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ANSWER
(i) Two electrons in the same orbital of an atom cannot have same values for all four quantum numbers/no two electrons can have the same four quantum numbers/ no two electrons in the same orbital of an atom can have the same spin.
(ii) Electrons occupy each orbital singly first before pairing takes place in a degenerate orbital/the most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins.
In (b)(i), only few candidates correctly wrote the electronic configuration of Cu+ and Cu2+
The expected response from candidates was as follows:
Cu+ – 1s2 2s2 2p6 3s2 3p6 3d10
Cu2+ – 1s2 2s2 2p6 3s2 3p6 3d 9
In (b)(ii), majority of the candidates did not know that Cu+ had no unpaired electron and Cu2+ had only one unpaired electrons.
In (b)(iii), most candidates wrote either “reduction” or “oxidation” instead of redox or disproportionation
In (b)(iv), candidates were able to write Cu2O/CuCl/Cu2Cl2 as the formula of one compound of Cu+.
In (c)(i), most candidates knew that gamma rays penetrate lead block and alpha particles will be stopped by thin paper. In (c)(ii), they were able to give the charge on each of the radiations as electrically neutral and positively charged respectively.
In (c)(iii), candidates correctly gave nuclear fusion, nuclear fission and half – life respectively, for each of the nuclear processes.
In (d), only few candidates correctly arranged the ions in order of increasing size with a reason as follows:
Li+, Na+, K+; Size increases down a group as more shell are being added.
F-, O2-, N3-; Size increases as nuclear charge decreases in the same period/isoelectronic.
In (e), candidates correctly determined the percentage composition of phosphorus and oxygen in phosphorus (V) oxide as follows:
Phosphorus (V) oxide = P4 O10 /P2O5
= (31 x 4) + (16 x 10) OR (31 x 2) + (16 x 5)
= 124 + 160 OR 62 + 80
= 284 OR 142
% by mass of phosphorus = 4P x 100% OR 2P x 100%
P4 O10 P2 O5
= 124 x 100% OR 62 x 100%
284 142
= 43.7%
or 100 – 43.7 = 56.3% Oxygen
[Any other correct method was accepted]
Question 2
(a) (i) Define in terms of electron transfer
I. oxidizing agent;
II. reducing agent.
(ii) Write a balanced equation to show that carbon is a reducing agent.
(iii) State the change in oxidation number of the specie that reacted with carbon in 2 (a)(ii). [5 marks]
(b) A gas X has a vapour density of 32. It reacts with sodium hydroxide solution to form salt and water only. It decolourizes acidified potassium tetraoxomanganate (VII) solution and reacts with H2S to form sulphur. Using the information provided:
identify gas X;
state two properties exhibited by X;
give two uses of X. [5 marks]
(c) Consider the following substances:
(i) sodium;
(ii) lead (II) iodide;
(iii) hydrogen;
(iv) magnesium;
(v) oxygen.
Which of the substances
(i) conducts electricity?
(ii) is produced at the cathode during electrolysis of H2SO4(aq)?
(iii) corresponds to the molecular formula AB2 ?
(iv) is an alkaline earth metal? [5 marks]
(d) (i) Define the term salt.
(ii) Mention two types of salt.
(iii) Give an example of each of the salts mentioned in 2(d)(ii) above. [6 marks]
(e) In a neutralization reaction, dilute tetraoxosulphate (VI) acid completely reacted with sodium hydroxide solution.
(i) Write a balanced equation for the reaction.
(ii) How many moles of sodium hydroxide would be required for the complete neutralization of 0.50 moles of tetraoxosulphate (VI) acid? [4 marks]
ANSWER
(i) Oxidizing agent is a substance which accepts electrons/is an electron acceptor.
(ii) Reducing agent is a substance which donates electrons/is an electron donor.
