WAEC SSCE - Chemistry - 1991

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Van der Waal's forces are weak intermolecular forces that exist between molecules. These forces are formed due to the attraction between the temporary dipoles that arise due to the fluctuations in the electron cloud of molecules. Out of the given options, only iodine is held together by van der Waal's forces only. Iodine exists as a diatomic molecule, and each iodine atom is held together by a covalent bond. However, the molecules are held together in the crystal lattice by weak van der Waal's forces. Sodium chloride, ice, diamond, and quartz are all held together by strong ionic or covalent bonds, and therefore, van der Waal's forces are not the primary forces holding these crystals together.

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The statement that is not correct about Group 7 elements is: "have relatively low ionization potentials." Explanation: Group 7 elements, also known as halogens, include fluorine, chlorine, bromine, iodine, and astatine. These elements are diatomic, meaning that they exist naturally as a molecule containing two atoms of the same element. They are highly electronegative, which means that they have a strong attraction for electrons. This high electronegativity makes them good oxidizing agents since they readily accept electrons from other atoms. However, the statement that Group 7 elements have relatively low ionization potentials is incorrect. Ionization potential refers to the amount of energy required to remove an electron from an atom or ion. Halogens have relatively high ionization potentials since they have a strong attraction for their outermost electrons. This means that it takes a significant amount of energy to remove an electron from a halogen atom or ion, making it less likely to form a cation. Therefore, the correct answer is "have relatively low ionization potentials."

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The formula of a dicarboxylic acid is a molecule with two carboxylic acid functional groups (-COOH) present in its structure. The correct option is (d) CH₂(COOH)₂, which is the formula for oxalic acid, a dicarboxylic acid commonly found in plants and used in industrial applications such as rust removal and bleaching.

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The particles of a solid exhibit vibrational motion only. A solid is one of the three states of matter, in which the particles are closely packed together and have a fixed shape and volume. The particles of a solid are not free to move around like those in a liquid or gas. Instead, they can only vibrate around their fixed positions. Therefore, the only mode of motion exhibited by the particles of a solid is vibrational motion, which means that the particles oscillate back and forth around their fixed positions. They do not move from one place to another, which is referred to as translational motion, nor do they move randomly, which is referred to as random motion.

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Brass is an alloy of copper and zinc. An alloy is a combination of two or more metals, or a metal and a non-metal. In the case of brass, copper is the main metal and zinc is the other component. The proportion of copper to zinc can vary, resulting in different types of brass with varying properties. Brass is known for its attractive golden appearance and is widely used in the production of decorative items, musical instruments, and plumbing fixtures, among other things.

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The balanced chemical equation is 2NaNO₃ → 2NaNO₂ + O₂. It shows that 2 moles of NaNO₃ produces 1 mole of O₂. We are given 8.5g of 2NaNO₃. To find the number of moles, we divide the mass by the molar mass. Molar mass of 2NaNO₃ = (2 x 23) + 2(14 + 3x16) = 46 + 2(14 + 48) = 46 + 124 = 170g/mol. Number of moles of 2NaNO₃ = 8.5g / 170g/mol = 0.05mol. Since 2 moles of NaNO₃ produces 1 mole of O₂, the number of moles of O₂ produced is also 0.05mol. The molar volume of gases at s.t.p. is 22.4dm³/mol. Therefore, the volume of O₂ produced at s.t.p. is: Volume of O₂ = Number of moles x Molar volume of gases at s.t.p. = 0.05mol x 22.4dm³/mol = 1.12dm³ Therefore, the volume of oxygen produced at s.t.p. when 8.5g of 2NaNO₃ is heated until no further gas is evolved is 1.12dm³. Hence, the correct option is 1.12dm³.

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Compounds that have the same molecular formula but different structures are said to be isomeric. Isomers are molecules that have the same number and types of atoms but different arrangements of those atoms. This means that they have different physical and chemical properties, even though they have the same molecular formula. For example, butane and isobutane both have the molecular formula C4H10, but they have different structures and therefore different properties. This property of compounds to have same molecular formula but different structures is called isomerism.

