WAEC SSCE - Chemistry - 1994 (Essay)
Question 1 Report
(1)(a) State Avogadro's law
(b) Which of the state of matter contains particles that are: (i) readily used
(ii) held firmly together by some forces of cohesion;
(iii) involved in rapid random motion?
(a) Avogadro's law
Equal volumes of all gases, measured at the same temperature and pressure, contain the same number of molecules.
(b) States of matter
- (i) Particles that can move readily and flow (but are still fairly close together): liquid.
- (ii) Particles held firmly together by strong forces of cohesion: solid.
- (iii) Particles in rapid, random motion and far apart: gas.
Answer Details
(a) Avogadro's law
Equal volumes of all gases, measured at the same temperature and pressure, contain the same number of molecules.
(b) States of matter
- (i) Particles that can move readily and flow (but are still fairly close together): liquid.
- (ii) Particles held firmly together by strong forces of cohesion: solid.
- (iii) Particles in rapid, random motion and far apart: gas.
Question 2 Report
(a) Define oxidation in terms of electron transfer.
(b) Consider the following: Cu\(_{(s)}\) + 2Ag\(^{+}_{(Ag)}\) \(\to\) Cu\(^{2+}_{(aq)}\) + 2Ag\(_{(s)}\)
(i) State the species that is reduced
(ii) Write half-cell equation for each of the species.
(a) Oxidation (in terms of electron transfer)
Oxidation is the loss of electrons by an atom, ion or molecule.
(b) \( Cu_{(s)} + 2Ag^+_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)} \)
(i) The species that is reduced is Ag+, because it gains electrons (its oxidation state falls from +1 to 0).
(ii) Half-cell equations:
Oxidation: \[ Cu \to Cu^{2+} + 2e^- \]
Reduction: \[ 2Ag^+ + 2e^- \to 2Ag \]
Answer Details
(a) Oxidation (in terms of electron transfer)
Oxidation is the loss of electrons by an atom, ion or molecule.
(b) \( Cu_{(s)} + 2Ag^+_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)} \)
(i) The species that is reduced is Ag+, because it gains electrons (its oxidation state falls from +1 to 0).
(ii) Half-cell equations:
Oxidation: \[ Cu \to Cu^{2+} + 2e^- \]
Reduction: \[ 2Ag^+ + 2e^- \to 2Ag \]
Question 3 Report
(a) Give one example each of a naturally occurring substance that;
(i) conforms to the general formula C(_x\)(H\(_2\)O)\(_y\);
(ii) contains the carboxyl group as its functional group.
(b) Name the process for obtaining: (i) paraffin oil from crude oil; (ii) benzene from ethyne.
(a) Examples of naturally occurring substances
- (i) Conforms to the general formula Cx(H2O)y (a carbohydrate): glucose (or starch, cellulose, sucrose).
- (ii) Contains the carboxyl group (-COOH) as its functional group: a fatty acid such as stearic acid (or citric acid found in citrus fruits).
(b) Names of the processes
- (i) Obtaining paraffin oil (kerosene) from crude oil: fractional distillation.
- (ii) Obtaining benzene from ethyne: polymerisation (cyclic trimerisation), \(3C_2H_2 \to C_6H_6\).
Answer Details
(a) Examples of naturally occurring substances
- (i) Conforms to the general formula Cx(H2O)y (a carbohydrate): glucose (or starch, cellulose, sucrose).
- (ii) Contains the carboxyl group (-COOH) as its functional group: a fatty acid such as stearic acid (or citric acid found in citrus fruits).
(b) Names of the processes
- (i) Obtaining paraffin oil (kerosene) from crude oil: fractional distillation.
- (ii) Obtaining benzene from ethyne: polymerisation (cyclic trimerisation), \(3C_2H_2 \to C_6H_6\).
Question 4 Report
(a) When MgSO\(_4\).XH\(_2\)O crystals were exposed to the atmosphere for several days, three was a loss in mass
(i) What name is given to this phenomenon?
(ii) Give another example of a compound that exhibits this phenomenon
(b) If 0.50 mole of MgSO\(_4\).XH\(_2\)O has a mass of 123g. calculate the value of X. (H = 1, O = 16, Mg = 24, S = 32)
(a)(i) The loss in mass of a hydrated crystal (loss of water of crystallisation) to the atmosphere is called efflorescence.
(ii) Another example: sodium trioxocarbonate(IV) decahydrate, Na2CO3·10H2O (washing soda).
(b) Finding X in MgSO4·XH2O
Molar mass of the hydrate \( = \dfrac{\text{mass}}{\text{moles}} = \dfrac{123}{0.50} = 246\ \text{g mol}^{-1} \)
Molar mass of anhydrous MgSO4 \( = 24 + 32 + (4\times16) = 120\ \text{g mol}^{-1} \)
Mass of water \( = 246 - 120 = 126\ \text{g} \)
\[ X = \frac{126}{18} = \mathbf{7} \]
The compound is MgSO4·7H2O.
