hey guys welcome back this is part three of 4.3 on soils this is the first science understanding silicon dioxide silicon sand alumina silicates are important components of rocks and soils you need to know how to write the formula of the anion given the formula of the silica or alumina silica in terms of the first compound mentioned silicon dioxide is present in sand there are silicates and alumina silicates are a large component of rocks caves and soils and they essentially comprise of 92 to 93 percent of all the mineral compounds that we find in the Earth's crust this diagram shows you the abundance of elements in the earth's crust and surprise surprise we can see that silicon and oxygen constitute majority of the elements found within the crust followed by aluminium so that helps correspond to the presence of these silicates and alumina silicates as well as silicon dioxide let's now consider the structure of silicates and alumina silicates we know that these are negatively charged ions silicates are based on this sio4 tetrahedra which essentially consists of one silicon atom which is covalently bonded to four oxygen atoms when oxygen atoms help bond different silicon atoms together this can help result in the formation of long stable chains and rings these structures can form quite readily and they're extremely stable in terms of alumina silicates they have some silicon atoms essentially substituted with aluminium we usually associate aluminum as a metal but in this case because it's involved in some covalent bonding we usually term a metalloid one thing to note is that for every aluminium that replaces one silicon atom this essentially increases the net negative charge of your silicon / aluminosilicate the reason for this is that aluminium and silicon have different oxidation numbers and this is something we're going to consider next before we do this diagram just shows you all the different structures that silicates and alumina silicates can take up so they could essentially form individual ions like this one here so this is what we call an ortho silicate which is just one SiO 4 you the next structure to the right we would consider this to be a die silica ion which essentially consists of two SiO 4 units which are joined together through one oxygen atom we have a hexa silicate structure here which is a cyclic structure we then have single chains double chains sheet structures and then we have a continuous Network structure which we find in silicon dioxide one key point from this is that no matter what type of structure it forms each of the ions or the repeating units essentially will consist of the same negative charge and so what we're going to do now is look at how we can determine that negative charge of these silicate iron or the repeating unit and its structure to do that we have to consider that silicates and alumina silicates form ionic compounds we know that they consists of negative charge and this negative charge is balanced by the presence of cations we've got a range of different cations here that are commonly present it's also important to note that anions as well as water may help contribute to the silicates structure in order to determine the charge of a silicon ion or alumina silica tile we're going to look at two methods that can be used so the first method is a charge balancing method and the second is oxidation numbers we're going to start off with method 1 which is charge balancing so the first example is for a silicate we have Lawson I in bold we can see the formula of the silica which means all the other components are ions or water we can see that last night is comprised of calcium ions aluminium ions silicon ions hydroxide ions as well as water and the idea behind charge balancing is in fact that the sum of the charges of each of these ions as well as water is going to cancel out so effectively it's going to equal zero so we know calcium ions are two positively charged aluminum three positive hydroxide one negative water is neutral so we can leave that out and so we're just going to sum up the number of those ions within this compound formula so we have here one calcium 2 plus with two aluminum 3 plus plus one silica ion with an unknown negative charge we two hydroxides with the one negative charge that's going to balance out and result in an overall charge of zero we're going to now just convert these islands into numbers so positive 2 plus 6 plus the silica ion minus 2 equals 0 we just need to rearrange that and solve for the charge of the silicate which in this case is negative 6 or 6 negative so therefore we could write the formula as SI 207 6 negative the second example is an aluminosilicate we have sodalite here with its formula and we can see again in bold this is the structure of the alumina silica Isle and there are six for every eight sodium ions and two chloride ions we're going to work this out in exactly the same way so we have eight and a plus with six lots of our Lumino silica ion with an unknown negative charge plus two chloride ions with one negative charge equals zero because the charges will balance out we're going to convert those into numbers so positive eight plus six lots of the alumina silica Island negative two equals zero let's just bring these over to the other side we get six lots of the alumina silica ion is equal to negative six so if we then divide the 6 over we would find that each alumina silica own should have a one negative charge for that to be true method two involves the use of oxidation numbers and these are the three oxidation numbers that you just need to know so aluminium being in group 3 has an oxidation number of plus 3 silicon in group 4 with plus 4 and oxygen in group 6 or 16 with negative 2 using this we can say that the sum of the oxidation numbers of each of these atoms within the silica or alumina silicon ion will essentially equal the charge of the ion itself so again with this first example we'll consider a silicate and in this case it is Beryl with this formula here I've put in bold the formula of the silica and just like that aluminium is present but this is just going to help balance negative charge of the silica having said that we won't need to consider the charges of these two oils here we're just going to look at the oxidation numbers of silicon and oxygen we could essentially say that six silicon's plus 18 oxygens is going to equal the charge of the ion so if we substitute in the oxidation numbers above here that equals six lots of positive 4 plus 18 lots of negative 2 and if we solve for that we should end up with a number equal to negative 12 that then suggests that the formula of this is SI 6 Oh 18 12 negative so again be some of the oxidation numbers of the silicon oxygens and possibly aluminium is going to equal the charge for our last example we've got an alumina silica again and we have a North ight with this formula here this is the formula of the alumina silica ion before we even look at observation numbers we could possibly suggest that this has a charge of 2 negative because there is one calcium ion for every one of these units here and calcium has a two positive charge so that means this must be a two negative but let's just confirm that's the case using oxidation numbers we could say that two aluminium x' plus two silicon's plus eight oxygens is going to equal the charge so that's two lots of positive three plus two lots of positive 4 plus 8 lots of negative 2 if we substitute those in we can then end up with an answer of negative 2 and as we said before the charge of the alumina silica ion will be al 2 SI 308 2 negative in class will look out working through some more examples of determining the formula of silicates and alumina silicates but for now that concludes part 3 of 4.3 I'll see you guys in the last video