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The molar mass of sodium trioxocarbonate (IV) is 2(23) + 12 + 3(16) = 106 g/mol (by adding the atomic masses of the elements present in one mole of the compound). To find the amount (in moles) of the compound in 5.3 g, we need to divide the given mass by the molar mass: Amount of Na₂CO₃ = Mass of Na₂CO₃ / Molar mass of Na₂CO₃ Amount of Na₂CO₃ = 5.3 g / 106 g/mol Amount of Na₂CO₃ = 0.05 mol Therefore, the amount (in moles) of sodium trioxocarbonate (IV) in 5.3 g of the compound is 0.05. The answer is.

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The oxidation number of an element is defined as the charge that it would carry if all the shared electrons are assigned to the more electronegative atom. The sum of the oxidation numbers of all atoms in a molecule must be equal to the charge of the molecule. In NH₃, the oxidation number of nitrogen is -3 because hydrogen has an oxidation number of +1 and the sum of oxidation numbers in the compound is zero. In NaNO₃, the oxidation number of nitrogen is +5 because sodium has an oxidation number of +1 and oxygen has an oxidation number of -2. In HNO₂, the oxidation number of nitrogen is +3 because hydrogen has an oxidation number of +1 and oxygen has an oxidation number of -2. In NOI₃, the oxidation number of nitrogen is +5 because oxygen has an oxidation number of -2 and iodine has an oxidation number of +7. The only option where nitrogen has an oxidation number of zero is N₂, because it is a diatomic molecule and the oxidation number of each nitrogen atom is zero.

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When air is passed through a heated tube containing finely divided copper, oxygen is the component that is absorbed in the process. This is because copper is a good catalyst for the oxidation of gases, especially oxygen. As air passes through the heated tube containing finely divided copper, the copper catalyses the oxidation of oxygen in the air to form copper oxide. The copper oxide is then absorbed onto the surface of the copper, leaving behind the other components of the air, such as nitrogen, carbon dioxide, water vapour, and noble gases. Therefore, the correct option is oxygen.

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Hydrogen is used for all of the following except extinguishing fire. Hydrogen gas itself is highly flammable and can even ignite explosively in the presence of air, oxygen or any other oxidizing gas. It does not work as an extinguisher and, in fact, can even make a fire worse by adding fuel to the fire. The other options listed are all valid uses of hydrogen. It is used in the manufacturing of ammonia, the synthesis of hydrochloric acid, the conversion of coal to petrol and in the manufacture of margarine.

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Entropy is a measure of the degree of randomness or disorder in a system. The greater the number of ways in which the particles in a system can be arranged, the greater the entropy of that system. The process that has the greatest positive entropy change is the one that involves the largest increase in the degree of randomness or disorder of the particles. This occurs when a solid or liquid is converted to a gas or when a gas is allowed to expand into a larger volume. Looking at the given equations, we can see that the equation (a) C₆H_3(l) → C₆H _{6 (g)} involves the conversion of a liquid to a gas, which leads to a large increase in the number of ways in which the particles can be arranged. Therefore, this process has the greatest positive entropy change among the given equations. Note: The answer may vary depending on the reference state and conditions under which the reactions occur.

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Mass spectrometer is suitable for determining different isotopes present in an element which exhibits isotopy. This is because mass spectrometer can separate different isotopes based on their mass-to-charge ratio. In a mass spectrometer, the sample is ionized and then accelerated in an electric field. The ions are then separated based on their mass-to-charge ratio using a magnetic field. The detector records the number of ions of each mass-to-charge ratio, which can be used to determine the relative abundance of each isotope present in the sample. Therefore, mass spectrometer is a very useful tool in the study of isotopes and can be used to identify and quantify different isotopes in a sample.

