
List-I: (a) CO₂ (b) BF₃ (c) CH₄ (d) NH₃
List-II: (i) Trigonal planar (ii) Linear (iii) Trigonal pyramidal (iv) Tetrahedral
A. a-ii, b-i, c-iii, d-iv B. a-ii, b-iii, c-iv, d-i C. a-i, b-ii, c-iv, d-iii D. a-ii, b-i, c-iv, d-iii
12. When an ammonia molecule donates its nitrogen lone pair to H⁺, which ion is formed?
A. NH₂⁻ B. NO₃⁻ C. NH₄⁺ D. N₂H₄
13. Match the molecule in List-I with its molecular shape in List-II.
List-I: (a) NH₃ (b) H₂O (c) CO₂ (d) SF₆
List-II: (i) Linear (ii) Octahedral (iii) Trigonal pyramidal (iv) Bent
A. a-iv, b-iii, c-i, d-ii B. a-iii, b-iv, c-ii, d-i C. a-iii, b-iv, c-i, d-ii D. a-i, b-iv, c-iii, d-ii
14. Which of these substances is best described by a lattice of oppositely charged ions rather than discrete covalent molecules?
A. Methane (CH₄) B. Hydrogen chloride (HCl) C. Sodium chloride (NaCl) D. Hydrogen gas (H₂)
15. How many hydrogen bonds hold a guanine–cytosine (G–C) base pair together in DNA?
A. 1 B. 4 C. 3 D. 2
16. In the trigonal bipyramidal phosphorus pentachloride (PCl₅) molecule, how many P–Cl bonds occupy axial positions?
A. 3 B. 2 C. 5 D. 1
17. How many sigma (σ) and pi (π) bonds are present in one molecule of ethene (H₂C=CH₂)?
A. 5 σ and 1 π B. 4 σ and 2 π C. 6 σ and 0 π D. 4 σ and 1 π
18. Hydrogen and chlorine have different electronegativities, and the bonding electron pair in H–Cl is unequally shared toward chlorine. How is the H–Cl bond classified?
A. Polar covalent B. Non-polar covalent C. Ionic D. Metallic
19. Which statement correctly describes the two oxygen–oxygen bonds in an ozone (O₃) molecule?
A. One bond is a pure single bond and the other a pure double bond B. Both bonds are pure double bonds C. Both bonds are identical, each with a bond order of 1.5 D. Both bonds are pure single bonds
20. In which unit is the dipole moment of a molecule commonly expressed?
A. Pascal (Pa) B. Angstrom (Å) C. Tesla (T) D. Debye (D)
21. Which of the following compounds does NOT form hydrogen bonds between its own molecules?
A. Methane (CH₄) B. Water (H₂O) C. Ammonia (NH₃) D. Hydrogen fluoride (HF)
22. Among the following, which compound has the shortest carbon–carbon bond?
A. Propane (C₃H₈) B. Ethane (C₂H₆) C. Ethene (C₂H₄) D. Ethyne (C₂H₂)
23. At approximately what temperature does liquid water have its maximum density?
A. 0 °C B. 4 °C C. 100 °C D. −4 °C
24. What is the empirical formula of hydrogen peroxide, whose molecular formula is H₂O₂?
A. HO B. H₂O C. H₂O₂ D. H₄O₄
25. Which of the following compounds shows intramolecular hydrogen bonding (a hydrogen bond within a single molecule)?
A. p-Nitrophenol B. Ethanol C. Hydrogen fluoride D. o-Nitrophenol
26. Three equivalent Lewis contributors for CO₃²⁻ each show one C=O bond and two C–O bonds. What average C–O bond order follows from this simplified resonance model?
