In the sulphate ion, SO42-, the sulphur atom is the central
atom ...

Question

In the sulphate ion, SO42-, the sulphur atom is the central
atom with the other 4 oxygen atoms attached to it. 
(d) How many electrons are in the p system of the ion?

Accepted Answer

welcome back everyone in this example, we need to calculate the number of pi electrons in the pi system of the per chlorate ion. So we're going to recall that per chlorate is represented by C l minus. And our first step is to draw out our lewis structure but we need to calculate total valence electrons before doing so. So beginning with our one chlorine atom, we're going to recognize that chlorine is located in Group seven A. On our periodic table corresponding to seven valence electrons that we multiply by moving on. We recognize that we have four oxygen atoms And we're going to multiply by the six films electrons that oxygen has because it's located in group six a. On the periodic table. So just to make sure everything is visible, let's scoot this over a bit. And lastly we recognize that we have that -1 anti in charge. We call that a negative charge means we gain electrons. So we have plus one extra electron from that an ion charge, so totaling everything together. This gives us a total of 32 valence electrons total. And so to draw a structure based on our formula, we recognize that we have one chlorine atom in the center surrounded by four oxygen atoms. Our next step is to recall the bonding preference of oxygen which is to have two bonds and two lone pairs. So we can give this first oxygen at the top two bonds to chlorine and two lone pairs on itself For the second oxygen, we can do the same. Sorry, let's just make this neater. And then for this third oxygen will also have two bonds and two lone pairs. So so far we can count a total of 245678, 10, 12, 14 6. Or sorry, 10 12, 13 14 15, 16, 18, 20 21 22 23. 24. Of our valence electrons that we view so far. So we can say minus 24 electrons. And this is going to leave us eight valence electrons left to fill in which will leave us with no choice but to make a single bond to this third oxygen atom where it's now going to have a total of three lone pairs because it can't form a double bond since chlorine should only have seven valence electrons and we want our central atom to remain stable right now. We can see that in each bond it has a total of one valence electron contributed. So we have one valence electron here, 23456 and seven. And recall that chlorine is capable of having all of these bonds attached to itself violating the octet rule due to the fact that it's a period three and below element we can find chlorine on the periodic table in group seven A across period three. So it's OK that it violates the octet rule. And because our oxygen adam has a total of three lone pairs around itself with one of its valence electrons being shared in a co valent bond with chlorine. We can count a total of 1234567 electrons around our oxygen atom here, which will give it a formal charge of minus one. Recall that we can calculate formal charge by taking our valence electrons for that atom subtracted from non bonding electrons, which is then subtracted from bonding electrons. And so we have a formal charge of minus one on this oxygen atom. And our next step is to determine whether our lewis structure is complete and we can see that we've used up the last eight of our valence electrons. And so we now have a complete lewis structure with a net formal charge of minus one contributed from this oxygen atom here. Now, based on the prompt, we need to determine pi electrons and we want to recall that one double bond contains one sigma bond And one pi bond. And so we can therefore count a total of three double bonds here where in each double bond we have one sigma bond and one pi bond which would give us a total of three pi bonds because one of these double bonds is a pi bond here. So we would say therefore we have three pi bonds in our per chlorate lower structure. And we also want to recognize that because we have this minus one charge of this extra loan pair on this third oxygen atom here. This is going to contribute to residents of our Lewis structure, meaning we can make an entirely different LeWIS structure where this minus one charge could be on any of these other oxygen atoms here. So because we have this negative charge contributing extra loan pair, two residents, we can say that therefore our number of pi electrons Increases by value of two. And so for our total pi electrons, we can say that we have three pi bonds, which corresponds to a total of because we know each bond has two electrons, we have 123456 pi electrons. So six pi electrons added to the two extra, which is contributed from the negative charge, making our extra loan pair on that third oxygen be contributed to residents. So that would give us an extra of two pi electrons from our oxygen atom. Or we can say from our oxygen and ion. This will give us a total of eight pi electrons total. And so this would actually be our final answer to complete this example. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video

In the sulphate ion, SO42-, the sulphur atom is the central atom with the other 4 oxygen atoms attached to it. (d) How many electrons are in the p system of the ion?

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