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The solution is saturated. A saturated solution is a solution that contains the maximum amount of solute that can be dissolved in the solvent at a given temperature and pressure. When a crystal is added to a saturated solution, it will not dissolve because there is no more room for additional solute to dissolve. The solution remains unchanged because the concentration of solute in the solution is already at its maximum limit, and no more solute can be dissolved.

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The property that is not characteristic of electrovalent compounds is "high vapour pressures". Electrovalent compounds, also known as ionic compounds, have high melting and boiling points due to the strong electrostatic forces of attraction between oppositely charged ions in the crystal lattice. They also tend to form crystalline solids and are typically soluble in polar solvents. In addition, electrovalent compounds can conduct electricity in aqueous solution because the ions are free to move and carry an electric charge. However, they do not have high vapour pressures because they are held together by strong ionic bonds and are not easily vaporized.

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The reaction illustrated in the equation above is an example of an addition reaction. Addition reactions are a type of organic reaction where two or more molecules combine to form a larger molecule with no elimination of any atoms. In this case, the double bond in ethene (CH₂=CH₂) reacts with the sulfuric acid (H₂SO₄) to form a single bond between carbon atoms, resulting in the formation of ethyl hydrogen sulfate (CH₃-CH₂-OSO₃H).

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When starch undergoes complete enzyme-catalyzed hydrolysis, it is broken down into its constituent glucose molecules. Starch is a polymer made up of glucose units, and enzymes such as amylase and amyloglucosidase can break the glycosidic bonds between these units, resulting in the formation of individual glucose molecules. Therefore, the resulting product of complete enzyme-catalyzed hydrolysis of starch is glucose.

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The organic compound which reacts readily with bromine to form a compound with the formula CH₃CHBrCH₂Br is propene. Propene is an unsaturated hydrocarbon with the chemical formula C₃H₆. It contains a carbon-carbon double bond that is readily attacked by bromine in the presence of an appropriate catalyst. The addition of two bromine atoms across the double bond of propene produces 2-bromopropane, which has the formula CH₃CHBrCH₂Br. This reaction is an example of electrophilic addition, in which an electrophile (in this case, Br₂) is added to an unsaturated molecule (propene) to form a saturated molecule (2-bromopropane). Therefore, propene is the organic compound that reacts readily with bromine to form 2-bromopropane.

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The compound which liberates carbon (IV) oxide from a hydrogen trioxocarbonate (IV) is an acid. The molecular formula of an acid is usually expressed as HX, where X is the anion of the acid. Therefore, the molecular formula of the compound could be HCOOH (formic acid). When formic acid is added to hydrogen trioxocarbonate (IV), it reacts to liberate carbon (IV) oxide gas, as shown by the following equation: 2HCOOH + H₂CO₃ → 2H₂O + 2CO₂ Therefore, formic acid is a possible compound that could liberate carbon (IV) oxide from a hydrogen trioxocarbonate (IV).

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The reaction that will take place at room temperature is (2Na + 2H₂O → 2NaOH + H₂). This is because sodium is a highly reactive metal and it readily reacts with water at room temperature to form sodium hydroxide and hydrogen gas. The other reactions require higher temperatures or catalysts for the reactions to occur.

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When the temperature of the reaction vessel is decreased, the equilibrium shifts to favor the exothermic direction, which is the forward reaction in this case. As a result, more hydrogen and iodine molecules react to form hydrogen iodide. The equilibrium constant (K) for this reaction indicates that the reaction favors the products, which means that at equilibrium, there is a greater concentration of hydrogen iodide molecules than hydrogen and iodine molecules. Therefore, a decrease in temperature will increase the yield of hydrogen iodide molecules, resulting in a shift towards the forward reaction. Therefore, the correct option is "the yield of hydrogen iodide increases".

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In the given redox reaction, the copper (ll) ions (Cu²⁺) are being reduced to copper atoms (Cu) while the zinc atoms (Zn) are being oxidized to zinc ions (Zn²⁺). Therefore, "copper (ll) ions are reduced to copper atoms" is correct.

