WAEC SSCE - Chemistry - 2003 (Essay)
Question 1 Report
(a) The electronic configuration of five elements represented by the letters P, Q, R, S and T are indicated below.
P - \(1s_2 2s_2 2p_2\)
Q \(1s_2 2s_2 2p_4\)
R \(1s_2 2s_2 2p_6\)
S - \(1s_2 2s_2 2p_6 3s_2\)
T - \(1s_2 2s_2 2p_6 3s_2 3p_5\)
without identifying the elements, state which of them
(i) belongs to group VI in the periodic table;
(ii) is strongly metallic in character;
(iii) readily ionizes by gaining one electron;
(iv) contains two unpaired electrons in the ground state atom
(v) readily loses two electrons during chemical bonding
(vi) Does not paricipates in chemical reactions?
(vii) is an s-block element.
(b)(i) Copy and complete the table below as appropriate
| Particle | Number of Protons | Number of Electrons | Number of Neutrons |
| \(^1_1H\) | 1 | 1 | |
| \(^{27}_{13}\)Al\(^{3+}\) | |||
| \(^{16}_{8}O^{2+}\) | 8 |
(ii) Give the reason why atomic radius increases down a group in the periodic table but decreases from left to right in a period.
(c)(i) What is meant by the half-life of a radioactive element?
(ii) The nuclide \(^{210}_{84}PO\) loses an alpha 4° particle to form lead. Write an equation for the reaction.
(d) State the type of chemical bonding which accounts for each of the following observations:
(i) Chlorine exists as discrete molecules
(ii) Sodium chloride dissolves readily in water;
(iii) \(\mathrm{CuSO_{4(aq)}}\) forms a deep blue complex ion with excess \(\mathrm{NH_{3(aq)}}\)
(a) Identifying the elements from configuration (not required, but shown for clarity)
- P: 1s\(^2\)2s\(^2\)2p\(^2\) (Z = 6, group IV)
- Q: 1s\(^2\)2s\(^2\)2p\(^4\) (Z = 8, group VI)
- R: 1s\(^2\)2s\(^2\)2p\(^6\) (Z = 10, group VIII, noble gas)
- S: 1s\(^2\)2s\(^2\)2p\(^6\)3s\(^2\) (Z = 12, group II)
- T: 1s\(^2\)2s\(^2\)2p\(^6\)3s\(^2\)3p\(^5\) (Z = 17, group VII)
(i) Belongs to group VI: Q (six outermost electrons, 2s\(^2\)2p\(^4\)).
(ii) Strongly metallic: S (a group II element that readily loses electrons).
(iii) Readily ionises by gaining one electron: T (needs one electron to complete its octet, forming a 1\(^-\) ion).
(iv) Two unpaired electrons in the ground-state atom: P and Q. In 2p\(^2\) the two electrons occupy separate orbitals (Hund's rule); in 2p\(^4\) one orbital is doubly filled leaving two singly-occupied orbitals - both give two unpaired electrons.
(v) Readily loses two electrons in bonding: S (3s\(^2\) outer electrons lost to form a 2\(^+\) ion).
(vi) Does not take part in chemical reactions: R (a stable, filled-shell noble gas).
(vii) An s-block element: S (its outermost electrons enter the 3s sub-shell).
(b)(i) Completed table
| Particle | Protons | Electrons | Neutrons |
|---|---|---|---|
| \(^{1}_{1}\)H | 1 | 1 | 0 |
| \(^{27}_{13}\)Al\(^{3+}\) | 13 | 10 | 14 |
| \(^{16}_{8}\)O\(^{2-}\) | 8 | 10 | 8 |
Neutrons = mass number - atomic number. For Al\(^{3+}\), three electrons are lost (13 - 3 = 10). For the oxide ion, two electrons are gained (8 + 2 = 10).
(b)(ii) Trend in atomic radius
Down a group the number of occupied electron shells increases and inner-shell shielding increases, so the outermost electrons lie farther from the nucleus and the radius increases. Across a period electrons are added to the same shell while the nuclear charge rises; the greater effective nuclear attraction pulls the shell inward, so the radius decreases.