Answer Details
(a)(i) The loss in mass of a hydrated crystal (loss of water of crystallisation) to the atmosphere is called efflorescence.
(ii) Another example: sodium trioxocarbonate(IV) decahydrate, Na2CO3·10H2O (washing soda).
(b) Finding X in MgSO4·XH2O
Molar mass of the hydrate \( = \dfrac{\text{mass}}{\text{moles}} = \dfrac{123}{0.50} = 246\ \text{g mol}^{-1} \)
Molar mass of anhydrous MgSO4 \( = 24 + 32 + (4\times16) = 120\ \text{g mol}^{-1} \)
Mass of water \( = 246 - 120 = 126\ \text{g} \)
\[ X = \frac{126}{18} = \mathbf{7} \]
The compound is MgSO4·7H2O.
Question 5 Report
(a)(i) Draw a labelled diagram for the laboratory preparation of a dry sample of chlorine
(ii) Give one chemical test for chlorine.
(b) Write equations to represent the reaction of chlorine gas with: (i) iron (II) chloride solution;
(ii) potassium iodide solution;
(iii) hot concentrated sodium hydroxide solution.
(c) State what is observed on:
(i) bubbling hydrogen chloride gas into an solution of lead (II) trioxonitrate (V);
(ii) heating the mixture from (c)(i) above to boiling and aging it to cool.
(d) A solution of bismuth chloride was prepared by adding the oxychloride which is a white powder to concentrated hydrochloric acid. The following equilibrium was set up: BiOCI\(_{(s)}\) + 2HCI\(_{(aq)}\) \(\rightleftharpoons\) BiCl\(_{3(aq)}\) + H\(_2\)O\(_{(q)}\). State what would be observed if some water is added to the system. Explain your answer
(a)(i) Laboratory preparation of a dry sample of chlorine
Heat manganese(IV) oxide with concentrated hydrochloric acid. The chlorine produced is passed through saturated brine (or aqueous potassium hydrogencarbonate) to remove hydrogen chloride, then through concentrated sulphuric acid to dry it. It is collected in an upright gas jar by downward displacement of air, since chlorine is denser than air.
\[\mathrm{MnO_2(s)+4HCl(conc)\rightarrow MnCl_2(aq)+Cl_2(g)+2H_2O(l)}\]
(ii) Test for chlorine
Hold moist starch-iodide paper in the gas. It turns blue-black because chlorine liberates iodine from iodide ions.
\[\mathrm{Cl_2+2I^-\rightarrow 2Cl^-+I_2}\]
(b) Equations for the reactions of chlorine
(i) With iron(II) chloride solution:
\[\mathrm{Cl_2+2FeCl_2\rightarrow 2FeCl_3}\]
(ii) With potassium iodide solution:
\[\mathrm{Cl_2+2KI\rightarrow 2KCl+I_2}\]
(iii) With hot concentrated sodium hydroxide solution:
\[\mathrm{3Cl_2+6NaOH\rightarrow 5NaCl+NaClO_3+3H_2O}\]
(c)
(i) A white precipitate of lead(II) chloride is formed when hydrogen chloride gas is bubbled into lead(II) trioxonitrate(V) solution.
\[\mathrm{Pb(NO_3)_2(aq)+2HCl(aq)\rightarrow PbCl_2(s)+2HNO_3(aq)}\]
(ii) The white precipitate dissolves on heating to boiling. On cooling, white crystals of lead(II) chloride reappear.
(d)
A white precipitate or milky suspension of bismuth oxychloride, \(\mathrm{BiOCl}\), forms when water is added.
Adding water dilutes the hydrochloric acid and shifts the equilibrium to the left. Soluble \(\mathrm{BiCl_3}\) is thereby converted into insoluble white \(\mathrm{BiOCl}\):
\[\mathrm{BiOCl(s)+2HCl(aq)\rightleftharpoons BiCl_3(aq)+H_2O(l)}\]
Answer Details
(a)(i) Laboratory preparation of a dry sample of chlorine
Heat manganese(IV) oxide with concentrated hydrochloric acid. The chlorine produced is passed through saturated brine (or aqueous potassium hydrogencarbonate) to remove hydrogen chloride, then through concentrated sulphuric acid to dry it. It is collected in an upright gas jar by downward displacement of air, since chlorine is denser than air.
\[\mathrm{MnO_2(s)+4HCl(conc)\rightarrow MnCl_2(aq)+Cl_2(g)+2H_2O(l)}\]
(ii) Test for chlorine
Hold moist starch-iodide paper in the gas. It turns blue-black because chlorine liberates iodine from iodide ions.