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The metal extracted from cassiterite is tin. Cassiterite is a mineral composed of tin dioxide (SnO2), and it is the primary ore of tin. Tin is a silvery-white metal that is commonly used to coat other metals to prevent corrosion. It is also used to make alloys, such as bronze and pewter, and is an important component of solders. Tin has a low toxicity, which makes it suitable for use in food packaging and other applications where contact with humans or animals is possible.

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The ∆H in the given chemical equation refers to the enthalpy change or heat change of the reaction. It is the difference between the enthalpy of the products and the enthalpy of the reactants. In this equation, energy is released in the form of heat (exothermic reaction), and the value of ∆H is 890KJ mo-1. Since the reactants in the equation include CH4 and O2, which are the typical fuels used in combustion reactions, it can be concluded that the reaction is a combustion reaction. Therefore, the correct answer is "combustion".

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Methane, Ethanol, Petrol, Kerosene are all direct petroleum products as they are obtained directly from the refining of crude oil. Vaseline, on the other hand, is not a direct petroleum product. It is produced from petroleum jelly, which is a semi-solid mixture of hydrocarbons obtained from petroleum. Vaseline is obtained from petroleum jelly through a refining process that removes impurities and increases its purity. Thus, Vaseline is a derivative of petroleum, but not a direct petroleum product.

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The representation of aminoethanoic acid (glycine) as NH⁺₃ CH₂COO indicates that it can function as a weak acid or a weak base. The NH⁺₃ group is a positively charged ion (cation) which can act as an acid by donating a hydrogen ion (proton) to a base. On the other hand, the CH₂COO group is a negatively charged ion (anion) which can act as a base by accepting a proton from an acid. Therefore, aminoethanoic acid (glycine) can function as both an acid and a base in solution, depending on the nature of the other species present in the solution.

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The electronic configuration of an atom shows the distribution of electrons in its various energy levels or shells. Each shell can hold a certain number of electrons. The first shell can hold up to two electrons, the second shell can hold up to eight electrons, and the third shell can hold up to 18 electrons. The electronic configuration given as 1s²2s²2p⁶3s¹ shows that there are electrons in the first, second, and third shells. The first shell has two electrons (1s²), the second shell has eight electrons (2s²2p⁶), and the third shell has one electron (3s¹). Therefore, the atom has three principal electronic shells. Answer: 3.

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The separation of a mixture of calcium trioxocarbonate (lV) and sodium trioxocarbonate (IV) is most easily carried out by using the differences in their solubility. When two or more substances are mixed together, they may have different solubility in a particular solvent. In this case, calcium trioxocarbonate (IV) and sodium trioxocarbonate (IV) are both soluble in water, but their solubility is different. Calcium trioxocarbonate (IV) is less soluble in water compared to sodium trioxocarbonate (IV). Therefore, by dissolving the mixture in water, the less soluble calcium trioxocarbonate (IV) can be separated from the more soluble sodium trioxocarbonate (IV) by filtration or centrifugation. This method is called fractional crystallization.

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(a) Equation

The soda-lime supplies sodium hydroxide, which reacts with the anhydrous sodium ethanoate to release methane: \[CH_3COONa_{(s)} + NaOH_{(s)} \rightarrow CH_{4(g)} + Na_2CO_{3(s)}\]

(b) Why soda-lime is preferred to sodium hydroxide

Soda-lime (a mixture of sodium hydroxide and calcium oxide) is preferred because solid sodium hydroxide alone is strongly deliquescent and corrosive: when heated it melts, absorbs moisture and attacks (corrodes) the glass test tube, which may crack. Soda-lime is a dry, granular, easier-to-handle solid that does not fuse or attack the glass in the same way, so it gives a safer and more convenient preparation.

