WAEC SSCE - Chemistry - 1998 (Essay)
Question 1 Report
(a)(i) What is meant by the activation energy of a reaction?
(ii) State the effect of a catalyst on activation energy.
(b) What substance serves as a catalyst in each of the following?
(I) Hydrogenation of oils
(ii) Biochemical reactions.
(a)(i) The activation energy of a reaction is the minimum amount of energy that the reacting particles must possess before a collision between them can lead to a reaction (i.e. the minimum energy needed for an effective collision).
(ii) A catalyst lowers (reduces) the activation energy of the reaction by providing an alternative reaction pathway of lower activation energy, so more colliding particles have the required energy and the reaction proceeds faster.
(b) Catalyst in each case:
- (i) Hydrogenation of oils: nickel (finely divided nickel).
- (ii) Biochemical reactions: enzymes (biological catalysts).
Answer Details
(a)(i) The activation energy of a reaction is the minimum amount of energy that the reacting particles must possess before a collision between them can lead to a reaction (i.e. the minimum energy needed for an effective collision).
(ii) A catalyst lowers (reduces) the activation energy of the reaction by providing an alternative reaction pathway of lower activation energy, so more colliding particles have the required energy and the reaction proceeds faster.
(b) Catalyst in each case:
- (i) Hydrogenation of oils: nickel (finely divided nickel).
- (ii) Biochemical reactions: enzymes (biological catalysts).
Question 2 Report
(a) What is the IUPAC name of Fe\(_2\)(SO\(_4\))\(_3\)?
(b)(i) Write an equation to represent the reaction of hydrogen sulphide with iron (III) chloride solution.
(ii) Mention one change observed during the reaction in (b)(i) above.
(a) IUPAC name of \(Fe_2(SO_4)_3\)
The compound contains the iron(III) cation, \(Fe^{3+}\), and the sulphate anion, \(SO_4^{2-}\). Because iron shows more than one oxidation state, the oxidation number of the metal must be shown in Roman numerals, and the anion is named in full as the tetraoxosulphate(VI) ion. The IUPAC name is therefore iron(III) tetraoxosulphate(VI) (commonly, iron(III) sulphate).
(b)(i) Reaction of hydrogen sulphide with iron(III) chloride solution
Hydrogen sulphide is a reducing agent. It reduces iron(III) to iron(II), and is itself oxidised, so that its sulphur is deposited as free (elemental) sulphur:
\[2FeCl_3(aq) + H_2S(g) \rightarrow 2FeCl_2(aq) + 2HCl(aq) + S(s)\]In ionic terms the essential change is \(2Fe^{3+} + H_2S \rightarrow 2Fe^{2+} + 2H^+ + S\), which shows clearly that iron is reduced from the \(+3\) to the \(+2\) state while sulphide sulphur (\(-2\)) is oxidised to sulphur (\(0\)).
(b)(ii) One observed change
The yellow-brown iron(III) chloride solution turns pale green, showing that iron(III) has been reduced to iron(II). At the same time a pale yellow deposit (turbidity) of sulphur forms in the mixture. Either of these observations is acceptable: the colour change from yellow-brown to pale green, or the appearance of the yellow solid sulphur.
Answer Details
(a) IUPAC name of \(Fe_2(SO_4)_3\)
The compound contains the iron(III) cation, \(Fe^{3+}\), and the sulphate anion, \(SO_4^{2-}\). Because iron shows more than one oxidation state, the oxidation number of the metal must be shown in Roman numerals, and the anion is named in full as the tetraoxosulphate(VI) ion. The IUPAC name is therefore iron(III) tetraoxosulphate(VI) (commonly, iron(III) sulphate).
(b)(i) Reaction of hydrogen sulphide with iron(III) chloride solution
Hydrogen sulphide is a reducing agent. It reduces iron(III) to iron(II), and is itself oxidised, so that its sulphur is deposited as free (elemental) sulphur:
\[2FeCl_3(aq) + H_2S(g) \rightarrow 2FeCl_2(aq) + 2HCl(aq) + S(s)\]In ionic terms the essential change is \(2Fe^{3+} + H_2S \rightarrow 2Fe^{2+} + 2H^+ + S\), which shows clearly that iron is reduced from the \(+3\) to the \(+2\) state while sulphide sulphur (\(-2\)) is oxidised to sulphur (\(0\)).
(b)(ii) One observed change
The yellow-brown iron(III) chloride solution turns pale green, showing that iron(III) has been reduced to iron(II). At the same time a pale yellow deposit (turbidity) of sulphur forms in the mixture. Either of these observations is acceptable: the colour change from yellow-brown to pale green, or the appearance of the yellow solid sulphur.
Question 3 Report
(a) List two uses of sodium trioxocarbonate (IV).
(b) Sodium trioxocarbonate (IV) solution is alkaline.
(i) What phenomenon is responsible for this observation?
(ii) Name the product obtained on passing carbon (IV) oxide into saturated sodium trioxocarbonate (IV) solution.