A. 1 B. 3/2 C. 4/3 D. 2
27. In chemistry, the theory used to predict molecular shapes is called VSEPR. What does VSEPR stand for?
A. Valence Shell Electron Pair Repulsion B. Valence Shell Electron Pair Resonance C. Variable Shell Electron Pair Repulsion D. Valence Subshell Electron Proton Repulsion
28. Why does solid sodium chloride conduct poorly while molten sodium chloride conducts electricity?
A. Its covalent molecules break into atoms on melting B. Only the solid has free electrons C. Its ions are fixed in the solid but mobile in the melt D. The sodium and chlorine nuclei move through the solid
29. Consider the following statements about hybridisation:
Statement I: sp² hybridisation produces three hybrid orbitals directed at 120° to one another.
Statement II: sp hybridisation produces two hybrid orbitals directed at 180° to each other.
Which of the above is/are correct?
A. Only I B. Only II C. Both I and II D. Neither I nor II
30. Consider the statements. I: An ordinary intermolecular hydrogen bond is typically stronger than a typical covalent bond within a molecule. II: Ice floats because its hydrogen-bonded structure has lower density than liquid water. Which statement(s) is/are correct?
A. Only I B. Only II C. Both I and II D. Neither I nor II
31. Consider the following statements:
Statement I: Free rotation is possible about the carbon–carbon double bond in ethene.
Statement II: A carbon–carbon triple bond consists of three sigma (σ) bonds.
Which of the above is/are correct?
A. Only I B. Only II C. Both I and II D. Neither I nor II
32. Which substance among the following shows unusually strong hydrogen bonding between its own molecules?
A. Hydrogen chloride (HCl) B. Hydrogen bromide (HBr) C. Hydrogen fluoride (HF) D. Hydrogen iodide (HI)
33. Which of the following is NOT a typical property of crystalline sodium chloride?
A. It dissolves readily in water B. It has high melting and boiling points C. It is hard but brittle D. It dissolves readily in non-polar solvents like kerosene
34. Which of the following molecules is NOT linear?
A. H₂O B. CO₂ C. C₂H₂ (ethyne) D. HCl
35. A reaction follows 2H₂ + O₂ → 2H₂O. A closed vessel initially contains 5 mol of H₂ and 2 mol of O₂. Assuming the reaction proceeds completely, what remains after the limiting reactant is used up?
A. 1 mol of H₂ and 4 mol of H₂O B. 1 mol of O₂ and 5 mol of H₂O C. 3 mol of H₂ and 2 mol of H₂O D. Only 5 mol of H₂O
36. Match the reaction (List-I) with its type (List-II):
List-I: (a) 2Na + Cl₂ → 2NaCl (b) 2NaCl (molten) → 2Na + Cl₂ (electrolysis) (c) Zn + CuSO₄ → ZnSO₄ + Cu (d) AgNO₃ + NaCl → AgCl + NaNO₃
List-II: (i) Decomposition (ii) Combination (iii) Single displacement (iv) Double displacement
A. a-i, b-ii, c-iii, d-iv B. a-ii, b-i, c-iii, d-iv C. a-ii, b-i, c-iv, d-iii D. a-ii, b-iii, c-i, d-iv
37. In Zn + CuSO₄ → ZnSO₄ + Cu, which species is oxidised?
A. Copper metal, Cu B. Sulfate ion, SO₄²⁻ C. Zinc metal, Zn D. Copper(II) ion, Cu²⁺
38. What is the oxidation state of iron in iron(III) oxide, Fe₂O₃?
A. +3 B. +2 C. +6 D. 0
39. According to CH₄ + 2O₂ → CO₂ + 2H₂O, what mass of carbon dioxide is produced by the complete combustion of 64 g of methane? (C = 12, H = 1, O = 16)
A. 88 g B. 176 g C. 64 g D. 44 g
40. What is the mass of 2.5 moles of sodium chloride? (Na = 23, Cl = 35.5)
A. 58.5 g B. 146.25 g C. 117 g D. 23.4 g
41. How many molecules are present in 2 moles of oxygen gas (O₂)? (Nₐ = 6.022 × 10²³ mol⁻¹)
A. 1.2044 × 10²⁴ B. 6.022 × 10²³ C. 2.4088 × 10²⁴ D. 3.011 × 10²³
42. How many moles of calcium carbonate are present in 50 g of CaCO₃? (Ca = 40, C = 12, O = 16)
A. 0.25 mol B. 5 mol C. 0.5 mol D. 2 mol
43. In a reaction, the theoretical yield of a product is 80 g, but only 60 g is actually obtained. What is the percent yield?