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(a) Two factors which affect the rate of a chemical reaction are:

1. Concentration of the reactants.
2. Temperature of the reacting mixture.

(b) The graph is plotted with volume of 2 mol dm⁻³ HCl on the horizontal axis and volume of hydrogen evolved on the vertical axis.

Volume of HCl used / cm3	5	15	25	35	45
Volume of H2 evolved / cm3	120	360	550	600	600

(c) Reading from the graph:

1. For 50 cm³ of acid, the volume of hydrogen produced is 600 cm³.
2. The volume of acid required to produce 480 cm³ of hydrogen is approximately 21 cm³.
3. The volume of acid needed to dissolve the 0.72 g of magnesium completely is 35 cm³.

(d) At 35 cm³ of acid, the graph becomes horizontal at 600 cm³ of hydrogen. This shows that all the magnesium has been used up. Addition of more acid cannot produce more hydrogen because magnesium is now the limiting reactant.

(e)

1. Mass of magnesium used = 0.72 g.

   \[n(\mathrm{Mg})=\frac{0.72}{24}=0.030\ \text{mol}\]

   0.030 mol of Mg is completely dissolved by 35 cm³ of the acid.

   \[\text{Volume for 1 mol Mg}=\frac{35}{0.030}=1166.7\ \text{cm}^3\]

   Therefore, 1.17 × 10³ cm³ of the acid will dissolve 1 mole of magnesium completely.

2. 0.030 mol of Mg produces 600 cm³ of hydrogen.

   \[\text{Volume of }\mathrm{H_2}\text{ for 1 mol Mg}=\frac{600}{0.030}=2.00\times10^4\ \text{cm}^3\]

   Therefore, 2.00 × 10⁴ cm³ of hydrogen is liberated when 1 mole of magnesium dissolves completely.

3. The balanced equation is:

   \[\mathrm{Mg_{(s)}+2HCl_{(aq)}\rightarrow MgCl_{2(aq)}+H_{2(g)}}\]

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(a) Products of electrolysis of dilute copper(II) tetraoxosulphate(VI) solution

• (i) Carbon (inert) electrodes: copper is deposited at the cathode; oxygen gas is evolved at the anode. The blue colour fades and the solution becomes acidic (tetraoxosulphate(VI) acid remains).
• (ii) Copper electrodes: copper is deposited at the cathode; the copper anode dissolves (loses mass) so no gas is evolved. The concentration and colour of the solution stay essentially the same.

(b) For each process

(i) Anodic half reaction

• Carbon anode: \[2H_2O_{(l)} \rightarrow O_{2(g)} + 4H^+_{(aq)} + 4e^-\]
• Copper anode: \[Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-\]

(ii) Effect on pH

• With carbon electrodes, oxygen is released and H⁺ ions accumulate in solution, so the pH decreases (the electrolyte becomes more acidic).
• With copper electrodes, copper dissolves at the anode at the same rate as it deposits at the cathode, so the Cu²⁺ concentration is unchanged and the pH remains essentially constant.

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(a) Paper chromatography.

(b)

(i) Three components are resolved.

(ii) Chromatography paper (filter paper/cellulose) is the adsorbent medium normally used in the laboratory.

(iii) Label III marks the point of application of the universal indicator.

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(a)

(i) Acetylene (ethyne).

(ii) Ethene.

(b)

The structural formula of acetylene is:

H—C≡C—H

It is converted to neoprene rubber by polymerisation, after conversion to chloroprene (2-chlorobuta-1,3-diene).

(c)(i) Complete dehydration of potatoes gives mainly starch.

(c)(ii) Conversion of potatoes to ethanol

1. Wash, crush and pressure-cook the potatoes to rupture the cells and release the starch granules.
2. Add malt. The enzyme diastase in malt hydrolyses starch to maltose.
3. Allow the mixture to cool, then add yeast. The enzyme maltase hydrolyses maltose to glucose.
4. Keep the mixture warm (about 30–35°C) in the absence of air. The enzyme zymase in yeast ferments glucose to ethanol and carbon(IV) oxide.