In (a)(ii) and (iii),most of the candidates could neither write a balanced equation to show that carbon is a reducing agent nor state the change in oxidation number of the specie that reacted with carbon. The expected answers from candidates were:
(ii) 2CuO(s) + C(s) → 2Cu(s) + CO2(g)
H2O(g) + C(s) → 2CO(g) + H2(g)
CO2 (g) + C(s) → 2CO(g)
(iii) Cu in CuO from +2 to O
H in H2O from +1 to O
C in CO2 from + 4 to +2
In (b), candidates correctly identified gas X, stated two properties exhibited by X and gave two uses of X as follows:
(i) X is sulphur (IV) oxide/sulphur dioxide
(ii) – heavier than air
– acidic
– reducing/oxidizing agent
– colourless (poisonous) gas with irritating smell
(iii) – used for bleaching
– for the manufacture of H2SO4
– as germicide and fumigant
– for preservation
– as refrigerant e.t.c.
In (c)(i) – (iv), most candidates were able to correctly give their responses as follows:
(i) sodium/magnesium
(ii) hydrogen
(iii) lead (II) iodide
(iv) magnesium
In (d)(i), candidates correctly defined salt as a compound formed when all or parts of the hydrogen of an acid is replaced by metal or ammonium ion.
They were also able to correctly mention two types of salt with corresponding example in (d)(ii) and (iii) as follows:
(ii) – normal salt
– acid salt
– basic salt
– double salt
– complex salt
(iii) NaCl/ZnSO4/ KHSO4/ NaH2PO4/ Zn(OH)Cl/Mg(OH)NO3/NH4Fe(SO4)2. 6H2O
In (e)(i), candidates correctly wrote a balanced equation for the reaction between dilute tetraoxosulphate (VI) acid with sodium hydroxide solution as follows:
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O (1)
In (e)(ii), candidates correctly determined the number of moles of sodium hydroxide that would be required for the complete neutralization of the given tetraoxosulphate (VI) acid thus:
From the reaction 1 mole of H2SO4 ≡ 2 moles of NaOH
∴ 0.50 moles of H2SO4 ≡ 2 x 0.5 of NaOH
1
= 1 mole of NaOH
Question 3 |
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(a)Consider the following reaction sequence.
(i) What process leads to the formation of K? (b) (i) What are carbohydrates? (ii) Give one example each of a
(c) Consider the following structure of a simple sugar. H (i) Which functional group makes the compound a reducing agent? (ii) State what would be observed when (iii) Write an equation for the reaction in 3(c)(ii)(II). [6marks] (d) A hydrocarbon Z with molecular mass 78 on combustion gave 3.385 g of CO2 and 0.692 g of H2O. Determine the molecular formula of Z.
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ANSWER
(i) Dehydration
(ii) C2H4
(iii) H H
| | ethane – 1,2-diol/ethylene glycol
H – C – C – H
| |
OH OH
(iii) Polyethene/polythene.
In (b)(i), candidates correctly stated that carbohydrates are (naturally occurring) organic compounds containing carbon, hydrogen and oxygen with the hydrogen and oxygen present in the ratio of 2:1 (as in water).
In (b)(ii), they correctly gave example of each of monosaccharide, disaccharide and polysaccharide thus:
I. monosaccharide: glucose/fructose
II. disaccharide: sucrose/lactose/maltose
III. polysaccharide: starch/glycogen/cellulose
H
|
In (c)(i), candidates gave C = O /CHO as the functional group that makes the compound a reducing sugar.
In (c)(ii), candidates correctly stated what would be observed in I and II thus:
- brick red precipitate
- black/charred mass (of carbon is observed)
In (c)(iii), most candidates correctly wrote an equation for the reaction in 3(c)(ii) (II) as follows:
C6H12O6 → 6C + 6H2O
In (d), majority of the candidates correctly determined the molecular formula of Z thus:
mass of hydrogen in Z = 2 x 0.692 = 0.077g
18
mass of carbon in Z = 12 x 3.385 = 0.923g
44
C H
0.923 0.077
12 1
0.077 0.077
0.077 0.077
0.077 0.077
1 1
Empirical formula of Z = CH.
