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. 
(c) Are there multiple equivalent resonance structures for the ion?

Accepted Answer

Hello everyone. So in this video we're going to talk about our value a little bit about R. S. 032 minus. So what we're gonna do here first is calculate the total number of electrons. So again we have our S. 032 minus. So from our sulfur atom we have one sulfur atom. Each sulfur item adam is going to give us six balance electrons. And from our oxygen we have three of those atoms and each will give us six valence electrons as well. So you do the math here three times six equal to 18 electrons. So adding our 18 and six, that's going to give us 24 advanced electrons. And let's not forget that we have a ion with a negative two charge. And because of that that means that we have two extra electrons. Let's go ahead and add two more electrons. Doing the math here 24-plus 2 is 26. So this ion has a total of 26 valence electrons. Second step let's go ahead and draw our lewis structure. So our central atom is going to be our sulfur Connected to three oxygen's. Then of course we have one of the oxygen's. And so for having a pi bond then the other two will have just a single bond. This means that Our sulfur root gets the remaining two valence electrons. So The oxygen's with a single bond connected to the sulfur will have a formal charge of negative one and collectively negative one plus negative one is negative two. And that is okay because our ion has a negative two charge as well. And that's exactly what we want to represent. And so from our lower structure, we can actually say that this lower structure here is going to be our first president structure. So just label that structure one. Alright. Our next resident structure. Then let's go ahead and draw our arrow pushing. Then we're gonna have one of the lone pairs on the oxygen. Go and form a double bond. And then we can have this double bond break and create a long pair. Then this will give us a new business structure. We're basically the pipe on is breaking here, the right to oxygen means the same. It's not personally the same as well. They left oxygen will now have a pi bond. All right, this will be our structure too. And now drawing our 3rd structure. Then again, drawing out our arrow pushing. We have this long pair make a pi bond and have this pipe on break and have a loan or make a lone pair. In that case we get a structure. Central atom still is going to be sulfur. Now the right side oxygen will have a pi bond spot. How do you fix this? So again on the right side we have a pi bond, send a single bond and another single bond. And now we have our last structure our third structure the structure three. And as you can see from all three shelters that we have drawn is that these three are going to be equivalent. The only thing that is moving around is just a pi bond breaking to make a long pair and a long pair coming down to make a bond. So our answer to this question is that we have three equivalent resonance structures, and that is going to be our final answer for this problem. Thank you all so much for watching.

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