A. 75% B. 80% C. 60% D. 133%
Answers and explanations
1. B. Each nitrogen atom has 5 valence electrons and needs 3 more, so the two atoms share three pairs, forming a triple bond N≡N. This very strong, short bond is why nitrogen gas is relatively unreactive.
2. A. In the electron-sea model of metallic bonding, valence electrons are delocalised across many atoms and the positive metal cores are held together by their attraction to this shared electron distribution. The mobility of electrons helps explain electrical conduction; nondirectional bonding helps explain malleability. The picture is a useful model, not a claim that every atom has permanently lost a particular electron.
3. C. The law of conservation of mass states that atoms are only rearranged in a reaction, so total mass is unchanged; it is why equations must be balanced. The law of definite proportions concerns fixed composition of a compound, and multiple proportions concerns ratios between compounds of the same elements.
4. D. Oxygen atoms on the right = 2 (in CO₂) + 2 (in 2H₂O) = 4, so 2 O₂ molecules are needed: CH₄ + 2O₂ → CO₂ + 2H₂O. Carbon (1 = 1) and hydrogen (4 = 4) are already balanced. Only coefficients are changed, never subscripts.
5. B. Na (2, 8, 1) loses its single 3s electron to become Na⁺ (2, 8), i.e. 1s² 2s² 2p⁶ — the configuration of neon. The chloride ion Cl⁻ (2, 8, 8), by contrast, matches argon.
6. C. Six equivalent S–F directions in octahedral SF₆ point along three perpendicular axes. Adjacent directions meet at 90°, while opposite directions meet at 180°. Shape and angle are the testable facts here; an sp³d² label is a traditional teaching model and should not be presented as proof of substantial sulphur d-orbital bonding.
7. B. Oxygen in water has 2 bond pairs and 2 lone pairs (tetrahedral electron geometry). Lone pairs repel more strongly, squeezing the angle from 109.5° to about 104.5°, giving a bent shape. NH₃ is about 107°.
8. D. PCl₅ has a trigonal-bipyramidal molecular geometry: three equatorial bonds lie about 120° apart and two axial bonds are opposite each other. This geometric statement does not require treating the traditional sp³d shorthand as a literal description of the bonding.
9. A. In O=C=O, carbon has only two electron groups (two double bonds) and no lone pairs, so it is sp hybridised and the molecule is linear with a 180° bond angle. The remaining p orbitals form the two π bonds.
10. D. Methane is perfectly tetrahedral (109.5°), so the four C–H bond dipoles cancel and the net dipole is zero. NH₃ (pyramidal) and H₂O (bent) have lone pairs and non-cancelling dipoles; HCl is a polar diatomic.
11. D. CO₂ – linear (2 groups); BF₃ – trigonal planar (3 bond pairs, no lone pair); CH₄ – tetrahedral (4 bond pairs); NH₃ – trigonal pyramidal (3 bond pairs + 1 lone pair). Hence a-ii, b-i, c-iv, d-iii.
12. C. NH₃ has a lone pair on nitrogen and can accept H⁺ to form ammonium, NH₄⁺. The newly formed N–H bond can be described as coordinate at formation, although the four N–H bonds in the ion are equivalent. NH₂⁻ is an amide ion and N₂H₄ is hydrazine.
