AS Unit F321: Atoms, Bonds and Groups

AS Unit F321: Atoms, bonds and groups

Module 3: The Periodic Table
Question 1 / Total marks: 15
(a) The melting points of sodium, magnesium and aluminium are relatively high and show an increase as atomic number increases. Explain these facts in terms of bonding and structure.
Marks available: 3
Student answer:
(a) All three have metallic bonding and therefore have giant structures. This means a great deal of energy is needed to overcome the strong attraction, so they all have relatively high melting points.
However, the attraction between the metal cations and the surrounding delocalised electrons is highest for aluminium as aluminium loses three electrons per atom to form Al3+ cations, compared with magnesium which loses two electrons per atom to form Mg2+ cations, and sodium which loses only one electron per atom.
Examiner comments:
(a) This answer gives a clear, thorough explanation of the bonding and the structure involved. It goes on to explain why there are differences in the strengths of the bonding.
(b) Describe what you would see if calcium, a grey metal, was ignited in air. Construct an equation for the reaction which takes place.
Marks available: 3
Student answer:
(b) The grey metal would burn with a red flame to form a white powder.
2Ca(s) + O2(g) 2CaO(s)
Examiner comments:
(b) The more important observations are the flame colour and the colour of the product. The equation has also been balanced correctly; don’t forget to do this yourself!
(c) The reaction between barium and air is far more vigorous than that for calcium and air. Explain why.
Marks available: 2
Student answer:
(c)  Both metals need to lose their two outermost electrons when reacting with oxygen. The outermost electrons of barium are further from the attraction of the positive nucleus and experience greater shielding so are lost far more easily than those of calcium.
Examiner comments:
(c) The relevant factors have been mentioned here – distance of electrons from nucleus, shielding and effect of nuclear charge.
(d) This question is about compounds of strontium.
Substance X was formed by adding 0.1 moles of water to 0.1 moles of strontium oxide, SrO.
Excess water was added to substance X to produce solution Y.
Carbon dioxide gas was bubbled through solution Y. This formed precipitate Z.
(i) Name substance X and describe its appearance.
(ii) Name solution Y and give its approximate pH.
(iii) Name solid Z.
(iv) How could substance Z be converted to strontium oxide, SrO?
Marks available:
(i) 2 (ii) 2 (iii) 1 (iv) 1
Student answer:
(d) (i) Strontium hydroxide. It is a white powder.
(ii) Strontium hydroxide. It has a pH of about 11.
(iii) Strontium carbonate.
(iv) By filtering out the solid and strongly heating it.
Examiner comments:
(d) (i) It is important to say that the substance is solid as the information stated an equal number of moles were added.
(ii) Most pH values above 7 would be acceptable here but do not opt for 14 as this is too strong an alkali.
(iii) Carbon dioxide is an acidic oxide that reacts with the base strontium hydroxide to form the salt, strontium carbonate which is insoluble.
(iv) The student has sensibly mentioned filtering, otherwise the surrounding water would simply boil.
(e) Suggest why magnesium hydroxide, Mg(OH)2, is used in preference to calcium hydroxide, Ca(OH)2, in indigestion tablets.
Marks available: 1
Student answer:
(e) Magnesium hydroxide is not as soluble as calcium hydroxide so not so many OH- ions are released into solution. Therefore the solution is not too alkaline.
Examiner comments:
(e) This question asks you to suggest an answer. Therefore, you have to use your chemistry knowledge and apply common sense.
Module 3: The Periodic Table
Question 2 / Total marks: 15
(a)  Describe the physical states of chlorine, bromine and iodine at room temperature and explain any trend.
Marks available: 3
Student answer:
(a)  Chlorine is a gas, bromine a liquid and iodine a solid. This is because they all form diatomic molecules which have no overall dipole. Therefore they are held together by relatively weak instantaneous dipoles. As iodine has more electrons than bromine, which has more than chlorine, the temporary intermolecular attractions arising from these dipoles are greatest for iodine. This means iodine is the least volatile.
Examiner comments:
(a) This is a very competent answer. Both the description and the explanation are accurate.
(b) The labels of aqueous potassium chloride, aqueous potassium iodide and aqueous potassium bromide have become removed from their containers. Ben has access to samples of chlorine water, bromine water, aqueous iodine and cyclohexane, an inert organic solvent. Explain how he could use these reagents to identify the solutions. Describe any colour changes he might see and write equations for any reactions which take place.
Marks available: 6
Student answer:
(b) He should place a sample of each unknown solution into a test tube. Then add chlorine water to all three solutions. Two will become coloured, one a pale orange solution and the other a yellowy-brown solution. The third solution will keep the very pale green colour of chlorine water.
As chlorine is the most reactive halogen, it wants to gain electrons most and so will remove electrons from both Br- and I- ions. So chlorine displaces both bromine and iodine from their halide solutions.
Bromine is orange when aqueous and iodine is yellowy-brown when aqueous. To confirm which halogen has been displaced, add cyclohexane and shake. Any halogen molecules will preferentially go to the organic solvent and a bright orange colour is seen if bromine has been formed, and a bright purple colour if iodine has been formed.
An equation for a displacement reaction is:
Cl2(aq) + 2KBr(aq) 2KCl(aq) + Br2(aq)
Examiner comments:
(b) It can be quite hard to score full marks on this type of question, mainly because it is easy to leave out an important fact. However, this is a sound, correct answer. Not all the reagents have been used, however, and although a good response has already been produced here, the reactions of bromine water with chloride solution (none) and iodide solution (yellow-brown colour forms) could also have been described.
Furthermore, only one equation has been given. The equation for Cl2 + KI should be included as well.
(c) Samples of aqueous potassium chloride, aqueous potassium iodide and aqueous potassium bromide could be detected by using aqueous silver ions. Describe the changes you would observe if aqueous silver nitrate was added to each of these solutions. Include an ionic equation, with state symbols, for one of these reactions.
Marks available: 4
Student answer:
(c) The colourless solution would form a white precipitate with aqueous potassium chloride.
The colourless solution would form a cream precipitate with aqueous potassium bromide.
The colourless solution would form a yellow precipitate with aqueous potassium iodide.
Ag+(aq) + Cl-(aq) AgCl(s)
Examiner comments:
(c) All observations are good and the student remembered to include state symbols into the ionic equation.
(d) Explain what is meant by the phrase, ‘chlorine undergoes disproportionation’. Write an equation to show this occurring.
Marks available: 2
Student answer:
(d) Disproportionation occurs when one substance undergoes oxidation and reduction at the same time.
An example of this is when chlorine reacts with water to form a mixture of acids.
Cl2(g) + H2O(l) HCl(aq) + HOCl(aq)
Both chlorine atoms start with an oxidation number of 0. The chlorine atom in HCl has an oxidation number of -1 and has been reduced, and the chlorine atom in HOCl has an oxidation number of +1 and has been oxidised.
Examiner comments:
(d) The explanation of disproportionation is spot-on.
The reaction between chlorine and sodium hydroxide could also be used as an example.
Also, it is sometimes easier to place the oxidation numbers above the elements within the equation.