Chemistry May/June 2008

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

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:

1. brick red precipitate
2. 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

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:

1. non- conductors of electricity.
2. insoluble in water/soluble in non-polar solvents.
3. 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.

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 
                       |      |     |      |
                       H     |      H     H
                       H –  C –  H                 
                              |                                                         
                              H       
2,3 -dimethylbutane                     1,4-dibromocyclohexane           

Not Available

Not Available

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:

1. Acidic oxides:  SO2/CO2/ NO2, etc.
2. Basic oxides:    CaO/Na2O/K2O, etc.
3. Amphoteric oxides: Al2O3/ZnO/PbO, etc.
4. 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)

In (d)(i), candidates correctly gave two characteristics of noble gases from the following:

1. they are monoatomic gases
2. they are chemically inert or unreactive
3. 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.         

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.