(c)(i) Half-life
The half-life of a radioactive element is the time taken for half the atoms (nuclei) in a given sample to decay (disintegrate).
(c)(ii) Alpha decay of polonium-210
\[ ^{210}_{84}\text{Po} \rightarrow\ ^{4}_{2}\text{He} +\ ^{206}_{82}\text{Pb} \](d) Type of bonding
- (i) Chlorine as discrete molecules: covalent bonding (shared electron pair within each Cl\(_2\) molecule).
- (ii) Sodium chloride dissolving readily in water: ionic (electrovalent) bonding (the polar water molecules hydrate the Na\(^+\) and Cl\(^-\) ions).
- (iii) Deep-blue complex ion with excess ammonia: co-ordinate (dative) covalent bonding (lone pairs on N donate into the Cu\(^{2+}\) ion).
Answer Details
(a) Identifying the elements from configuration (not required, but shown for clarity)
- P: 1s\(^2\)2s\(^2\)2p\(^2\) (Z = 6, group IV)
- Q: 1s\(^2\)2s\(^2\)2p\(^4\) (Z = 8, group VI)
- R: 1s\(^2\)2s\(^2\)2p\(^6\) (Z = 10, group VIII, noble gas)
- S: 1s\(^2\)2s\(^2\)2p\(^6\)3s\(^2\) (Z = 12, group II)
- T: 1s\(^2\)2s\(^2\)2p\(^6\)3s\(^2\)3p\(^5\) (Z = 17, group VII)
(i) Belongs to group VI: Q (six outermost electrons, 2s\(^2\)2p\(^4\)).
(ii) Strongly metallic: S (a group II element that readily loses electrons).
(iii) Readily ionises by gaining one electron: T (needs one electron to complete its octet, forming a 1\(^-\) ion).
(iv) Two unpaired electrons in the ground-state atom: P and Q. In 2p\(^2\) the two electrons occupy separate orbitals (Hund's rule); in 2p\(^4\) one orbital is doubly filled leaving two singly-occupied orbitals - both give two unpaired electrons.
(v) Readily loses two electrons in bonding: S (3s\(^2\) outer electrons lost to form a 2\(^+\) ion).
(vi) Does not take part in chemical reactions: R (a stable, filled-shell noble gas).
(vii) An s-block element: S (its outermost electrons enter the 3s sub-shell).
(b)(i) Completed table
| Particle | Protons | Electrons | Neutrons |
|---|---|---|---|
| \(^{1}_{1}\)H | 1 | 1 | 0 |
| \(^{27}_{13}\)Al\(^{3+}\) | 13 | 10 | 14 |
| \(^{16}_{8}\)O\(^{2-}\) | 8 | 10 | 8 |
Neutrons = mass number - atomic number. For Al\(^{3+}\), three electrons are lost (13 - 3 = 10). For the oxide ion, two electrons are gained (8 + 2 = 10).
(b)(ii) Trend in atomic radius
Down a group the number of occupied electron shells increases and inner-shell shielding increases, so the outermost electrons lie farther from the nucleus and the radius increases. Across a period electrons are added to the same shell while the nuclear charge rises; the greater effective nuclear attraction pulls the shell inward, so the radius decreases.
(c)(i) Half-life
The half-life of a radioactive element is the time taken for half the atoms (nuclei) in a given sample to decay (disintegrate).
(c)(ii) Alpha decay of polonium-210
\[ ^{210}_{84}\text{Po} \rightarrow\ ^{4}_{2}\text{He} +\ ^{206}_{82}\text{Pb} \](d) Type of bonding
- (i) Chlorine as discrete molecules: covalent bonding (shared electron pair within each Cl\(_2\) molecule).
- (ii) Sodium chloride dissolving readily in water: ionic (electrovalent) bonding (the polar water molecules hydrate the Na\(^+\) and Cl\(^-\) ions).