\[\mathrm{Cl_2+2I^-\rightarrow 2Cl^-+I_2}\]
(b) Equations for the reactions of chlorine
(i) With iron(II) chloride solution:
\[\mathrm{Cl_2+2FeCl_2\rightarrow 2FeCl_3}\]
(ii) With potassium iodide solution:
\[\mathrm{Cl_2+2KI\rightarrow 2KCl+I_2}\]
(iii) With hot concentrated sodium hydroxide solution:
\[\mathrm{3Cl_2+6NaOH\rightarrow 5NaCl+NaClO_3+3H_2O}\]
(c)
(i) A white precipitate of lead(II) chloride is formed when hydrogen chloride gas is bubbled into lead(II) trioxonitrate(V) solution.
\[\mathrm{Pb(NO_3)_2(aq)+2HCl(aq)\rightarrow PbCl_2(s)+2HNO_3(aq)}\]
(ii) The white precipitate dissolves on heating to boiling. On cooling, white crystals of lead(II) chloride reappear.
(d)
A white precipitate or milky suspension of bismuth oxychloride, \(\mathrm{BiOCl}\), forms when water is added.
Adding water dilutes the hydrochloric acid and shifts the equilibrium to the left. Soluble \(\mathrm{BiCl_3}\) is thereby converted into insoluble white \(\mathrm{BiOCl}\):
\[\mathrm{BiOCl(s)+2HCl(aq)\rightleftharpoons BiCl_3(aq)+H_2O(l)}\]
Question 6 Report
Potassium trioxochlorate (V) undergoes thermal decomposition according to the fotpowing equation: 2KCIO\(_3\) \(\to\) 2KCI + 3O\(_2\)
(a) What substance could be used in the laboratory to in ease the rate of the reaction
(ii) absorb the oxygen produced
(b) Give the reason why an aqueous solution of silver trioxonitrate (V) gives a white precipitate with KCI but not with KClO\(_3\).
Thermal decomposition: \( 2KClO_3 \to 2KCl + 3O_2 \)
(a)(i) Substance to increase the rate of the reaction (a catalyst): manganese(IV) oxide, MnO2.
(ii) Substance to absorb the oxygen produced: alkaline pyrogallol solution (pyrogallol in KOH).
(b) An aqueous solution of silver trioxonitrate(V) gives a white precipitate (AgCl) with KCl because KCl contains free chloride ions, Cl-:
\[ Ag^+ + Cl^- \to AgCl\downarrow \]
KClO3 gives no precipitate because its chlorine is locked in the chlorate ion, ClO3- (not free Cl-), and silver chlorate, AgClO3, is soluble in water. Only free chloride ions form the insoluble AgCl.
Answer Details
Thermal decomposition: \( 2KClO_3 \to 2KCl + 3O_2 \)
(a)(i) Substance to increase the rate of the reaction (a catalyst): manganese(IV) oxide, MnO2.
(ii) Substance to absorb the oxygen produced: alkaline pyrogallol solution (pyrogallol in KOH).
(b) An aqueous solution of silver trioxonitrate(V) gives a white precipitate (AgCl) with KCl because KCl contains free chloride ions, Cl-:
\[ Ag^+ + Cl^- \to AgCl\downarrow \]
KClO3 gives no precipitate because its chlorine is locked in the chlorate ion, ClO3- (not free Cl-), and silver chlorate, AgClO3, is soluble in water. Only free chloride ions form the insoluble AgCl.
Question 7 Report
(a)(i) What is meant by hydrocarbons?
(ii) A hydrocarbon consists of 92.3% carbon. If it's vapour density is 39, determine its molecular formula. (H = 1; C = 12)
(b)(i) Outline a suitable laboratory procedure for obtaining ethanol from cassava tubers
(ii) List two laboratory reagents used for oxidizing ethanol to ethanoic acid.
(c) What name is given to each of the following processes?:
(i) Conversion of alkanols to alkanoates;
(ii) Breakdown of proteins to amino acids;
(iii) Conversion of oils to fats
(iv) Alkaline hydrolysis of fats and oils.
(a)(i) Hydrocarbons are compounds that contain only carbon and hydrogen.
(ii) Molecular formula
Carbon = 92.3%, so hydrogen = 100 - 92.3 = 7.7%.
C: \( \dfrac{92.3}{12} = 7.69 \) H: \( \dfrac{7.7}{1} = 7.70 \) ratio \(1:1\), so empirical formula = CH (mass 13).
Molar mass \( = 2 \times \text{vapour density} = 2 \times 39 = 78 \)
\( n = \dfrac{78}{13} = 6 \), so molecular formula \( = \mathbf{C_6H_6} \) (benzene).
(b)(i) Ethanol from cassava
Cassava is rich in starch. Grate the tubers and mix with water; hydrolyse the starch to glucose (using malt/diastase enzyme or by boiling with dilute acid then neutralising). Cool, add yeast and allow to ferment anaerobically for a few days at about 35°C; the enzyme zymase in yeast converts glucose to ethanol and carbon dioxide. Finally distil (fractional distillation) to concentrate the ethanol.
\[ C_6H_{12}O_6 \xrightarrow{\text{yeast}} 2C_2H_5OH + 2CO_2 \]
(ii) Reagents for oxidising ethanol to ethanoic acid: acidified potassium dichromate(VI) (K2Cr2O7/H2SO4); acidified potassium tetraoxomanganate(VII) (KMnO4/H2SO4).