The reactions are:

Starch hydrolysis:

\[\left(C_6H_{10}O_5\right)_n+\frac{n}{2}H_2O\xrightarrow{\text{diastase}}\frac{n}{2}C_{12}H_{22}O_{11}\]

Maltose hydrolysis:

\[C_{12}H_{22}O_{11}+H_2O\xrightarrow{\text{maltase}}2C_6H_{12}O_6\]

Final fermentation stage:

\[C_6H_{12}O_6\xrightarrow{\text{zymase}}2C_2H_5OH+2CO_2\]

(c)(iii) The fermented liquid is distilled using the apparatus shown below. Ethanol, with a boiling point of about 78°C, distils before most of the water.

The technique is fractional distillation.

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(a) Why fused alumina is mixed with cryolite

• Cryolite dissolves the alumina and lowers its melting point from about 2050 °C to about 950 °C, greatly reducing the energy and cost of the extraction.
• The molten cryolite increases the electrical conductivity of the melt, allowing electrolysis to proceed efficiently.

(b) Reaction of sodium hydroxide with bauxite

The amphoteric alumina in bauxite dissolves in hot concentrated sodium hydroxide to form soluble sodium aluminate: \[Al_2O_{3(s)} + 2NaOH_{(aq)} \rightarrow 2NaAlO_{2(aq)} + H_2O_{(l)}\]

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At constant pressure, the solubility of a gas in a liquid decreases as the temperature increases.

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(a)(i) Two uses of chlorine are:

1. sterilising drinking water and swimming-pool water;
2. manufacture of bleaching agents and for bleaching wood pulp and textiles.

(a)(ii) Chlorine first turns moist blue litmus paper red and then bleaches it white.

(b) A dry sample of chlorine is prepared by warming manganese(IV) oxide with concentrated hydrochloric acid. The gas is washed with saturated sodium chloride solution to remove hydrogen chloride, dried with concentrated tetraoxosulphate(VI) acid, and collected by downward displacement of air.

\[\mathrm{MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)}\]

(c)(i) Chlorine reacts with aqueous iron(II) chloride by a redox reaction. Iron(II) ions are oxidised to iron(III) ions.

\[\mathrm{Cl_2(aq)+2Fe^{2+}(aq)\rightarrow 2Cl^-(aq)+2Fe^{3+}(aq)}\]

Equivalently:

\[\mathrm{Cl_2(aq)+2FeCl_2(aq)\rightarrow 2FeCl_3(aq)}\]

(c)(ii) Chlorine reacts with propane in sunlight or ultraviolet light by a substitution reaction (free-radical substitution).

\[\mathrm{C_3H_8(g)+Cl_2(g)\xrightarrow{UV\ light}C_3H_7Cl(g)+HCl(g)}\]

The organic product is chloropropane, formed as a mixture of 1-chloropropane and 2-chloropropane.

(d) The increasing order of oxidising ability is:

\[\mathrm{Br_2 < Cl_2 < F_2}\]

Chlorine oxidises bromide ions to bromine, so chlorine is a stronger oxidising agent than bromine. Fluorine oxidises chloride ions to chlorine, so fluorine is a stronger oxidising agent than chlorine.

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(a) Two possible gases

A colourless, odourless gas that burns in oxygen with a pale blue flame could be hydrogen (H₂) or carbon(II) oxide (CO). Both are colourless and odourless, and both burn with a pale blue flame.

(b) Chemical test to distinguish them

Burn a sample of X and test the product of combustion:

• If X is hydrogen, it burns to form water only, which turns white anhydrous copper(II) tetraoxosulphate(VI) blue (or blue cobalt chloride paper pink) and gives no reaction with limewater.
• If X is carbon(II) oxide, it burns to form carbon(IV) oxide, which turns limewater milky. \[2CO + O_2 \rightarrow 2CO_2\]

Therefore, passing the combustion product into limewater identifies the gas: a milky precipitate confirms carbon(II) oxide, while the formation of water confirms hydrogen.