(CH)n = 78
13n = 78
n = 78 = 6
13
Molecular formula of Z = C6H6
Question 4 |
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(a) (i) Define covalent bond.
(b) (i) Write three subatomic particles with their corresponding relative masses. (c) (i) State Graham’s law of diffusion. [ H = 1, He = 4, C = 12, N = 14 ] [4 marks] (d) Draw the structures of the following compounds: |
ANSWER
covalent bond is a bond between two atoms in which each of the atoms contributes to the shared pair of electrons.
In (a) (ii), candidates correctly gave two properties of covalent compounds from the following:
- non- conductors of electricity.
- insoluble in water/soluble in non-polar solvents.
- have low melting/boiling point.
In (a)(iii), most candidates could not correctly show with the aid of a diagram how ammonia molecule is formed. The required answer to the question was
The candidates were also unable to correctly illustrate the formation of ammonium ion with a diagram in (a)(iv).
The expected diagram was as follows:
In (a)(v), candidates correctly stated that the type of bond which exists in each of ammonia and ammonium ion were covalent bond and covalent bond/dative or coordinate covalent bond respectively.
In (b)(i), candidates could not correctly write the three subatomic particles with their corresponding relative masses as they did not understand the demand of the question.
The expected answers were as shown in the table below.
PARTICLE | RELATIVE MASSES |
Proton | 1 |
Neutron | 1 |
Electron | 1 |
In (b)(ii), candidates correctly named the possible states in which water can exist as, solid, liquid and gaseous.
In (c)(i), most candidates correctly stated Graham’s law of diffusion however, some of them lost marks as they left the conditions under which the law holds while others wrote density instead of vapour density. The expected response from candidates was at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its vapour density/molecular mass.
In (c)(ii) only few candidates correctly arranged the gases in increasing order of rates of diffusion with a reason as follows: N2, CH4, He: the smaller the molar mass, the faster the rate at which the gas diffuses or the larger the molar mass, the slower the rate at which gas diffuses.
In (d), most candidates could not draw correctly the structures of the organic compounds. The expected structures from candidate were as drawn below.
H
|
H – C – H
H – H | H
| | | |
H -C – C – C – C – H
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H | H H
H – C – H
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H
2,3 -dimethylbutane 1,4-dibromocyclohexane
Not Available
Not Available
Question 7 |
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(a) (i) What are acidic oxides? (b) (i) Define each of the following terms: (ii) Write an equation to illustrate each of the terms in 7(b)(i) above. (iii) Given that the standard heat of combustion of butane (C4H10) is + 5877 kJmol-1, calculate the heat of combustion of 14.5 g butane. [H = 1, C = 12] [10 marks] (c) (i) Name two allotropes of sulphur. (d) (i) Give two characteristics of noble gases. (e) State what is observed on warming ammonium trioxonitrate (V) with sodiumhydroxide. [2 marks] |
ANSWER
In (a)(i), most candidates gave an incomplete definition of acidic oxide by stating that they are oxides of non metals without stating that they dissolve in water to form acidic solutions hence lost the marks. However, they correctly gave one example each of acidic oxide, basic oxide, amphoteric oxide and neutral oxide as follows:
- Acidic oxides: SO2/CO2/ NO2, etc.
- Basic oxides: CaO/Na2O/K2O, etc.
- Amphoteric oxides: Al2O3/ZnO/PbO, etc.
- Neutral oxides: CO/N2O, etc.
In (b)(i), most candidates correctly defined each term thus:
(I) Heat of combustion is the heat change when one mole of a substance is completely burnt in oxygen/burnt in excess oxygen.
(II) Heat of neutralization is the heat change when one mole of water is produced as a result of a reaction between an acid and alkali in dilute solution/the heat change when one mole of H+ from an acid reacts with one mole of OH- from an alkali to form one mole of H2O.