(c) Names of processes
- (i) Conversion of alkanols to alkanoates: esterification.
- (ii) Breakdown of proteins to amino acids: hydrolysis.
- (iii) Conversion of oils to fats: hydrogenation.
- (iv) Alkaline hydrolysis of fats and oils: saponification.
Answer Details
(a)(i) Hydrocarbons are compounds that contain only carbon and hydrogen.
(ii) Molecular formula
Carbon = 92.3%, so hydrogen = 100 - 92.3 = 7.7%.
C: \( \dfrac{92.3}{12} = 7.69 \) H: \( \dfrac{7.7}{1} = 7.70 \) ratio \(1:1\), so empirical formula = CH (mass 13).
Molar mass \( = 2 \times \text{vapour density} = 2 \times 39 = 78 \)
\( n = \dfrac{78}{13} = 6 \), so molecular formula \( = \mathbf{C_6H_6} \) (benzene).
(b)(i) Ethanol from cassava
Cassava is rich in starch. Grate the tubers and mix with water; hydrolyse the starch to glucose (using malt/diastase enzyme or by boiling with dilute acid then neutralising). Cool, add yeast and allow to ferment anaerobically for a few days at about 35°C; the enzyme zymase in yeast converts glucose to ethanol and carbon dioxide. Finally distil (fractional distillation) to concentrate the ethanol.
\[ C_6H_{12}O_6 \xrightarrow{\text{yeast}} 2C_2H_5OH + 2CO_2 \]
(ii) Reagents for oxidising ethanol to ethanoic acid: acidified potassium dichromate(VI) (K2Cr2O7/H2SO4); acidified potassium tetraoxomanganate(VII) (KMnO4/H2SO4).
(c) Names of processes
- (i) Conversion of alkanols to alkanoates: esterification.
- (ii) Breakdown of proteins to amino acids: hydrolysis.
- (iii) Conversion of oils to fats: hydrogenation.
- (iv) Alkaline hydrolysis of fats and oils: saponification.
Question 8 Report
(a) (i) Define heat of combustion.
(ii) What name is given to the container used for determining tne reaction?
(b) The heat of combustion of carbon in excess air is - 3935 kJ.
(i) Sketch an energy profile diagram for the reaction.
(ii) Calculate the heat change when 60 g of carbon undergoes complete combustion to produce carbon (IV) oxide. (C = 12)
(iii) Explain why the value of the heat of neutralization of strong acids by strong bases is constant.
(c) Give reason for the following:
(i) rusting of iron is regarded as a slow combustion process;
(ii) iron filings rust much more faster than iron nails when exposed to the same atmospheric condition;
(iii) iron is better protected from corrosion by plating it with.zinc than with tin
(a)(i) The heat of combustion is the heat evolved when one mole of a substance is completely burnt in oxygen.
(a)(ii) The container used to determine the heat change is a calorimeter. A bomb calorimeter is a type of calorimeter used for combustion reactions.
(b)(i) Combustion of carbon is exothermic:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \]
The products are at a lower energy level than the reactants because heat is released. Therefore, \( \Delta H \) is negative.
(b)(ii) The stated heat of combustion is understood as \( -3935\ \text{kJ mol}^{-1} \).
\[ \text{Moles of carbon}=\frac{60\ \text{g}}{12\ \text{g mol}^{-1}}=5\ \text{mol} \]
\[ \text{Heat change}=5\ \text{mol}\times(-3935\ \text{kJ mol}^{-1}) \]
\[ \boxed{\text{Heat change}=-19675\ \text{kJ}} \]
The negative sign shows that \(19675\ \text{kJ}\) of heat is released to the surroundings.
(b)(iii) Strong acids and strong bases completely ionise in water. Thus, every neutralisation between them has the same net ionic equation:
\[ H_3O^+(aq)+OH^-(aq)\rightarrow 2H_2O(l) \]
Since the same reaction occurs in each case, the same heat is released per mole of water formed. Therefore, the heat of neutralisation of a strong acid by a strong base is constant.
(c)(i) Rusting is regarded as slow combustion because both processes involve oxidation: iron combines with oxygen to form an oxide, and heat is released. Rusting occurs much more slowly than ordinary burning.
(c)(ii) Iron filings rust faster than iron nails because filings have a much larger surface area exposed to oxygen and water. A larger surface area increases the rate of oxidation.
(c)(iii) Zinc protects iron better than tin because zinc is more reactive than iron. If a zinc coating is scratched, zinc oxidises in preference to iron, giving sacrificial protection. Tin is less reactive than iron; therefore, if a tin coating is broken, the exposed iron can rust.