- (iii) Deep-blue complex ion with excess ammonia: co-ordinate (dative) covalent bonding (lone pairs on N donate into the Cu\(^{2+}\) ion).
Question 2 Report
(a) State the role of each of the following substances in the treatment of river water for town supply (i) Sand bed. (ii) Alum (iii) Chlorine
(b)(i) Give three major uses of H\(_2\)SO\(_4\)
(ii) Explain the following observation: A strip of blue litmus paper dropped into concentrated H\(_2\)SO\(_4\) becomes charred whereas in dilute H\(_2\)SO\(_4\), it turns red and is not charred.
(iii) Write an equation to show how concentrated H\(_2\)SO\(_4\) reacts with zinc.
(c)(i) List two gaseous pollutants that can be generated by burning coal.
(ii) Explain why coal burns more easily when it is broken into pieces than when it is in lump form.
(ii) What gas is responsible for most of the explosions in coal mines?
(iv) Name: the non-volatile residue left behind after the destructive distlillation of coal
(d)(i) What is meant by allotropy?
(ii) Name two crystalline allotropes of carbon
(iii) Name two elements apart from carbon, which exhibit allotropy
(iv) It is now known that carbon has an allotropic form called fullerene, containing molecules of formula C\(_{60}\). Calculate the mass of one mole of these molecules [C = 12]
(a) In the treatment of river water for town supply:
(i) Sand bed: used for filtration to remove suspended solid particles.
(ii) Alum: used as a coagulant to cause fine suspended and colloidal particles to clump together and settle.
(iii) Chlorine: used for sterilization; it kills harmful bacteria and other micro-organisms.
(b)(i) Three major uses of H2SO4:
- Manufacture of fertilizers.
- As an electrolyte in lead-acid batteries.
- Manufacture of detergents, pigments and plastics.
(ii) Concentrated H2SO4 is a strong dehydrating agent. It removes water from the blue litmus paper, leaving carbon behind; hence, the paper becomes black and charred. Dilute H2SO4 produces H3O+ ions in water, which turn blue litmus red. It does not char the paper because it is not sufficiently dehydrating.
(iii)
Zn + 2H2SO4(conc.) → ZnSO4 + SO2 + 2H2O
(c)(i) Two gaseous pollutants generated by burning coal are sulphur(IV) oxide, SO2, and carbon(II) oxide, CO.
(ii) When coal is broken into smaller pieces, it has a larger surface area exposed to oxygen in the air. This increases the rate of combustion, so it burns more easily than coal in lump form.
(iii) Methane, CH4 (firedamp), is responsible for most explosions in coal mines.
(iv) The non-volatile residue left after the destructive distillation of coal is coke.
(d)(i) Allotropy is the existence of an element in two or more different forms in the same physical state.
(ii) The two crystalline allotropes of carbon are diamond and graphite.
(iii) Two elements apart from carbon that exhibit allotropy are sulphur and phosphorus.
(iv)
Mass of one mole of C60 = 60 × 12 g = 720 g mol−1.
Answer Details
(a) In the treatment of river water for town supply:
(i) Sand bed: used for filtration to remove suspended solid particles.
(ii) Alum: used as a coagulant to cause fine suspended and colloidal particles to clump together and settle.
(iii) Chlorine: used for sterilization; it kills harmful bacteria and other micro-organisms.
(b)(i) Three major uses of H2SO4:
- Manufacture of fertilizers.
- As an electrolyte in lead-acid batteries.
- Manufacture of detergents, pigments and plastics.
(ii) Concentrated H2SO4 is a strong dehydrating agent. It removes water from the blue litmus paper, leaving carbon behind; hence, the paper becomes black and charred. Dilute H2SO4 produces H3O+ ions in water, which turn blue litmus red. It does not char the paper because it is not sufficiently dehydrating.
(iii)
Zn + 2H2SO4(conc.) → ZnSO4 + SO2 + 2H2O
(c)(i) Two gaseous pollutants generated by burning coal are sulphur(IV) oxide, SO2, and carbon(II) oxide, CO.