(a) Two uses of sodium trioxocarbonate(IV), \(Na_2CO_3\) (washing soda): in the manufacture of glass and in softening hard water. (It is also used in the manufacture of soap, detergents and paper.)
(b)(i) The phenomenon responsible for the alkalinity of sodium trioxocarbonate(IV) solution is salt hydrolysis. Being the salt of a strong base (NaOH) and a weak acid (trioxocarbonate(IV) acid), the carbonate ion reacts with water to release hydroxide ions, making the solution alkaline:
\[CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-\]
(ii) When carbon(IV) oxide is passed into a saturated sodium trioxocarbonate(IV) solution, the product obtained is sodium hydrogen trioxocarbonate(IV) (sodium hydrogencarbonate), \(NaHCO_3\):
\[Na_2CO_3 + CO_2 + H_2O \rightarrow 2NaHCO_3\]
Answer Details
(a) Two uses of sodium trioxocarbonate(IV), \(Na_2CO_3\) (washing soda): in the manufacture of glass and in softening hard water. (It is also used in the manufacture of soap, detergents and paper.)
(b)(i) The phenomenon responsible for the alkalinity of sodium trioxocarbonate(IV) solution is salt hydrolysis. Being the salt of a strong base (NaOH) and a weak acid (trioxocarbonate(IV) acid), the carbonate ion reacts with water to release hydroxide ions, making the solution alkaline:
\[CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-\]
(ii) When carbon(IV) oxide is passed into a saturated sodium trioxocarbonate(IV) solution, the product obtained is sodium hydrogen trioxocarbonate(IV) (sodium hydrogencarbonate), \(NaHCO_3\):
\[Na_2CO_3 + CO_2 + H_2O \rightarrow 2NaHCO_3\]
Question 4 Report
(a) State two postulates of the kinetic theory of gases.
(b) Write two chemical properties that are common to both carbon (IV) oxide and sulphur (IV) oxide.
(a) Two postulates of the kinetic theory of gases:
- A gas consists of a very large number of tiny particles (molecules) which are in continuous, rapid, random motion.
- The actual volume of the gas particles is negligible compared with the total volume of the container (there is a lot of empty space between them). (Also acceptable: there are negligible forces of attraction between the particles, and collisions between them are perfectly elastic; the average kinetic energy is proportional to the absolute temperature.)
(b) Two chemical properties common to both carbon(IV) oxide, \(CO_2\), and sulphur(IV) oxide, \(SO_2\):
- Both are acidic oxides: they dissolve in water to form acids (\(CO_2\) forms trioxocarbonate(IV) acid, \(SO_2\) forms trioxosulphate(IV) acid) and turn moist blue litmus red.
- Both react with alkalis (bases) to form salt and water, e.g. \(CO_2 + 2NaOH \rightarrow Na_2CO_3 + H_2O\) and \(SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O\). (Both are also non-supporters of combustion.)
Answer Details
(a) Two postulates of the kinetic theory of gases:
- A gas consists of a very large number of tiny particles (molecules) which are in continuous, rapid, random motion.
- The actual volume of the gas particles is negligible compared with the total volume of the container (there is a lot of empty space between them). (Also acceptable: there are negligible forces of attraction between the particles, and collisions between them are perfectly elastic; the average kinetic energy is proportional to the absolute temperature.)
(b) Two chemical properties common to both carbon(IV) oxide, \(CO_2\), and sulphur(IV) oxide, \(SO_2\):
- Both are acidic oxides: they dissolve in water to form acids (\(CO_2\) forms trioxocarbonate(IV) acid, \(SO_2\) forms trioxosulphate(IV) acid) and turn moist blue litmus red.
- Both react with alkalis (bases) to form salt and water, e.g. \(CO_2 + 2NaOH \rightarrow Na_2CO_3 + H_2O\) and \(SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O\). (Both are also non-supporters of combustion.)
Question 5 Report
(a)(i) What type of reaction is involved in each of the conversion processes indicated as I to V below?
(ii) Name one isomer of glucose
(iii) Explain why palmwine becomes sour on prolonged exposure to air.
(b)(i) List the reagents and the reaction condition necessary for ethanoic acid to form an alkanoate.
(ii) Give two uses of alkanoates
(c)(i) What is the lUPAC name of the following compound?
(ii) Outline one chemical test to distinguish between methane and the compound in (c)(i) above.
(iii) Write an equation for the combustion of ethene in excess oxygen.
(a)(i) Type of reaction in each conversion (I to V)
| Step | Type of reaction |
|---|---|
| I | Hydrolysis |
| II | Fermentation (decomposition) |
| III | Dehydration |
| IV | Polymerization |
| V | Oxidation |
(a)(ii) One isomer of glucose: fructose (galactose is also acceptable), each of formula C6H12O6.
(a)(iii) Why palm wine becomes sour on prolonged exposure to air. Palm wine contains ethanol. On exposure to air, bacterial/atmospheric oxidation converts the ethanol to ethanoic (acetic) acid, and it is this acid that gives the drink its sour taste:
\[ CH_3CH_2OH + O_2 \rightarrow CH_3COOH + H_2O \]
(b)(i) Reagents and conditions for ethanoic acid to form an alkanoate
- An alkanol (e.g. ethanol).