Examination reminder: For corrosion questions, compare the reactivity of the coating metal with iron. A metal more reactive than iron can protect it even when the coating is damaged.
Answer Details
(a)(i) The heat of combustion is the heat evolved when one mole of a substance is completely burnt in oxygen.
(a)(ii) The container used to determine the heat change is a calorimeter. A bomb calorimeter is a type of calorimeter used for combustion reactions.
(b)(i) Combustion of carbon is exothermic:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \]
The products are at a lower energy level than the reactants because heat is released. Therefore, \( \Delta H \) is negative.
(b)(ii) The stated heat of combustion is understood as \( -3935\ \text{kJ mol}^{-1} \).
\[ \text{Moles of carbon}=\frac{60\ \text{g}}{12\ \text{g mol}^{-1}}=5\ \text{mol} \]
\[ \text{Heat change}=5\ \text{mol}\times(-3935\ \text{kJ mol}^{-1}) \]
\[ \boxed{\text{Heat change}=-19675\ \text{kJ}} \]
The negative sign shows that \(19675\ \text{kJ}\) of heat is released to the surroundings.
(b)(iii) Strong acids and strong bases completely ionise in water. Thus, every neutralisation between them has the same net ionic equation:
\[ H_3O^+(aq)+OH^-(aq)\rightarrow 2H_2O(l) \]
Since the same reaction occurs in each case, the same heat is released per mole of water formed. Therefore, the heat of neutralisation of a strong acid by a strong base is constant.
(c)(i) Rusting is regarded as slow combustion because both processes involve oxidation: iron combines with oxygen to form an oxide, and heat is released. Rusting occurs much more slowly than ordinary burning.
(c)(ii) Iron filings rust faster than iron nails because filings have a much larger surface area exposed to oxygen and water. A larger surface area increases the rate of oxidation.
(c)(iii) Zinc protects iron better than tin because zinc is more reactive than iron. If a zinc coating is scratched, zinc oxidises in preference to iron, giving sacrificial protection. Tin is less reactive than iron; therefore, if a tin coating is broken, the exposed iron can rust.
Examination reminder: For corrosion questions, compare the reactivity of the coating metal with iron. A metal more reactive than iron can protect it even when the coating is damaged.
Question 9 Report
(a) Write an equation to show the action of strong heat on:
(i) potassium trioxonitrate (V):
(ii) sucrose.
(b) State two observations in respect of the reaction between granulated zinc and dilute tetraoxosulphate (VI) acid.
(a) Action of strong heat
(i) Potassium trioxonitrate(V) (potassium nitrate) decomposes to the nitrite and oxygen:
\[ 2KNO_3 \to 2KNO_2 + O_2 \]
(ii) Sucrose chars (is dehydrated) to carbon and water vapour:
\[ C_{12}H_{22}O_{11} \to 12C + 11H_2O \]
(b) Granulated zinc with dilute tetraoxosulphate(VI) acid
- Effervescence occurs as bubbles of a colourless gas (hydrogen) are given off and the zinc gradually dissolves/gets smaller.
- The mixture warms up (the reaction is exothermic), leaving a colourless solution of zinc sulphate.
\[ Zn + H_2SO_4 \to ZnSO_4 + H_2 \]
Answer Details
(a) Action of strong heat
(i) Potassium trioxonitrate(V) (potassium nitrate) decomposes to the nitrite and oxygen:
\[ 2KNO_3 \to 2KNO_2 + O_2 \]
(ii) Sucrose chars (is dehydrated) to carbon and water vapour:
\[ C_{12}H_{22}O_{11} \to 12C + 11H_2O \]
(b) Granulated zinc with dilute tetraoxosulphate(VI) acid
- Effervescence occurs as bubbles of a colourless gas (hydrogen) are given off and the zinc gradually dissolves/gets smaller.
- The mixture warms up (the reaction is exothermic), leaving a colourless solution of zinc sulphate.
\[ Zn + H_2SO_4 \to ZnSO_4 + H_2 \]
Question 10 Report
A shortened form of the Periodic Table is shown below. Use it to answer questions (a) and (b)
(a) Which of the elements represented as A to E in the table, above is:
(i) transition metal;
(ii) an alkaline earth meta
(iii) the least reactive;
(iv) the most electronegative?
(b)(i) What type of bond would exist in a compound formed when element D reacts with oxygen?
(ii) Write the formula of the compound formed in (b)(i) above.
Reading the shortened Periodic Table. The columns are the groups, labelled I, II (on the far left), then the wide transition-metal block, then III, IV, V, VI, VII, 0 (on the far right). The rows 1, 2, 3, 4 are the periods. From the diagram the five elements sit as follows:
- A: far top-right, in Group 0 (period 1) - a noble gas.
- B: on the right, in the Group VII region (period 3) - a non-metal.
- C: far left, in Group I (period 3) - an alkali metal.
- D: on the left, in Group II (period 4) - an alkaline earth metal.