(ii) When coal is broken into smaller pieces, it has a larger surface area exposed to oxygen in the air. This increases the rate of combustion, so it burns more easily than coal in lump form.
(iii) Methane, CH4 (firedamp), is responsible for most explosions in coal mines.
(iv) The non-volatile residue left after the destructive distillation of coal is coke.
(d)(i) Allotropy is the existence of an element in two or more different forms in the same physical state.
(ii) The two crystalline allotropes of carbon are diamond and graphite.
(iii) Two elements apart from carbon that exhibit allotropy are sulphur and phosphorus.
(iv)
Mass of one mole of C60 = 60 × 12 g = 720 g mol−1.
Question 3 Report
(a)Give two reasons why aluminium is preferred to copper for making overhead electric cables.
(ii) Describe briefly the electrolytic extraction of aluminium from purified bauxite.
(b) The diagram below represents an electrolytic cell used for the purification of copper.
(i) Which of the electrodes I and II increases in mass during the electrolysis? Give reasons for your answers
(ii) State with reason the site of oxidation
(iii) Identify III and explain why its colour does not change in intensity during the electrolysis.
(c) Calculate the current in amperes that will deposit 8.00 g of calcium from used CaCl\(_2\) in 1 hour 15 minutes. [Ca = 40.0; 1 Faraday = 96500C]
(a)(i) Two reasons aluminium is preferred to copper for overhead cables: aluminium is much lighter (lower density), so the cables sag less and need fewer supports; aluminium is cheaper and more abundant than copper (while still being a good conductor).
(ii) Electrolytic extraction of aluminium: Purified alumina (Al2O3) is dissolved in molten cryolite (Na3AlF6) to lower the melting point and improve conductivity. The molten mixture is electrolyzed in a steel tank lined with graphite (the cathode), with carbon anodes.
Cathode: Al3+ + 3e- → Al (molten aluminium collects at the bottom and is tapped off).
Anode: 2O2- → O2 + 4e-; the oxygen burns the carbon anodes to CO2, so they are replaced from time to time.
(b) In copper purification the impure copper is the anode and pure copper is the cathode, in copper(II) tetraoxosulphate(VI) solution.
(i) The cathode (pure copper) increases in mass, because Cu2+ ions are discharged and deposited on it (Cu2+ + 2e- → Cu).
(ii) Oxidation occurs at the anode, because there copper atoms lose electrons to form ions (Cu → Cu2+ + 2e-).
(iii) III is the electrolyte, copper(II) sulphate solution. Its blue colour does not fade because the rate at which Cu2+ ions leave the solution at the cathode equals the rate at which Cu2+ ions enter the solution from the dissolving anode, so [Cu2+] stays constant.
(c) Deposit 8.00 g Ca in 1 h 15 min = 4500 s, Ca2+ + 2e- → Ca.
\[ n(\text{Ca})=\frac{8.00}{40}=0.200\ \text{mol};\quad n(e^-)=0.400\ \text{mol} \] \[ Q=0.400\times96500=38600\ \text{C};\quad I=\frac{Q}{t}=\frac{38600}{4500}=8.58\ \text{A} \]Answer Details
(a)(i) Two reasons aluminium is preferred to copper for overhead cables: aluminium is much lighter (lower density), so the cables sag less and need fewer supports; aluminium is cheaper and more abundant than copper (while still being a good conductor).
(ii) Electrolytic extraction of aluminium: Purified alumina (Al2O3) is dissolved in molten cryolite (Na3AlF6) to lower the melting point and improve conductivity. The molten mixture is electrolyzed in a steel tank lined with graphite (the cathode), with carbon anodes.
Cathode: Al3+ + 3e- → Al (molten aluminium collects at the bottom and is tapped off).
Anode: 2O2- → O2 + 4e-; the oxygen burns the carbon anodes to CO2, so they are replaced from time to time.