13. C. NH₃ has 3 bond pairs and 1 lone pair (trigonal pyramidal); H₂O has 2 bond pairs and 2 lone pairs (bent); CO₂ has two electron domains around carbon (linear, 180°); SF₆ has six bond pairs (octahedral, 90°).
14. C. Solid sodium chloride consists of Na⁺ and Cl⁻ ions in an extended electrostatic lattice. Methane, hydrogen chloride and hydrogen gas are molecular substances with covalent bonds. A fixed electronegativity-difference cutoff is only a rough guide and does not by itself determine every bond type.
15. C. A G–C pair is held by three hydrogen bonds, while an A–T pair has two. This is why DNA regions rich in G–C pairs need more energy (higher temperature) to separate.
16. B. A trigonal bipyramid has three equatorial and two axial positions. Therefore two P–Cl bonds in PCl₅ are axial, opposite each other (180°), and each makes about 90° with an equatorial bond. A literal sp³d bonding assignment is unnecessary to answer the geometry question.
17. A. Each sp² carbon forms σ bonds with two H atoms and with the other carbon: 4 C–H σ + 1 C–C σ = 5 σ bonds. The unhybridised p orbitals on the two carbons overlap sideways to give 1 π bond, so the C=C double bond is one σ plus one π.
18. A. H–Cl is a polar covalent bond: the atoms share electrons, but chlorine attracts the pair more strongly, leaving partial charges Hδ+ and Clδ−. It is neither a non-polar bond nor a metallic bond. The conclusion does not require a rigid numerical electronegativity cutoff.
19. C. Ozone is a resonance hybrid of two Lewis structures. The real molecule has two equivalent O–O bonds, each with bond order 1.5 and a length (about 128 pm) between that of an O–O single and an O=O double bond; it does not flip between the two structures.
20. D. Dipole moment is commonly expressed in debye (1 D ≈ 3.34 × 10⁻³⁰ C·m). For example HCl is about 1.08 D and H₂O about 1.85 D, while symmetric molecules such as CO₂ and CH₄ have zero dipole moment. Pascal measures pressure, angstrom length and tesla magnetic field.
21. A. Hydrogen bonding needs H attached to a highly electronegative atom — F, O or N — as in HF, H₂O and NH₃. In methane, H is bonded to carbon, which is not electronegative enough, so CH₄ has only weak van der Waals forces.
22. D. Ethyne has a C≡C triple bond; more shared electron pairs pull the atoms closer, so triple bonds are the shortest and strongest. Ethene has a C=C double bond and ethane/propane have C–C single bonds.
23. B. Water is densest at about 4 °C. Below this, hydrogen bonds begin arranging molecules into an open, cage-like structure, so the liquid expands on further cooling and ice at 0 °C is less dense than water — which is why ice floats and lakes freeze from the top.
24. A. An empirical formula gives the simplest whole-number ratio of atoms. H₂O₂ has H:O = 2:2, which reduces to 1:1, or HO. H₂O₂ remains its molecular formula; H₄O₄ has the same 1:1 ratio but is not simplest.
25. D. In o-nitrophenol the –OH and –NO₂ groups are adjacent, so the phenolic H bonds to a nitro oxygen within the same molecule. In p-nitrophenol the groups are too far apart and it forms intermolecular H-bonds, as do ethanol and HF — which is why o-nitrophenol is more volatile.
26. C. The sum of the C–O bond orders in each contributor is 2 + 1 + 1 = 4. Sharing this over three equivalent C–O positions gives 4/3. The ion has delocalised bonding; it does not repeatedly switch between three distinct molecules. An individual Lewis contributor alone does not represent three different observed bond lengths.
27. A. VSEPR = Valence Shell Electron Pair Repulsion. Bond pairs and lone pairs around the central atom repel each other and arrange themselves as far apart as possible, which fixes the molecule's shape.
28. C. NaCl contains ions in both states. In the solid, the ions are constrained to lattice positions and cannot carry charge across the material. When molten, mobile Na⁺ and Cl⁻ ions carry current. This differs from metallic conduction by delocalised electrons.