- E: in the central transition block (period 4), between Groups II and III.
(a)(i) Transition metal. Transition metals occupy the block wedged between Group II and Group III. The only element sitting in that central block is E.
(a)(ii) Alkaline earth metal. The alkaline earth metals are the Group II elements. The element placed in the second column (Group II) is D.
(a)(iii) Least reactive. The least reactive elements are the noble gases of Group 0, which have stable, filled outer electron shells and rarely form compounds. The Group 0 element is A.
(a)(iv) Most electronegative. Electronegativity (the pull an atom exerts on a shared electron pair) increases across a period towards Group VII and decreases down a group; the halogens of Group VII are the most electronegative reactive elements (noble gases are excluded as they scarcely bond). The Group VII element here is B.
(b)(i) Bond type in D's oxide. Element D is a Group II metal, so it readily loses its two outer electrons to form a
D2+ cation, while oxygen (Group VI) gains two electrons to form the O2- anion. A metal reacting with a non-metal by electron transfer gives an ionic (electrovalent) bond.
(b)(ii) Formula of the compound. Balancing the charges of the ions:
\[ \text{D}^{2+} \;+\; \text{O}^{2-} \;\longrightarrow\; \text{DO} \]One D2+ exactly balances one O2-, so the formula of the oxide is DO.
Answer Details
Reading the shortened Periodic Table. The columns are the groups, labelled I, II (on the far left), then the wide transition-metal block, then III, IV, V, VI, VII, 0 (on the far right). The rows 1, 2, 3, 4 are the periods. From the diagram the five elements sit as follows:
- A: far top-right, in Group 0 (period 1) - a noble gas.
- B: on the right, in the Group VII region (period 3) - a non-metal.
- C: far left, in Group I (period 3) - an alkali metal.
- D: on the left, in Group II (period 4) - an alkaline earth metal.
- E: in the central transition block (period 4), between Groups II and III.
(a)(i) Transition metal. Transition metals occupy the block wedged between Group II and Group III. The only element sitting in that central block is E.
(a)(ii) Alkaline earth metal. The alkaline earth metals are the Group II elements. The element placed in the second column (Group II) is D.
(a)(iii) Least reactive. The least reactive elements are the noble gases of Group 0, which have stable, filled outer electron shells and rarely form compounds. The Group 0 element is A.
(a)(iv) Most electronegative. Electronegativity (the pull an atom exerts on a shared electron pair) increases across a period towards Group VII and decreases down a group; the halogens of Group VII are the most electronegative reactive elements (noble gases are excluded as they scarcely bond). The Group VII element here is B.
(b)(i) Bond type in D's oxide. Element D is a Group II metal, so it readily loses its two outer electrons to form a
D2+ cation, while oxygen (Group VI) gains two electrons to form the O2- anion. A metal reacting with a non-metal by electron transfer gives an ionic (electrovalent) bond.
(b)(ii) Formula of the compound. Balancing the charges of the ions:
\[ \text{D}^{2+} \;+\; \text{O}^{2-} \;\longrightarrow\; \text{DO} \]One D2+ exactly balances one O2-, so the formula of the oxide is DO.
Question 11 Report
(a) State two differences in the chemical properties of metals and non-metals.
(b) List two general methods of extract metals from their ores.
(a) Chemical differences between metals and non-metals
- Metals form basic oxides (and hydroxides), whereas non-metals form acidic oxides.
- Metals are electropositive and lose electrons to form positive ions (acting as reducing agents), whereas non-metals are electronegative and gain electrons to form negative ions (acting as oxidising agents).
(b) Two general methods of extracting metals from their ores
- Reduction of the metal oxide with carbon (coke), used for moderately reactive metals such as iron and zinc.
- Electrolysis of the molten ore or compound, used for very reactive metals such as sodium and aluminium.
Answer Details
(a) Chemical differences between metals and non-metals
- Metals form basic oxides (and hydroxides), whereas non-metals form acidic oxides.
- Metals are electropositive and lose electrons to form positive ions (acting as reducing agents), whereas non-metals are electronegative and gain electrons to form negative ions (acting as oxidising agents).
(b) Two general methods of extracting metals from their ores
- Reduction of the metal oxide with carbon (coke), used for moderately reactive metals such as iron and zinc.
- Electrolysis of the molten ore or compound, used for very reactive metals such as sodium and aluminium.
Question 12 Report
When ethane - 1,2-dioic acid is heated with concentrated tetraoxosulphate (VI) acid, a reaction represented by the following equation occurs:
(a) State the type of process involved in the reaction
(b) What is the basicity of ethane-1, s-dioic acid?