- Concentrated H2SO4 (or dry HCl) as catalyst/dehydrating agent.
- Heat, about \(40^\circ\text{C}\) to \(110^\circ\text{C}\), or refluxing.
\[ CH_3COOH + C_2H_5OH \underset{\Delta}{\overset{conc.\,H_2SO_4}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O \]
(b)(ii) Two uses of alkanoates (esters)
- In the manufacture of perfumes, flavours and cosmetics.
- As solvents in the paint, pharmaceutical and food industries; also for making polyester fibres.
(c)(i) IUPAC name of the compound: but-2-yne, \( CH_3-C{\equiv}C-CH_3 \).
(c)(ii) Chemical test to distinguish methane from but-2-yne. Pass each gas in turn into bromine water (or acidified KMnO4). But-2-yne, being unsaturated, rapidly decolourizes the reagent; methane, being saturated, gives no visible reaction and the colour remains.
(c)(iii) Combustion of ethene in excess oxygen
\[ C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(g)} \]
Answer Details
(a)(i) Type of reaction in each conversion (I to V)
| Step | Type of reaction |
|---|---|
| I | Hydrolysis |
| II | Fermentation (decomposition) |
| III | Dehydration |
| IV | Polymerization |
| V | Oxidation |
(a)(ii) One isomer of glucose: fructose (galactose is also acceptable), each of formula C6H12O6.
(a)(iii) Why palm wine becomes sour on prolonged exposure to air. Palm wine contains ethanol. On exposure to air, bacterial/atmospheric oxidation converts the ethanol to ethanoic (acetic) acid, and it is this acid that gives the drink its sour taste:
\[ CH_3CH_2OH + O_2 \rightarrow CH_3COOH + H_2O \]
(b)(i) Reagents and conditions for ethanoic acid to form an alkanoate
- An alkanol (e.g. ethanol).
- Concentrated H2SO4 (or dry HCl) as catalyst/dehydrating agent.
- Heat, about \(40^\circ\text{C}\) to \(110^\circ\text{C}\), or refluxing.
\[ CH_3COOH + C_2H_5OH \underset{\Delta}{\overset{conc.\,H_2SO_4}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O \]
(b)(ii) Two uses of alkanoates (esters)
- In the manufacture of perfumes, flavours and cosmetics.
- As solvents in the paint, pharmaceutical and food industries; also for making polyester fibres.
(c)(i) IUPAC name of the compound: but-2-yne, \( CH_3-C{\equiv}C-CH_3 \).
(c)(ii) Chemical test to distinguish methane from but-2-yne. Pass each gas in turn into bromine water (or acidified KMnO4). But-2-yne, being unsaturated, rapidly decolourizes the reagent; methane, being saturated, gives no visible reaction and the colour remains.
(c)(iii) Combustion of ethene in excess oxygen
\[ C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(g)} \]
Question 6 Report
(a) Write an equation for the reaction of chlorine with
(i) potassium iodide solution;
(ii) zinc on heating
(b) Identify the product Q in the following reaction: Cl\(_2\) + 2NaOH \(\to\) NaCl + H\(_2\)O + Q.
(a) Reactions of chlorine:
- (i) With potassium iodide solution (chlorine displaces iodine): \[Cl_2 + 2KI \rightarrow 2KCl + I_2\]
- (ii) With zinc on heating: \[Zn + Cl_2 \rightarrow ZnCl_2\]
(b) In \(Cl_2 + 2NaOH \rightarrow NaCl + H_2O + Q\) (cold, dilute sodium hydroxide), the product Q is sodium chlorate(I) (sodium oxochlorate(I) / sodium hypochlorite), \(NaOCl\).
The balanced equation is: \[Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O\]
Answer Details
(a) Reactions of chlorine:
- (i) With potassium iodide solution (chlorine displaces iodine): \[Cl_2 + 2KI \rightarrow 2KCl + I_2\]
- (ii) With zinc on heating: \[Zn + Cl_2 \rightarrow ZnCl_2\]
(b) In \(Cl_2 + 2NaOH \rightarrow NaCl + H_2O + Q\) (cold, dilute sodium hydroxide), the product Q is sodium chlorate(I) (sodium oxochlorate(I) / sodium hypochlorite), \(NaOCl\).
The balanced equation is: \[Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O\]
Question 7 Report
(a)(i) Arrange the following elements in the order of increasing reactivity. Iron, Lead, Magnessium, Aluminium.
(ii) Which of the following elements in (a)(i) above reacts with sodium hydroxide to give hydrogen?
(b) What property of tetraoxosulphate (VI) acid does each of the following reactions illustrate?