29. C. Both are correct. sp² (one s + two p) gives three trigonal-planar orbitals at 120°, as in ethene; sp (one s + one p) gives two linear orbitals at 180°, as in ethyne. sp³ gives four at 109.5°.
30. B. II is correct: the open hydrogen-bonded arrangement of ordinary ice makes it less dense than liquid water. I is false for an ordinary intermolecular hydrogen bond compared with a typical covalent bond. The word “typically” matters because bond strengths vary across chemical systems.
31. D. Neither is correct. The π bond in C=C locks the atoms and prevents free rotation (this is why cis–trans isomers exist). A triple bond is one σ bond plus two π bonds, not three σ bonds.
32. C. HF molecules form prominent intermolecular hydrogen bonds because H is bonded to highly electronegative, small fluorine. HCl, HBr and HI have other intermolecular attractions but do not show comparably strong classical H-bond networks. This also helps explain HF’s anomalously high boiling point relative to neighbouring hydrogen halides.
33. D. Solid NaCl has a high melting point, is hard but brittle, and dissolves in water to a useful extent. It does not dissolve readily in non-polar kerosene. The wording names NaCl because water solubility varies considerably across ionic compounds; it would be wrong to assert that every ionic solid readily dissolves in water.
34. A. Water is bent (about 104.5°) because oxygen carries two lone pairs. CO₂ (O=C=O) and ethyne (H–C≡C–H) are linear with sp-hybridised carbon, and any diatomic molecule such as HCl is necessarily linear.
35. A. Each mole of O₂ consumes 2 mol of H₂ and produces 2 mol of H₂O. Thus 2 mol of O₂ consumes 4 of the 5 mol of H₂, leaving 1 mol of H₂ and forming 4 mol of water. Oxygen is the limiting reactant. Do not use all 5 mol of hydrogen when the available oxygen cannot support that amount of reaction.
36. B. (a) Two reactants form one product – combination. (b) One compound splits by electricity – decomposition. (c) More reactive zinc displaces copper – single displacement. (d) Ions are exchanged, AgCl precipitates – double displacement. Hence a-ii, b-i, c-iii, d-iv.
37. C. Zinc changes from oxidation state 0 to +2, losing two electrons, so it is oxidised. Copper(II) changes from +2 to 0 and is reduced. Sulfate remains a spectator in this simplified ionic account.
38. A. Oxygen is −2, so three O contribute −6. For a neutral compound, 2x − 6 = 0, giving x = +3. In the reaction 4Fe + 3O₂ → 2Fe₂O₃, iron goes from 0 to +3 and is oxidised.
39. B. Moles of CH₄ = 64 ÷ 16 = 4 mol. The equation gives 1 mol CO₂ per mol CH₄, so 4 mol CO₂ forms. Mass = 4 × 44 = 176 g.
40. B. Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol. Mass = moles × molar mass = 2.5 × 58.5 = 146.25 g. 58.5 g is one mole, 117 g is two moles, and 23.4 g comes from wrongly dividing 58.5 by 2.5.
41. A. Number of molecules = moles × Nₐ = 2 × 6.022 × 10²³ = 12.044 × 10²³ = 1.2044 × 10²⁴. The value 2.4088 × 10²⁴ would be the number of oxygen atoms (each O₂ has 2 atoms), not molecules.
42. C. Molar mass of CaCO₃ = 40 + 12 + 3 × 16 = 100 g/mol. Moles = mass ÷ molar mass = 50 ÷ 100 = 0.5 mol.
43. A. Percent yield = (actual pure product ÷ theoretical pure product) × 100 = 60/80 × 100 = 75%. The 133% distractor reverses the fraction. A measured crude mass can exceed a theoretical pure-product mass if the sample contains solvent or impurities, so a value over 100% is a signal to check the sample and calculation rather than a logical impossibility.