(b) In copper purification the impure copper is the anode and pure copper is the cathode, in copper(II) tetraoxosulphate(VI) solution.
(i) The cathode (pure copper) increases in mass, because Cu2+ ions are discharged and deposited on it (Cu2+ + 2e- → Cu).
(ii) Oxidation occurs at the anode, because there copper atoms lose electrons to form ions (Cu → Cu2+ + 2e-).
(iii) III is the electrolyte, copper(II) sulphate solution. Its blue colour does not fade because the rate at which Cu2+ ions leave the solution at the cathode equals the rate at which Cu2+ ions enter the solution from the dissolving anode, so [Cu2+] stays constant.
(c) Deposit 8.00 g Ca in 1 h 15 min = 4500 s, Ca2+ + 2e- → Ca.
\[ n(\text{Ca})=\frac{8.00}{40}=0.200\ \text{mol};\quad n(e^-)=0.400\ \text{mol} \] \[ Q=0.400\times96500=38600\ \text{C};\quad I=\frac{Q}{t}=\frac{38600}{4500}=8.58\ \text{A} \]Question 4 Report
(a) Mention one oxide in each case, which
(i) used in bleaching
(ii) is a redish-brown gas
(iii) reacts with NaOH and also with HCI;
(iv) dissolves in water to give a solution with pH greater than 7;
(v) oxidizes hot, concentrated HCI to chlorine
(b)(i) State three methods that can be used to removo hardness in a sample of water that contains calcium hydrogentrioxocarbonate (IV).
(ii) Explain with the aid of appropriate equation, why it is not advisable to build a house with limestone in an environment polluted by sulphur (IV) oxide.
(c)(i) List two compounds of potassium which yield oxygen when heated strongly
(ii) Calculate the amount (in moles) of gas which occupies 250 cm\(^3\) at s.t.p. [1 mole of gas occupies 22.4 dm\(^3\) at s.t.p]
(iii) If 250 cm\(^3\) of a gas at s.t.p. is heated to 27°C at constant pressure, calculate its new volume.
(iv) Explain in terms of the collision theory what happens as a gas is heated at constant pressure.
(a) (i) Used in bleaching: sulphur(IV) oxide, SO2. (ii) Reddish-brown gas: nitrogen(IV) oxide, NO2. (iii) Amphoteric (reacts with NaOH and HCl): aluminium oxide, Al2O3 (or ZnO). (iv) Dissolves in water to give pH greater than 7: sodium oxide, Na2O (or CaO). (v) Oxidizes hot concentrated HCl to chlorine: manganese(IV) oxide, MnO2.
(b)(i) Three methods to remove hardness due to Ca(HCO3)2 (temporary hardness): boiling; adding slaked lime (Clark's method); adding sodium trioxocarbonate(IV) (washing soda).
(ii) Sulphur(IV) oxide pollution forms acid with rain water (SO2 + H2O → H2SO3, and this is oxidized to H2SO4). This acid attacks limestone (CaCO3), wearing the building away: CaCO3 + H2SO3 → CaSO3 + H2O + CO2. The limestone is eroded, so it is unwise to build with it there.
(c)(i) Two potassium compounds that yield oxygen on strong heating: potassium trioxochlorate(V), KClO3, and potassium trioxonitrate(V), KNO3 (or KMnO4).
(ii) Amount of gas in 250 cm3 at s.t.p.:
\[ n=\frac{0.250}{22.4}=0.0112\ \text{mol} \](iii) 250 cm3 at s.t.p. (273 K) heated to 27°C (300 K) at constant pressure:
\[ V_2=250\times\frac{300}{273}=274.7\approx 275\ \text{cm}^3 \](iv) On heating at constant pressure the molecules gain kinetic energy and move faster, colliding more frequently and more forcefully; the gas expands (volume increases) so that the number of collisions per unit area of the walls, and hence the pressure, stays constant.