(I) S + 2H\(_2\)SO\(_4\) ---> 3SO\(_4\) + 2H\(_2\)O
(ii) MgO + H\(_2\)SO\(_4\) ---> MgSO\(_4\) + H\(_2\)O
(iii) C\(_{12}\)H\(_{22}\)C\(_{11}\) + H\(_2\)SO\(_4\) ---> 12C + H\(_2\)SO\(_4\) + 11H\(_2\)O
(a)(i) Arranged in order of increasing reactivity (least reactive first):
\[Lead < Iron < Aluminium < Magnesium\]
i.e. Lead, Iron, Aluminium, Magnesium (magnesium is the most reactive of the four).
(ii) The element that reacts with sodium hydroxide to give hydrogen is aluminium (it is amphoteric and reacts with alkalis to liberate hydrogen).
(b) Property of tetraoxosulphate(VI) acid illustrated by each reaction:
- (i) \(S + 2H_2SO_4 \rightarrow 3SO_2 + 2H_2O\): hot concentrated \(H_2SO_4\) is acting as an oxidizing agent (it oxidizes sulphur, and is itself reduced to \(SO_2\)). This shows its oxidizing property.
- (ii) \(MgO + H_2SO_4 \rightarrow MgSO_4 + H_2O\): the acid reacts with a base to give salt and water only. This shows its acidic property (neutralization).
- (iii) \(C_{12}H_{22}O_{11} + H_2SO_4 \rightarrow 12C + H_2SO_4 + 11H_2O\): the acid removes the elements of water from sugar, leaving carbon. This shows its dehydrating property.
Answer Details
(a)(i) Arranged in order of increasing reactivity (least reactive first):
\[Lead < Iron < Aluminium < Magnesium\]
i.e. Lead, Iron, Aluminium, Magnesium (magnesium is the most reactive of the four).
(ii) The element that reacts with sodium hydroxide to give hydrogen is aluminium (it is amphoteric and reacts with alkalis to liberate hydrogen).
(b) Property of tetraoxosulphate(VI) acid illustrated by each reaction:
- (i) \(S + 2H_2SO_4 \rightarrow 3SO_2 + 2H_2O\): hot concentrated \(H_2SO_4\) is acting as an oxidizing agent (it oxidizes sulphur, and is itself reduced to \(SO_2\)). This shows its oxidizing property.
- (ii) \(MgO + H_2SO_4 \rightarrow MgSO_4 + H_2O\): the acid reacts with a base to give salt and water only. This shows its acidic property (neutralization).
- (iii) \(C_{12}H_{22}O_{11} + H_2SO_4 \rightarrow 12C + H_2SO_4 + 11H_2O\): the acid removes the elements of water from sugar, leaving carbon. This shows its dehydrating property.
Question 8 Report
(a)(i) List the quantum number that are assigned to an electron in an atom.
(ii) What is the maximum number of electrons that can occupy the 3d orbital?
(b) An element represented as P has the following electronic configuration: 1s\(^2\)2s\(^2\)2p\(^6\)2s\(^2\)
(i) Write the electronic configuration of the ion of P.
(ii) Without identifying P, write the likely formula of its chloride.
(iii) State with reason, whether P will be a good oxidizing or reducing agent.
(c)(i) What is electron affinity?
(ii) Explain briefly why ammonia can precipitate in dative bonding.
(d)(i) If an element in Group IV loses an alpha particle, to which group would the product belong?
(ii) Two equally toxic substances X and Y which decay to non-toxic products, were absorbed through the skin. If their half-lives are 8 minutes and 2 months respectively, which of them constitutes the greater health hazard? Explain your answer
(a)(i) Quantum numbers assigned to an electron
- Principal quantum number, n (shell/energy level)
- Azimuthal (angular momentum) quantum number, l (subshell/shape)
- Magnetic quantum number, ml (orientation of orbital)
- Spin quantum number, ms (direction of electron spin, \(+\tfrac{1}{2}\) or \(-\tfrac{1}{2}\))
(ii) The 3d subshell contains 5 orbitals and each orbital holds 2 electrons, so the maximum number of electrons is 10.
(b) Taking the configuration as \(1s^2 2s^2 2p^6 3s^2\), element P has 12 electrons (a Group II metal).
(i) P loses its two outer electrons to form P2+: \(1s^2 2s^2 2p^6\).
(ii) As a divalent metal, its chloride is PCl2.
(iii) P is a good reducing agent. It has a low ionization energy and readily loses (donates) its two outer electrons to other species, thereby reducing them while itself being oxidized.
(c)(i) Electron affinity is the energy change that occurs when one mole of electrons is added to one mole of gaseous atoms of an element to form one mole of gaseous negative ions.
(ii) The nitrogen atom in ammonia carries a lone pair of electrons. This lone pair can be donated to an electron-deficient species (for example H+ to give NH4+), so ammonia readily forms a dative (coordinate) bond.
(d)(i) Loss of an alpha particle decreases the atomic number by 2, so the product moves two places to the left in the period. A Group IV element therefore gives a product in Group II.
(ii) X (half-life 8 minutes) constitutes the greater health hazard. A short half-life means X disintegrates very rapidly, so within the short time it is absorbed it emits radiation at a much higher rate (high activity) than Y, whose 2-month half-life means it decays very slowly and releases only a small amount of radiation over the same period.