(c) List three differences in the chemical properties of the two oxides of carbon produced during the reaction
The equation shown on the diagram is the action of hot concentrated tetraoxosulphate(VI) acid on ethanedioic (oxalic) acid:
\[ \underset{\text{(COOH)}_2}{\text{HOOC-COOH}} \;\xrightarrow[\;\Delta\;]{\text{conc. } H_2SO_4}\; CO_2 + CO + H_2O \]
(a) Type of process involved
The process is dehydration. The concentrated tetraoxosulphate(VI) acid acts as a dehydrating agent: it removes the elements of water (\(H_2O\)) from each mole of ethanedioic acid, leaving behind the two oxides of carbon, \(CO_2\) and \(CO\). Notice that the atoms balance exactly, \((COOH)_2 = C_2H_2O_4\) splits into \(CO_2 + CO + H_2O\), so no oxidation state change is imposed by the acid; it simply strips out water.
(b) Basicity of ethanedioic acid
The basicity of an acid is the number of replaceable (ionisable) hydrogen atoms it releases per molecule in aqueous solution. Ethanedioic acid, \((COOH)_2\), contains two carboxyl (-COOH) groups, each supplying one ionisable hydrogen:
\[ (COOH)_2 \rightleftharpoons 2H^+ + (COO)_2^{2-} \]
Therefore its basicity is 2 (it is a dibasic / diprotic acid).
(c) Three differences in the chemical properties of the two oxides of carbon (\(CO_2\) and \(CO\))
| Property | Carbon(IV) oxide, \(CO_2\) | Carbon(II) oxide, \(CO\) |
|---|---|---|
| Acid/base nature | Acidic oxide: turns moist blue litmus red and reacts with alkalis, e.g. \(CO_2 + 2NaOH \rightarrow Na_2CO_3 + H_2O\) | Neutral oxide: has no effect on litmus and does not react with alkalis |
| Reducing action | Not a reducing agent; it is a stable, fully oxidised product and cannot reduce heated metal oxides | Strong reducing agent: reduces heated metal oxides to the metal, e.g. \(CuO + CO \rightarrow Cu + CO_2\) |
| Combustibility | Does not burn and does not support combustion (used to put out fires) | Combustible: burns in air with a blue flame, \(2CO + O_2 \rightarrow 2CO_2\) |
A further valid distinction: \(CO_2\) turns lime water milky (forming \(CaCO_3\)), whereas \(CO\) does not; and \(CO\) is highly poisonous because it combines with haemoglobin, while \(CO_2\) is not poisonous in that way.
Answer Details
The equation shown on the diagram is the action of hot concentrated tetraoxosulphate(VI) acid on ethanedioic (oxalic) acid:
\[ \underset{\text{(COOH)}_2}{\text{HOOC-COOH}} \;\xrightarrow[\;\Delta\;]{\text{conc. } H_2SO_4}\; CO_2 + CO + H_2O \]
(a) Type of process involved
The process is dehydration. The concentrated tetraoxosulphate(VI) acid acts as a dehydrating agent: it removes the elements of water (\(H_2O\)) from each mole of ethanedioic acid, leaving behind the two oxides of carbon, \(CO_2\) and \(CO\). Notice that the atoms balance exactly, \((COOH)_2 = C_2H_2O_4\) splits into \(CO_2 + CO + H_2O\), so no oxidation state change is imposed by the acid; it simply strips out water.
(b) Basicity of ethanedioic acid
The basicity of an acid is the number of replaceable (ionisable) hydrogen atoms it releases per molecule in aqueous solution. Ethanedioic acid, \((COOH)_2\), contains two carboxyl (-COOH) groups, each supplying one ionisable hydrogen:
\[ (COOH)_2 \rightleftharpoons 2H^+ + (COO)_2^{2-} \]
Therefore its basicity is 2 (it is a dibasic / diprotic acid).
(c) Three differences in the chemical properties of the two oxides of carbon (\(CO_2\) and \(CO\))
| Property | Carbon(IV) oxide, \(CO_2\) | Carbon(II) oxide, \(CO\) |
|---|---|---|
| Acid/base nature | Acidic oxide: turns moist blue litmus red and reacts with alkalis, e.g. \(CO_2 + 2NaOH \rightarrow Na_2CO_3 + H_2O\) | Neutral oxide: has no effect on litmus and does not react with alkalis |
| Reducing action | Not a reducing agent; it is a stable, fully oxidised product and cannot reduce heated metal oxides | Strong reducing agent: reduces heated metal oxides to the metal, e.g. \(CuO + CO \rightarrow Cu + CO_2\) |
| Combustibility | Does not burn and does not support combustion (used to put out fires) | Combustible: burns in air with a blue flame, \(2CO + O_2 \rightarrow 2CO_2\) |
A further valid distinction: \(CO_2\) turns lime water milky (forming \(CaCO_3\)), whereas \(CO\) does not; and \(CO\) is highly poisonous because it combines with haemoglobin, while \(CO_2\) is not poisonous in that way.
Question 13 Report
(a)(i) Determine the maximum number of electrons that can occupy the principal energy level M of an atom.
