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Chemistry Paper 2 (Essay) ,May/June 2008  
Questions:   1 2 3 4 5 6 7 8   Main
General Comments

Question 1

(a) State
(i) Pauli’s Exclusion principle;
(ii) Hund’s rule of maximum multiplicity.  [4 marks]

(b)  Write the electronic configuration of each of the following  ions of copper:

       I.  Cu+;
      II.  Cu2+.

(ii)   Give the number of unpaired electrons in each of the ions in
        1(b) (i) above.

(iii)  State the type of reaction represented by the following equation:
                        2Cu+(aq)  →   Cu2+(aq) + Cu(s)

(iv)   Write the formula of one compound of Cu+.    [6 marks]

(c) (i)  Name the type of radiation that will

I.    penetrate lead block;
II.   be stopped by thin paper.

(ii) Give the charge on each of the radiations mentioned in
        1(c) (i) above.

(iii) What term is used to describe each of the following nuclear processes?

I.   Combination of two lighter nuclei to form a heavy nucleus
II.  Splitting of a heavy nucleus into two or more lighter nuclei
III. Time required for one-half of the atoms of a radioactive  substance to decay   [7marks]

(d) Arrange the following ions in order of increasing size.  Give a reason for your answer in each case.

I.   Li+, K+ , Na+;
II.  O2-,  F-,  N3-.  [4 marks]

(e) Determine the percentage composition of phosphorus and oxygen in phosphorus (V) oxide.
                                    [P  =  31, O  =  16]   [4 marks]


This question was attempted by majority of the candidates and the performance was good.

In (a)(i) and (ii), candidates correctly stated Pauli’s Exclusion principle and Hund’s rule of maximum multiplicity as follows:

(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]
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