Answer Details
(a)(i) Quantum numbers assigned to an electron
- Principal quantum number, n (shell/energy level)
- Azimuthal (angular momentum) quantum number, l (subshell/shape)
- Magnetic quantum number, ml (orientation of orbital)
- Spin quantum number, ms (direction of electron spin, \(+\tfrac{1}{2}\) or \(-\tfrac{1}{2}\))
(ii) The 3d subshell contains 5 orbitals and each orbital holds 2 electrons, so the maximum number of electrons is 10.
(b) Taking the configuration as \(1s^2 2s^2 2p^6 3s^2\), element P has 12 electrons (a Group II metal).
(i) P loses its two outer electrons to form P2+: \(1s^2 2s^2 2p^6\).
(ii) As a divalent metal, its chloride is PCl2.
(iii) P is a good reducing agent. It has a low ionization energy and readily loses (donates) its two outer electrons to other species, thereby reducing them while itself being oxidized.
(c)(i) Electron affinity is the energy change that occurs when one mole of electrons is added to one mole of gaseous atoms of an element to form one mole of gaseous negative ions.
(ii) The nitrogen atom in ammonia carries a lone pair of electrons. This lone pair can be donated to an electron-deficient species (for example H+ to give NH4+), so ammonia readily forms a dative (coordinate) bond.
(d)(i) Loss of an alpha particle decreases the atomic number by 2, so the product moves two places to the left in the period. A Group IV element therefore gives a product in Group II.
(ii) X (half-life 8 minutes) constitutes the greater health hazard. A short half-life means X disintegrates very rapidly, so within the short time it is absorbed it emits radiation at a much higher rate (high activity) than Y, whose 2-month half-life means it decays very slowly and releases only a small amount of radiation over the same period.
Question 9 Report
(a)(i) What is the general formula for alkanoic acids?
(ii) State two chemical properties of ethanoic acid.
(b) Which of propene, butane and pentane
(i) will decolorize acidified KMnO\(_4\) solution?
(ii) can be easily polymerized?
(iii) is an isomer of methylpropane?
(iv) can be obtained from an alkanol by dehydration?
(a)(i) The general formula for alkanoic acids is \(C_nH_{2n+1}COOH\) (equivalently \(C_nH_{2n}O_2\)).
(ii) Two chemical properties of ethanoic acid, \(CH_3COOH\):
- It reacts with reactive metals to liberate hydrogen, e.g. \(2CH_3COOH + Mg \rightarrow (CH_3COO)_2Mg + H_2\).
- It reacts with alkanols in the presence of concentrated \(H_2SO_4\) to form an ester (esterification). (It also reacts with bases and trioxocarbonate(IV) salts, turning blue litmus red as a weak acid.)
(b) Of propene, butane and pentane:
- (i) The one that decolorizes acidified \(KMnO_4\) solution is propene (it is unsaturated, containing a C=C double bond).
- (ii) The one that can be easily polymerized is propene (alkenes polymerize; propene gives polypropene).
- (iii) The isomer of methylpropane is butane (both have the molecular formula \(C_4H_{10}\)).
- (iv) The one that can be obtained from an alkanol by dehydration is propene (dehydration of propanol gives propene).
Answer Details
(a)(i) The general formula for alkanoic acids is \(C_nH_{2n+1}COOH\) (equivalently \(C_nH_{2n}O_2\)).
(ii) Two chemical properties of ethanoic acid, \(CH_3COOH\):
- It reacts with reactive metals to liberate hydrogen, e.g. \(2CH_3COOH + Mg \rightarrow (CH_3COO)_2Mg + H_2\).
- It reacts with alkanols in the presence of concentrated \(H_2SO_4\) to form an ester (esterification). (It also reacts with bases and trioxocarbonate(IV) salts, turning blue litmus red as a weak acid.)
(b) Of propene, butane and pentane:
- (i) The one that decolorizes acidified \(KMnO_4\) solution is propene (it is unsaturated, containing a C=C double bond).
- (ii) The one that can be easily polymerized is propene (alkenes polymerize; propene gives polypropene).
- (iii) The isomer of methylpropane is butane (both have the molecular formula \(C_4H_{10}\)).
- (iv) The one that can be obtained from an alkanol by dehydration is propene (dehydration of propanol gives propene).
Question 10 Report
Copy and complete the following table
| Element | Number of Neutrons | Electronic Configuration | Group in the periodic Table |
| \(^{23}_{11}Na\) | --- | \(1s^{2}2s^{2}2p^{6}3s^{1}\) | 1 |
| \(^4_2He\) | 2 | ---- | --- |
| ---- | 7 | \(1s^{2}2s^{2}2p^{2}\) | ---- |
Completed table
Reasoning for each entry:
- \( ^{23}_{11}Na \): number of neutrons \( = 23 - 11 = 12 \). The configuration \( 1s^2\,2s^2\,2p^6\,3s^1 \) ends in \( 3s^1 \) (one outer electron), so it is in Group 1.