(ii) Show the changes in the electronic structures of atoms of sodium and fluorine (\(^{23}_{11}\)Na: \(^{19}_{9}F\)) when they combine to form sodium fluoride.
(iii) State: three properties that sodium fluoride would have, based a-n the bond type present in the compound.
(b) Write equations to show when:
(i) sodium metal burns in limited supply of oxygen;
(ii) water is added to the product in (b)(i) above
(c) Describe a suitable laboratory procedure for comparing the conductance of 1 mol dm\(^{-3}\) aqueous solutions of sodium hydroxide and ethanoic acid.
(a)(i) Maximum electrons in the M shell (principal quantum number n = 3): \( 2n^2 = 2 \times 3^2 = \mathbf{18} \).
(ii) Formation of sodium fluoride
Sodium (2, 8, 1) loses its one outer electron to become Na+ (2, 8); fluorine (2, 7) gains that electron to become F- (2, 8). Both attain the stable octet:
\[ Na \to Na^+ + e^- \qquad F + e^- \to F^- \qquad Na^+ + F^- \to NaF \]
A dot-and-cross diagram showing the single electron transferring from Na to F is expected.
(iii) Properties of NaF (ionic compound): high melting and boiling points; conducts electricity when molten or dissolved in water; a hard, crystalline solid that is soluble in water.
(b)
(i) \[ 4Na + O_2 \to 2Na_2O \]
(ii) \[ Na_2O + H_2O \to 2NaOH \]
(c) Comparing the conductance of 1 mol dm-3 NaOH and ethanoic acid
Set up a simple conductivity cell: two clean carbon (or platinum) electrodes connected through a battery and a bulb (or ammeter). Dip the electrodes into a fixed volume of 1 mol dm-3 NaOH and note the brightness of the bulb (or the ammeter reading). Rinse the electrodes and repeat with the same volume of 1 mol dm-3 ethanoic acid. The bulb glows brightly (large current) with NaOH, which is a strong electrolyte (fully ionised), but only dimly (small current) with ethanoic acid, which is a weak electrolyte (only partially ionised). Hence NaOH conducts better.
Answer Details
(a)(i) Maximum electrons in the M shell (principal quantum number n = 3): \( 2n^2 = 2 \times 3^2 = \mathbf{18} \).
(ii) Formation of sodium fluoride
Sodium (2, 8, 1) loses its one outer electron to become Na+ (2, 8); fluorine (2, 7) gains that electron to become F- (2, 8). Both attain the stable octet:
\[ Na \to Na^+ + e^- \qquad F + e^- \to F^- \qquad Na^+ + F^- \to NaF \]
A dot-and-cross diagram showing the single electron transferring from Na to F is expected.
(iii) Properties of NaF (ionic compound): high melting and boiling points; conducts electricity when molten or dissolved in water; a hard, crystalline solid that is soluble in water.
(b)
(i) \[ 4Na + O_2 \to 2Na_2O \]
(ii) \[ Na_2O + H_2O \to 2NaOH \]
(c) Comparing the conductance of 1 mol dm-3 NaOH and ethanoic acid
Set up a simple conductivity cell: two clean carbon (or platinum) electrodes connected through a battery and a bulb (or ammeter). Dip the electrodes into a fixed volume of 1 mol dm-3 NaOH and note the brightness of the bulb (or the ammeter reading). Rinse the electrodes and repeat with the same volume of 1 mol dm-3 ethanoic acid. The bulb glows brightly (large current) with NaOH, which is a strong electrolyte (fully ionised), but only dimly (small current) with ethanoic acid, which is a weak electrolyte (only partially ionised). Hence NaOH conducts better.
Question 14 Report
(a) State whether entropy increases or decreases during each of the following processes
(i) condensation of steam;
(ii) melting of wax;
(iii) dissolution of sugar in water;
(iv) abscas on charcoal.
(b) What deduction can be made in each case given that the value of the free energy change for a particular reaction is:
(i) zero;
ii) negative
(a) Entropy change
- (i) Condensation of steam (gas → liquid): entropy decreases.
- (ii) Melting of wax (solid → liquid): entropy increases.
- (iii) Dissolution of sugar in water: entropy increases.
- (iv) Adsorption of a gas on charcoal: entropy decreases (gas particles become ordered on the surface).
(b) Free energy change, ΔG
- (i) ΔG = zero: the reaction is at equilibrium.
- (ii) ΔG negative: the reaction is spontaneous (feasible) in the forward direction.
Answer Details
(a) Entropy change
- (i) Condensation of steam (gas → liquid): entropy decreases.
- (ii) Melting of wax (solid → liquid): entropy increases.
- (iii) Dissolution of sugar in water: entropy increases.
- (iv) Adsorption of a gas on charcoal: entropy decreases (gas particles become ordered on the surface).
(b) Free energy change, ΔG
- (i) ΔG = zero: the reaction is at equilibrium.
- (ii) ΔG negative: the reaction is spontaneous (feasible) in the forward direction.