Answer Details
(a) (i) Used in bleaching: sulphur(IV) oxide, SO2. (ii) Reddish-brown gas: nitrogen(IV) oxide, NO2. (iii) Amphoteric (reacts with NaOH and HCl): aluminium oxide, Al2O3 (or ZnO). (iv) Dissolves in water to give pH greater than 7: sodium oxide, Na2O (or CaO). (v) Oxidizes hot concentrated HCl to chlorine: manganese(IV) oxide, MnO2.
(b)(i) Three methods to remove hardness due to Ca(HCO3)2 (temporary hardness): boiling; adding slaked lime (Clark's method); adding sodium trioxocarbonate(IV) (washing soda).
(ii) Sulphur(IV) oxide pollution forms acid with rain water (SO2 + H2O → H2SO3, and this is oxidized to H2SO4). This acid attacks limestone (CaCO3), wearing the building away: CaCO3 + H2SO3 → CaSO3 + H2O + CO2. The limestone is eroded, so it is unwise to build with it there.
(c)(i) Two potassium compounds that yield oxygen on strong heating: potassium trioxochlorate(V), KClO3, and potassium trioxonitrate(V), KNO3 (or KMnO4).
(ii) Amount of gas in 250 cm3 at s.t.p.:
\[ n=\frac{0.250}{22.4}=0.0112\ \text{mol} \](iii) 250 cm3 at s.t.p. (273 K) heated to 27°C (300 K) at constant pressure:
\[ V_2=250\times\frac{300}{273}=274.7\approx 275\ \text{cm}^3 \](iv) On heating at constant pressure the molecules gain kinetic energy and move faster, colliding more frequently and more forcefully; the gas expands (volume increases) so that the number of collisions per unit area of the walls, and hence the pressure, stays constant.
Question 5 Report
(a) i) An organic compound X contains 40% carbon, 6.67% hydrogen, the rest being oxygen. If X has a relative molecular mass of 60, determine its
(i) empirical formula (ii) molecular formula. [H = 1 ; C = 12; O = 16]
(b) An alkanoic acid Y has a relative molecular mass of 74.
(i) State the functional group of Y
(ii) What t of reaction is involved when Y is converted to an alkanoate?
(iii) Determine the structural formula of Y.
(iv) Write an equation for the reaction between Y and sodium
(v) If X in (a) above boils at 118°C and belongs to the same homologous series as Y, state with reason, whether the boiling point of Y will be equal to, higher or lower than 118°C.
(c)(i) What is fermentation?
(ii) Write an equation for the fermentation of glucose.
(iii) What must be added to glucose solution to make it ferment?
(iv) Explain why a tightly corked grass bottle filled to the brim with fresh palm wine shatters on standing.
(a) Empirical and molecular formula of X (40% C, 6.67% H, O = 100 - 40 - 6.67 = 53.33%)
| Element | C | H | O |
|---|---|---|---|
| Moles | 40/12 = 3.33 | 6.67/1 = 6.67 | 53.33/16 = 3.33 |
| Ratio (÷ 3.33) | 1 | 2 | 1 |
(i) Empirical formula = CH2O (mass = 30).
(ii) \(\dfrac{60}{30}=2\), so molecular formula = C2H4O2.
(b) Alkanoic acid Y, Mr = 74. For CnH2n+1COOH, propanoic acid C2H5COOH has Mr = 74.
(i) Functional group: carboxyl group, -COOH. (ii) Converting Y to an alkanoate (salt) is a neutralization reaction. (iii) Structural formula of Y: CH3CH2COOH. (iv) 2CH3CH2COOH + 2Na → 2CH3CH2COONa + H2.
(v) X (C2H4O2) is ethanoic acid; Y is propanoic acid, in the same homologous series. Y has a larger relative molecular mass (74 > 60) and longer chain, so stronger van der Waals forces; its boiling point will be higher than 118°C.