- \( ^{4}_{2}He \): neutrons \( = 4 - 2 = 2 \); with 2 electrons the configuration is \( 1s^2 \). Its outer shell is full, so it is a noble gas in Group 0.
- Third element: the configuration \( 1s^2\,2s^2\,2p^2 \) has 6 electrons, so the atomic number is 6 (carbon). With 7 neutrons the mass number is \( 6 + 7 = 13 \). The two \( p \) electrons make it a Group IV (Group 4) element.
| Element | Number of neutrons | Electronic configuration | Group in the periodic table |
|---|---|---|---|
| \( ^{23}_{11}Na \) | 12 | \( 1s^2\,2s^2\,2p^6\,3s^1 \) | 1 |
| \( ^{4}_{2}He \) | 2 | \( 1s^2 \) | 0 |
| \( ^{13}_{6}C \) (carbon) | 7 | \( 1s^2\,2s^2\,2p^2 \) | IV (4) |
The group number for a main-group element equals the number of electrons in its outermost shell (helium being the exception, placed in Group 0 with the noble gases).
Answer Details
Completed table
Reasoning for each entry:
- \( ^{23}_{11}Na \): number of neutrons \( = 23 - 11 = 12 \). The configuration \( 1s^2\,2s^2\,2p^6\,3s^1 \) ends in \( 3s^1 \) (one outer electron), so it is in Group 1.
- \( ^{4}_{2}He \): neutrons \( = 4 - 2 = 2 \); with 2 electrons the configuration is \( 1s^2 \). Its outer shell is full, so it is a noble gas in Group 0.
- Third element: the configuration \( 1s^2\,2s^2\,2p^2 \) has 6 electrons, so the atomic number is 6 (carbon). With 7 neutrons the mass number is \( 6 + 7 = 13 \). The two \( p \) electrons make it a Group IV (Group 4) element.
| Element | Number of neutrons | Electronic configuration | Group in the periodic table |
|---|---|---|---|
| \( ^{23}_{11}Na \) | 12 | \( 1s^2\,2s^2\,2p^6\,3s^1 \) | 1 |
| \( ^{4}_{2}He \) | 2 | \( 1s^2 \) | 0 |
| \( ^{13}_{6}C \) (carbon) | 7 | \( 1s^2\,2s^2\,2p^2 \) | IV (4) |
The group number for a main-group element equals the number of electrons in its outermost shell (helium being the exception, placed in Group 0 with the noble gases).
Question 11 Report
Xg of a pure sample of iron (II) sulphide reacted completely with excess dilute hydrochloric acid to give 3.20g of iron (II) chloride according to the following equation: FeS\(_{(s)}\) + 2HCI\(_{(aq)}\) ---> FeCl\(_{2(aq)}\) + H\(_2\)S\(_{(g)}\).
(a) Mention one method apart from heating by which the reaction can be made to proceed faster
(b) Calculate the value of X. [CI = 35.5, Fe = 56; FeS = 88g mol\(^{-1}\)]
(a) One method (apart from heating) by which the reaction can be made faster: increase the surface area of the iron(II) sulphide by using it in a finely powdered form (using a more concentrated acid is also acceptable).
(b) Calculating X. Equation: \[FeS_{(s)} + 2HCl_{(aq)} \rightarrow FeCl_{2(aq)} + H_2S_{(g)}\]
Molar mass of \(FeCl_2 = 56 + (2 \times 35.5) = 56 + 71 = 127\,g\,mol^{-1}\)
\[\text{moles of } FeCl_2 = \frac{3.20}{127} = 0.0252\,mol\]
From the equation, 1 mole \(FeS\) gives 1 mole \(FeCl_2\), so moles of \(FeS = 0.0252\,mol\).
\[X = \text{mass of } FeS = 0.0252 \times 88 = 2.22\,g\]
So X = 2.22 g (approximately).
Answer Details
(a) One method (apart from heating) by which the reaction can be made faster: increase the surface area of the iron(II) sulphide by using it in a finely powdered form (using a more concentrated acid is also acceptable).
(b) Calculating X. Equation: \[FeS_{(s)} + 2HCl_{(aq)} \rightarrow FeCl_{2(aq)} + H_2S_{(g)}\]
Molar mass of \(FeCl_2 = 56 + (2 \times 35.5) = 56 + 71 = 127\,g\,mol^{-1}\)
\[\text{moles of } FeCl_2 = \frac{3.20}{127} = 0.0252\,mol\]
From the equation, 1 mole \(FeS\) gives 1 mole \(FeCl_2\), so moles of \(FeS = 0.0252\,mol\).
\[X = \text{mass of } FeS = 0.0252 \times 88 = 2.22\,g\]
So X = 2.22 g (approximately).
Question 12 Report
(a)(i) Define heat of neutralization
(ii) Give the reason why copper (II) chloride can be prepared by neutralization, unlike lead (II) chloride.