In (b)(ii) majority of the candidates could not write an equation to illustrate each of the terms hence lost the marks. The correct equation were as follows:
I. C(s) + O2(g) → CO2 (g) ∆HcӨ
II. NaOH (aq) + HCI(aq) → NaCI(aq) + H2O(1) ∆HnӨ
In (b)(iii) candidates correctly calculated the standard heat of combustion of butane thus:
Molar mass of = 12 x 4 + 1 x 10 = 48 + 10 = 58gmol-1
58 g of C4H10 5877kJmol-1
∴ 14.5 of C4H10 = 5877 x 14.5 kJmol-1
58
= + 1469.25 kJ
In (c)(i) and (ii) candidates correctly named two allotropes of sulphur and stated one difference between the two allotropes.
(1) Rhombic
Monoclinic
(II)
Rhombic | Monoclinic |
Yellow translucent crystal | Transparent amber crystal |
Stable at temperature below 900 | Unstable at temperature below 900 |
Higher density than monoclinic | Lower density than rhombic |
Lower melting point | Higher melting point |
Octahedral shaped | Needle shaped. |
In (d)(i), candidates correctly gave two characteristics of noble gases from the following:
- they are monoatomic gases
- they are chemically inert or unreactive
- have low melting/boiling points.
However in (d)(ii), candidates could not correctly give one use of each of Helium and argon.
The expected answers from candidates were
He – filling balloons
Ar – arc welding
In (e), most candidates correctly stated that ammonia gas was given off on warming ammonium trioxonitrate (V) with sodium hydroxide.
Question 8 |
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(a) (i) Define hard water. (c) (i) Give two uses of sodium tetraoxosulphate (VI) exposure to air (d) (i) Define the term activated complex. Draw an energy profile diagram for the reaction showing the I. activated complex II. enthalpy of reactants. [7 marks] |
ANSWER
ANSWER
In (a)(i), candidates correctly defined hard water as water which will not readily form lather with soap.
In (a)(ii) candidates correctly named two substances responsible for hardness in water from among the following:
- calcium tetraoxosulphate (VI)
- calcium hydrogentetraoxosulphate (IV)
- magnesium tetraoxosulphate (VI)
- magnesium hydrogentetraoxosulphate (IV)
In (a) (iii), candidates gave two methods for the removal of hardness in water from among the following:
- heating the water to boiling/distillation
- addition of calculated amount of calcium hydroxide [Ca(OH) 2]
- addition of washing soda crystals to the water [Na2CO3.10H2O]
- ion exchange resin/permutit method.
In (b)(i), candidates correctly gave the raw materials for the manufacture of tetraoxosulphate (VI) acid by the contact process as sulphur and air/oxygen.
In (b)(ii) and (iii), candidates correctly wrote an equation for the reaction that requires a catalyst in the contact process and stated the catalyst used.
The equation is
2SO2 + O2 » 2SO3(g)
and the catalyst employed is V2O5/ vanadium (V) oxide.
In (c)(i) most candidates were able to correctly give two uses of sodium tetraoxosulphate (VI) from among the following:
- water softener
- in glass/paper manufacture
- as analytical reagent
- as purgative
- in the manufacture of detergents
In (c)(ii) most candidates correctly gave the name of the type of reaction that was represented by the given equation.
In (c)(iii) candidates correctly calculated the solubility of Na2SO3 at 25°C.
In (d)(i), candidates did not know that activated complex is the intermediate (complex) that reactants must attain in order to form a product or a temporary species formed by reactant molecules as a result collision before the products are formed, instead they gave the definition of activation energy.
In (d)(ii), most candidates could not state one reason why collision may not produce a chemical reaction. They did not know that if the energy of the colliding particles is less than the activation energy or the reactants are not properly aligned, chemical reaction may not occur inspite of collision.
In (d)(iii), candidates drew the energy profile diagram for the endothermic reaction but most of them lost marks because they indicated the enthalpy of reaction instead of the enthalpy of reactants. The correct energy profile diagram is as drawn below.