CAPE Chemistry Unit 1 May/June 2025 Paper 2, Past Paper Solution Question 3

Added: 2026-05-14

[00:00:08]Welcome back to Chem Exam explained where the aim is chemistry clarity exam mastery. In this video we will be looking at Cape Chemistry paper 2 unit 1 module 3 2025.

[00:00:20]Let's go.

[00:00:22]Three part A. Table three shows the variation in some physical properties of group four elements. Now all we are required to do is to look at the table and observe the trends and then describe what each of the following trend is saying. So part one describe the trend in each of the following electrical conductivity from silicon to lead. So in looking at silicon to lead in electrical conductivity we're observing the trend here. And so you can see that from that trend electrical conductivity increases from silicon to tin then decreases from tin to lead.

[00:01:10]The melting point from silicon to lead.

[00:01:14]Again you come here in the table and you look at the melting point from silicon to lead and you can see that the melting point decreases from silicon to tin then increases from tin to lead. That is all the question is asking you for to to describe what you're seeing with respect to the trend.

[00:01:39]3A part two account for the variation in melting point from carbon to tin in terms of structure and bonding. Well, carbon has giant atomic structure where the carbon atoms are bonded covealently giving the structure a high melting point. So this is this is strong atomic or strong giant atomic bonds and these bonds are very strong and therefore the melting point would be very high there.

[00:02:11]Going from silicon to germanmanium these are metaloids and these have larger atomic radi with weaker bonds. Now if you have larger atomic radi your bond length increases which makes the bond weaker and thus lower melting point.

[00:02:32]Going from or looking at now tin and lead they have metallic bonds with exceptionally low melting point due to only two of the four electrons being involved in bonding. So two of the electrons cannot two of the electrons cannot move because they are held onto by the nucleus very strongly and this is called the inert pure effect and the other two will be lost to the sea of electrons and flow throughout the cations and so that is why only two veence electrons are involved in bonding and so the attraction now between Those cations and those two electrons are weaker than the bonds in the giant structures and therefore the the lower melting point for tin and lead.

[00:03:31]B part one describe the reaction of lead 4 chloride with water. So this is this is our tetrachloride. So PBCL4 reacts with water through hydrarolysis to form a white precipitate of lead 4 oxide or lead dioxide and an acidic fumes of HCl gas.

[00:03:54]B part two write the balance equation for the reaction described in B part one. So here we have PBCL4 reacting with water and through hydraysis you form our lead 4 oxide aka lead dioxide and our HCl gas which is our acidic fume and of course the lead 4 oxide is our white precipitate.

[00:04:20]Three part C table four shows some physical properties of group two elements and here we have the various properties density atomic radi ionic radi part one state the trend in each of the following from berium to berium density just checking out the density you'll see that the density decreases from berium to calcium then increase to barium atomic radi the atomic radi increases is down the group from berium to berium.

[00:04:53]C part two account for the trend in atomic radi in terms of the nuclear charge and the screening effect which is same as the shielding effect.

[00:05:06]The number of filled shells increases down the group which leads to more screening of outer electrons. The increasing nuclear charge has little effect on the outer electrons due to the screening effect. So the atomic radi increases.

[00:05:24]D part one.

[00:05:26]Complete the table below by recording the observations that would be made when a one molar solution of sodium sulfate is added to 0.1 mole per dm cq of each of the group two elements. So the group two cations would be magnesium ion, calcium ion, strontium ion and berium ion. And these are all two plus because they belong to group two. We should know that the solubility of the group two elements decreases down the group. And so when you react sodium sulfate which is aquous and soluble with our berium ions, our magnesium ions, the strontium ions SR and the calcium ion. You'll see that these ions are combining with the sulfate from the sodium sulfate.

[00:06:27]And so you should observe precipitate of various size or no precipitate.

[00:06:35]So we know that the solubility of sulfates decreases down the group. So the most soluble sulfate would be the ones at the top and the least soluble ones would be the ones at at the bottom. So magnesium sulfate would not produce any precipitate. But as we go down the group, due to the fact that solubility decreases, you'll now have a thin white precipitate. And then you'd have a medium white precipitate.

[00:07:05]We could put the term medium just to compare the sizes of of the precipitate. And here you have a dense or large amount of precipitate as you produce berium sulfate. So going from magnesium sulfate, no precipitate to small amount of calcium sulfate to more than calcium sulfate for strontium sulfate and of course berium sulfate being the least soluble would produce the largest amount of precipitate and that would answer our question for this table knowing the fact that the solubility of sulfate decreases down the group.

[00:07:41]Part two, account for the observation in D. Part one. As the cations get larger, it has a lower charge density.

[00:07:53]Therefore, it becomes less attracted to the polar water molecule. The ionic latises.

[00:07:59]The ionic latice is harder to break.

[00:08:02]That is latis enthalpy is greater than hydration enthalpy. Therefore, the solution is more endothermic. that is less soluble. Part three, write the ionic equation for the reaction between the berium ion Ba2+ and one more sodium sulfate solution. And as you saw the examples up here, you simply write berium 2+ plus SO42 minus to produce Ba SO4 solid. And this is a perfect example of how you can test for the presence of berium ions using sulfate ions.

[00:08:40]Part E.

[00:08:45]Part E. Some compounds of aluminium, calcium, and magnesium have various uses. For any two of the metals, complete the following table by naming one compound of the metal and the use of the compound. So here we have a few of them. Even though they require one, here we have a few of them. So we have magnesium in the form of magnesium oxide can be used as the lining of furnaces because it is ionic and therefore would require a large amount of heat for it to be broken down. Calcium in the form of calcium oxide can be used to raise the acidity of soil because when the soil is acidic the plants cannot grow in in the low pH. So it can be used to increase the pH of the soil. Aluminum oxide can be used as an abrasive in the vehicle bodywork and of course it can be used in window frames.

[00:09:44]We have other examples. Magnesium hydroxide can be used in toothpaste or an antacid specifically the milk of magnesia. Magnesium sulfate is a laxative and it can be used as epsom salt. Calcium carbonate can be used in cement and of course calcium sulfate can be used in mortar or plaster of pare.

[00:10:09]And the plaster of pare is the when you get your hand broken or your legs broken the plaster they put on it is called plaster of pare and it is calcium sulfate. This ends module 3 question three of unit one paper 2 2025.