(b)(i) Describe in outline, the manufacture of trioxonitrate (V) acid by the catalytic oxidation of ammonia, giving equations where appropriate.
(ii) What are the products obtained when sodium tioxonitrate (V) is heated strongly?
(c) When powdered magnesium is heated to redness in a stream of nitrogen, magnesium nitride (Mg\(_3\)N\(_2\)) is formed.
(i) Write an equation for the reaction
(ii) Hence, calculate the amount (in mole) of magnesium nitride that can be obtained from 3.0g of magnesium [Mg = 24].
(a)(i) Heat of neutralization is the heat change when one mole of hydrogen ions (H+) from an acid reacts completely with one mole of hydroxide ions (OH-) from a base to form one mole of water under standard conditions.
(ii) Copper(II) chloride is soluble in water, so it can be made in solution by neutralizing an acid with a base and then crystallized. Lead(II) chloride is insoluble in cold water, so a neutralization in solution would simply precipitate it; it is therefore prepared by precipitation (double decomposition), not by neutralization.
(b)(i) Manufacture of trioxonitrate(V) acid (Ostwald process)
- Ammonia is mixed with excess air and passed over a platinum-rhodium gauze catalyst at about 900 °C:
\[4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O\] - The nitrogen(II) oxide is cooled and oxidized by more air:
\[2NO + O_2 \rightarrow 2NO_2\] - The nitrogen(IV) oxide is dissolved in water in the presence of excess air:
\[4NO_2 + O_2 + 2H_2O \rightarrow 4HNO_3\]
(ii) Sodium trioxonitrate(V) on strong heating gives sodium trioxonitrate(III) and oxygen:
\[2NaNO_3 \rightarrow 2NaNO_2 + O_2\]
(c)(i) \[3Mg + N_2 \rightarrow Mg_3N_2\]
(ii) Amount of Mg \(= \dfrac{3.0}{24} = 0.125\ \text{mol}\).
From the equation, 3 mol Mg give 1 mol Mg3N2, so
amount of Mg3N2 \(= \dfrac{0.125}{3} = 0.0417\ \text{mol} \approx 4.17 \times 10^{-2}\ \text{mol}\).
Answer Details
(a)(i) Heat of neutralization is the heat change when one mole of hydrogen ions (H+) from an acid reacts completely with one mole of hydroxide ions (OH-) from a base to form one mole of water under standard conditions.
(ii) Copper(II) chloride is soluble in water, so it can be made in solution by neutralizing an acid with a base and then crystallized. Lead(II) chloride is insoluble in cold water, so a neutralization in solution would simply precipitate it; it is therefore prepared by precipitation (double decomposition), not by neutralization.
(b)(i) Manufacture of trioxonitrate(V) acid (Ostwald process)
- Ammonia is mixed with excess air and passed over a platinum-rhodium gauze catalyst at about 900 °C:
\[4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O\] - The nitrogen(II) oxide is cooled and oxidized by more air:
\[2NO + O_2 \rightarrow 2NO_2\] - The nitrogen(IV) oxide is dissolved in water in the presence of excess air:
\[4NO_2 + O_2 + 2H_2O \rightarrow 4HNO_3\]
(ii) Sodium trioxonitrate(V) on strong heating gives sodium trioxonitrate(III) and oxygen:
\[2NaNO_3 \rightarrow 2NaNO_2 + O_2\]
(c)(i) \[3Mg + N_2 \rightarrow Mg_3N_2\]
(ii) Amount of Mg \(= \dfrac{3.0}{24} = 0.125\ \text{mol}\).
From the equation, 3 mol Mg give 1 mol Mg3N2, so
amount of Mg3N2 \(= \dfrac{0.125}{3} = 0.0417\ \text{mol} \approx 4.17 \times 10^{-2}\ \text{mol}\).
Question 13 Report
(a) If a steel spoon were to be plated with silver, state what would be suitable for use as the;
(i) anode. (ii) cathode; (iii) electrolyte
(b)(i) Write an equation for one of the reactions involved in the purification of bauxite.
(ii) Give the reason why the carbon anodes are changed at intervals during the electrolysis of pure alumina solution in molten cryolite.
(a) To electroplate a steel spoon with silver:
- (i) Anode: a bar (plate) of pure silver metal (which dissolves to replenish the silver ions).
- (ii) Cathode: the steel spoon itself (the article to be plated).
- (iii) Electrolyte: a solution of a soluble silver salt, e.g. silver trioxonitrate(V) solution, \(AgNO_3\) (a solution of potassium argentocyanide is used industrially).
(b)(i) An equation for one reaction in the purification of bauxite (dissolving the amphoteric aluminium oxide in hot sodium hydroxide, leaving impurities behind):
\[Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O\]
(ii) The carbon anodes must be changed (replaced) at intervals because the oxygen liberated at the anode reacts with the hot carbon to form carbon(IV) oxide (and carbon(II) oxide), so the carbon anodes gradually burn away (are oxidized) and become too small to use.
Answer Details
(a) To electroplate a steel spoon with silver:
- (i) Anode: a bar (plate) of pure silver metal (which dissolves to replenish the silver ions).