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(a) Differences between electrovalent and covalent compounds

Electrovalent (ionic) compounds	Covalent compounds
They consist of oppositely charged ions held by strong electrostatic forces.	They consist of molecules formed by sharing electrons.
They conduct electricity when molten or in aqueous solution because their ions are mobile.	They generally do not conduct electricity because they have no mobile ions or electrons.
They are generally soluble in water and have high melting and boiling points.	They are generally insoluble in water, soluble in organic solvents, and have relatively low melting and boiling points.

(b)

1. Metals are good conductors of electricity because they contain mobile, delocalised electrons.
2. Metals are malleable and ductile because layers of positive metal ions can slide over one another while metallic bonding is maintained.

(c)

(i) The process is radioactive beta decay, involving emission of a beta particle.

(ii) The mass number is unchanged, while the atomic number increases from 90 to 91. Therefore,

\[{}^{234}_{90}\mathrm{Th}\ \longrightarrow\ {}^{0}_{-1}\mathrm{e}+{}^{234}_{91}\mathrm{Pa}\]

Thus, \(X\) is a beta particle, \({}^{0}_{-1}\mathrm{e}\).

(iii) A Geiger-Müller tube or Geiger-Müller counter can be used to detect the beta radiation.

(iv) The mass of thorium decreases exponentially with time, becoming half of its previous value in each successive half-life.

(d)

1. Crush and concentrate the carbonate ore, \(\mathrm{YCO_3}\).
2. Heat the concentrated ore strongly (calcination) to decompose the carbonate to the oxide:
   \[\mathrm{YCO_3(s)\ \xrightarrow{\Delta}\ YO(s)+CO_2(g)}\]
3. Reduce the oxide with carbon in a furnace to obtain the metal:
   \[\mathrm{YO(s)+C(s)\ \xrightarrow{\Delta}\ Y(s)+CO(g)}\]
4. Tap off the metal and purify it where necessary.

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The diagram shows the same chemical symbol X written twice, each time with a subscript on the lower left and a superscript on the upper left:

• Left atom: \( {}^{q}_{p}\text{X} \)
• Right atom: \( {}^{r}_{p}\text{X} \)

In the standard nuclide notation \( {}^{A}_{Z}\text{X} \), the lower figure is the atomic number \(Z\) (number of protons) and the upper figure is the mass number \(A\) (protons plus neutrons). Reading the two atoms off the diagram:

Both atoms carry the same lower figure \(p\), so both have the same number of protons and are therefore atoms of the same element X. They differ only in the upper figure (\(q\) against \(r\)).

(a) Relationship between the two atoms

They are isotopes of the same element. Isotopes are atoms of the same element that have the same atomic number but different mass numbers.

(b) The difference between them

They have different mass numbers (\(q\) for the left atom and \(r\) for the right atom). Since the mass number is protons plus neutrons and both have the same number of protons \(p\), the difference lies wholly in the number of neutrons.

Number of neutrons is given by \( N = A - Z \):

Their number of protons (\(p\)) and, in the neutral state, their number of electrons are the same, so they have identical chemical properties and differ only in relative atomic mass and physical properties that depend on mass.

(c) Two examples of elements that exhibit this phenomenon (isotopy)

1. Chlorine: \( {}^{35}_{17}\text{Cl} \) and \( {}^{37}_{17}\text{Cl} \)
2. Carbon: \( {}^{12}_{6}\text{C} \) and \( {}^{14}_{6}\text{C} \)

(Hydrogen, \( {}^{1}_{1}\text{H} \), \( {}^{2}_{1}\text{H} \) and \( {}^{3}_{1}\text{H} \), and oxygen, \( {}^{16}_{8}\text{O} \) and \( {}^{18}_{8}\text{O} \), are also acceptable examples.)