(c)(i) Fermentation is the slow chemical breakdown of a complex organic substance (such as sugar) into simpler substances (ethanol and carbon(IV) oxide) by the action of enzymes from micro-organisms such as yeast. (ii) C6H12O6 → 2C2H5OH + 2CO2. (iii) Yeast (which supplies the enzyme zymase) must be added. (iv) The palm wine keeps fermenting and produces carbon(IV) oxide gas; because the bottle is full to the brim and tightly corked, the gas cannot escape, so the pressure builds up until it exceeds the strength of the bottle and it shatters.
Answer Details
(a) Empirical and molecular formula of X (40% C, 6.67% H, O = 100 - 40 - 6.67 = 53.33%)
| Element | C | H | O |
|---|---|---|---|
| Moles | 40/12 = 3.33 | 6.67/1 = 6.67 | 53.33/16 = 3.33 |
| Ratio (÷ 3.33) | 1 | 2 | 1 |
(i) Empirical formula = CH2O (mass = 30).
(ii) \(\dfrac{60}{30}=2\), so molecular formula = C2H4O2.
(b) Alkanoic acid Y, Mr = 74. For CnH2n+1COOH, propanoic acid C2H5COOH has Mr = 74.
(i) Functional group: carboxyl group, -COOH. (ii) Converting Y to an alkanoate (salt) is a neutralization reaction. (iii) Structural formula of Y: CH3CH2COOH. (iv) 2CH3CH2COOH + 2Na → 2CH3CH2COONa + H2.
(v) X (C2H4O2) is ethanoic acid; Y is propanoic acid, in the same homologous series. Y has a larger relative molecular mass (74 > 60) and longer chain, so stronger van der Waals forces; its boiling point will be higher than 118°C.
(c)(i) Fermentation is the slow chemical breakdown of a complex organic substance (such as sugar) into simpler substances (ethanol and carbon(IV) oxide) by the action of enzymes from micro-organisms such as yeast. (ii) C6H12O6 → 2C2H5OH + 2CO2. (iii) Yeast (which supplies the enzyme zymase) must be added. (iv) The palm wine keeps fermenting and produces carbon(IV) oxide gas; because the bottle is full to the brim and tightly corked, the gas cannot escape, so the pressure builds up until it exceeds the strength of the bottle and it shatters.
Question 6 Report
(a) (i) State three characteristic properties of transition metals.
(ii) What is the oxidation state of manganese In each of the following species? (1) \(\mathrm{MnCl_2}\) (II) \(\mathrm{MnO_2}\) (III) \(\mathrm{MnO_4^-}\)
(iii) Explain why manganese conducts electricity in the solid state but manganese chloride conducts only when molten or in solution.
(b)(i) The collision theory suggests that for two particles to react, they must collide. What two factors determine whether or not the collision would lead to formation of products?
(ii) Use an energy profile diagram to illustrate what is meant by the enthalpy change (\(\Delta H\)) and the activation energy (\(E_A\)) of a reaction.
(c) When few drops of aqueous KSCN are added to a solution of iron (III) salt the following equilibrium is set up:
| \(\mathrm{Fe^{3+}_{(aq)}}\) | + | \(3\mathrm{SCN^-_{(aq)}}\) | \(\rightleftharpoons\) | \(\mathrm{Fe(SCN)_{3(aq)}}\) |
| yellow | colourless | deep red |
The equilibrium mixture has a pale red colour.
(i) Explain what would happen if more \(\mathrm{KSCN_{(aq)}}\) were added to the equilibrium mixture.
(ii) Which of the ions in the equilibrium mixture forms an insoluble hydroxide with \(\mathrm{NaOH_{(aq)}}\)? Write an equation for the reaction
(iii) State two changes observed on adding \(\mathrm{NaOH_{(aq)}}\) to the equilibrium mixture.
(a)(i) Three characteristic properties of transition metals are:
- They exhibit variable oxidation states.
- They form coloured ions or compounds.
- They form complex ions.
(ii)
- In \(\mathrm{MnCl_2}\): \(x+2(-1)=0\), hence \(x=+2\).