- (ii) Cathode: the steel spoon itself (the article to be plated).
- (iii) Electrolyte: a solution of a soluble silver salt, e.g. silver trioxonitrate(V) solution, \(AgNO_3\) (a solution of potassium argentocyanide is used industrially).
(b)(i) An equation for one reaction in the purification of bauxite (dissolving the amphoteric aluminium oxide in hot sodium hydroxide, leaving impurities behind):
\[Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O\]
(ii) The carbon anodes must be changed (replaced) at intervals because the oxygen liberated at the anode reacts with the hot carbon to form carbon(IV) oxide (and carbon(II) oxide), so the carbon anodes gradually burn away (are oxidized) and become too small to use.
Question 14 Report
(a)(i) What is an electrolyte?
(ii) Classify each of the following as strong electrolyte/weak, electrolyte/non-electrolyte. Potassium chloride; sodium ethanoate, aqueous ammonia; cane sugar
(b)(i) Write half-cell equations for the reactions in the Daniel cell .
(ii) Why is the Daniel cell classified as an electrochemical cell?
(iii) Give two other examples of electrochemical cell.
(c) Explain the following observations.
(i) Graphite conducts electricity, unlike most non-metals
(ii) In the electrolysis of copper (II) tetraoxosulphate (VI) solution, the blue colour fades with platinum electrodes while the colour intensity is unaffected with copper electrodes (equations required)
(iii) A solution of dry hydrogen chloride in methylbenzene (toluene) does not conduct electricity whereas hydrochloric acid does.
(a)(i) An electrolyte is a substance which, in the molten state or in aqueous solution, conducts electricity and is chemically decomposed by it because it contains free mobile ions.
(ii)
- Potassium chloride - strong electrolyte
- Sodium ethanoate - strong electrolyte
- Aqueous ammonia - weak electrolyte
- Cane sugar - non-electrolyte
(b)(i) Daniell cell half-cell reactions
Anode (oxidation): \[Zn \rightarrow Zn^{2+} + 2e^-\]
Cathode (reduction): \[Cu^{2+} + 2e^- \rightarrow Cu\]
(ii) It is an electrochemical cell because it converts chemical energy (from the spontaneous redox reaction) into electrical energy.
(iii) Other examples: the dry (Leclanche) cell and the lead-acid accumulator.
(c)(i) In graphite, each carbon atom bonds to only three others, using three of its four valence electrons. The fourth electron of every atom is delocalized between the layers and is free to move, so graphite conducts electricity.
(ii) With platinum electrodes, Cu2+ ions are discharged at the cathode and removed from solution, so the blue colour fades:
\[Cu^{2+} + 2e^- \rightarrow Cu\]
at the anode oxygen is evolved: \[2H_2O \rightarrow O_2 + 4H^+ + 4e^-\]
With copper electrodes, the copper anode dissolves and replaces the Cu2+ removed at the cathode, so the concentration and blue colour stay the same:
\[Cu \rightarrow Cu^{2+} + 2e^-\]
(iii) In methylbenzene, hydrogen chloride exists as un-ionized covalent molecules, so there are no mobile ions and it does not conduct. In water, HCl ionizes into mobile H+ and Cl- ions, so hydrochloric acid conducts electricity.
Answer Details
(a)(i) An electrolyte is a substance which, in the molten state or in aqueous solution, conducts electricity and is chemically decomposed by it because it contains free mobile ions.
(ii)
- Potassium chloride - strong electrolyte
- Sodium ethanoate - strong electrolyte
- Aqueous ammonia - weak electrolyte
- Cane sugar - non-electrolyte
(b)(i) Daniell cell half-cell reactions
Anode (oxidation): \[Zn \rightarrow Zn^{2+} + 2e^-\]
Cathode (reduction): \[Cu^{2+} + 2e^- \rightarrow Cu\]
(ii) It is an electrochemical cell because it converts chemical energy (from the spontaneous redox reaction) into electrical energy.
(iii) Other examples: the dry (Leclanche) cell and the lead-acid accumulator.
(c)(i) In graphite, each carbon atom bonds to only three others, using three of its four valence electrons. The fourth electron of every atom is delocalized between the layers and is free to move, so graphite conducts electricity.
(ii) With platinum electrodes, Cu2+ ions are discharged at the cathode and removed from solution, so the blue colour fades:
\[Cu^{2+} + 2e^- \rightarrow Cu\]
at the anode oxygen is evolved: \[2H_2O \rightarrow O_2 + 4H^+ + 4e^-\]
With copper electrodes, the copper anode dissolves and replaces the Cu2+ removed at the cathode, so the concentration and blue colour stay the same:
\[Cu \rightarrow Cu^{2+} + 2e^-\]
(iii) In methylbenzene, hydrogen chloride exists as un-ionized covalent molecules, so there are no mobile ions and it does not conduct. In water, HCl ionizes into mobile H+ and Cl- ions, so hydrochloric acid conducts electricity.
