(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%
or 100 – 43.7 = 56.3% Oxygen
[Any other correct method was accepted]