- In \(\mathrm{MnO_2}\): \(x+2(-2)=0\), hence \(x=+4\).
- In \(\mathrm{MnO_4^-}\): \(x+4(-2)=-1\), hence \(x=+7\).
(iii) Manganese is metallic and contains mobile, delocalised electrons which carry electric current in the solid state. Manganese(II) chloride is ionic. In the solid state its ions are fixed in a lattice, but when molten or dissolved in water, the ions are free to move and conduct electricity.
(b)(i) A collision produces products only when:
- the colliding particles possess energy equal to or greater than the activation energy; and
- the particles collide with the correct orientation.
(ii) The energy profile below is for an exothermic reaction. \(E_A\) is the energy difference between the reactants and the activated complex, while \(\Delta H\) is the energy difference between products and reactants.
For this exothermic reaction, the products are at a lower energy than the reactants; therefore, \(\Delta H\) is negative.
(c)(i) Addition of \(\mathrm{KSCN(aq)}\) increases the concentration of \(\mathrm{SCN^-}\) ions. The equilibrium shifts to the right to use up some of the added \(\mathrm{SCN^-}\):
\[\mathrm{Fe^{3+}(aq)+3SCN^-(aq)\rightleftharpoons Fe(SCN)_3(aq)}\]
More deep-red \(\mathrm{Fe(SCN)_3}\) is formed, so the pale red solution becomes deeper red.
(ii) The ion that forms an insoluble hydroxide is \(\mathrm{Fe^{3+}}\).
\[\mathrm{Fe^{3+}(aq)+3OH^-(aq)\rightarrow Fe(OH)_3(s)}\]
(iii) On adding \(\mathrm{NaOH(aq)}\):
- a reddish-brown precipitate of \(\mathrm{Fe(OH)_3}\) is formed; and
- the red colour fades or disappears, leaving a colourless solution above the precipitate.
Answer Details
(a)(i) Three characteristic properties of transition metals are:
- They exhibit variable oxidation states.
- They form coloured ions or compounds.
- They form complex ions.
(ii)
- In \(\mathrm{MnCl_2}\): \(x+2(-1)=0\), hence \(x=+2\).
- In \(\mathrm{MnO_2}\): \(x+2(-2)=0\), hence \(x=+4\).
- In \(\mathrm{MnO_4^-}\): \(x+4(-2)=-1\), hence \(x=+7\).
(iii) Manganese is metallic and contains mobile, delocalised electrons which carry electric current in the solid state. Manganese(II) chloride is ionic. In the solid state its ions are fixed in a lattice, but when molten or dissolved in water, the ions are free to move and conduct electricity.
(b)(i) A collision produces products only when:
- the colliding particles possess energy equal to or greater than the activation energy; and
- the particles collide with the correct orientation.
(ii) The energy profile below is for an exothermic reaction. \(E_A\) is the energy difference between the reactants and the activated complex, while \(\Delta H\) is the energy difference between products and reactants.
For this exothermic reaction, the products are at a lower energy than the reactants; therefore, \(\Delta H\) is negative.
(c)(i) Addition of \(\mathrm{KSCN(aq)}\) increases the concentration of \(\mathrm{SCN^-}\) ions. The equilibrium shifts to the right to use up some of the added \(\mathrm{SCN^-}\):
\[\mathrm{Fe^{3+}(aq)+3SCN^-(aq)\rightleftharpoons Fe(SCN)_3(aq)}\]
More deep-red \(\mathrm{Fe(SCN)_3}\) is formed, so the pale red solution becomes deeper red.
(ii) The ion that forms an insoluble hydroxide is \(\mathrm{Fe^{3+}}\).
\[\mathrm{Fe^{3+}(aq)+3OH^-(aq)\rightarrow Fe(OH)_3(s)}\]
(iii) On adding \(\mathrm{NaOH(aq)}\):
- a reddish-brown precipitate of \(\mathrm{Fe(OH)_3}\) is formed; and
- the red colour fades or disappears, leaving a colourless solution above the precipitate.
