What noble-gas configurations and charge are the following elements likely to attain in reactions in which they form ions? (c) S

Video transcript

Hello everyone today, we have been given the following problem provide the noble gas configuration and the charge of the ion most likely formed by selenium. So the first thing you wanna do is you want to find the atomic number for selenium And according to the periodic table, that is 34. And since we have a neutral selenium, that means that we are going to have 30 for electrons to work with. And it's also important to note that selenium is in period four or the 4th row of the periodic table. And so in constructing the noble gas configuration, we must note what the noble gas is before selenium and that's going to be our gone. And we're going to put that in brackets. And so therefore a neutral selenium will have the configuration of that noble gas or a noble gas configuration of that noble gas in brackets. Since it is in the fourth period, we're going to have four. And then the first period that we encounter in the fourth period is going to be s too because we are filling in those first two electrons, we then dip into three D. Because our d orbital's start with three, We're filling in all of those electrons. So that gets an expert of 10. And then lastly we cross over on the right hand side of the periodic table with our four P. Since we are still in the fourth row, that's going to be four P. And selenium is the fourth element in that. So that's why it gets four electrons. So now we have the noble gas configuration for neutral selenium. We already noted The selenium is in period four. Now it's important to note that selenium is in group six A or the 6th column. And so because it is in the 6th column, it is going to gain two electrons from when it becomes neutral. And so you can think of it in a way that there were eight groups and that when you get to the sixth group it's going to be reflective of if you were in the second group, but it will be the reverse process. So the left hand side of transition metals will lose electrons but on the right hand side of transition metals you will gain electrons. And so essentially what this means is that you have to add electrons. So you have to gain those two electrons in the highest orbital that you have. And so an easy way to do this is to figure out what is the highest number that we have before our orbital. So we have four S and then we have four P four P is going to be higher energy than four S. It goes S. P. D and F and P has a higher energy level than S. So we're going to add electrons to our four p orbital. So our four p orbital. And so we have to gain those two electrons as we said. And so that's going to look like our argon in brackets are noble gas followed by our forests 23 D 10 and instead of four P four, it's going to be four P six since we gained six since we gained two electrons were simply just adding two to that four. And so what element is this going to correspond to? This is going to correspond to krypton and we can represent this with our selenium symbol And a 2 - to show that we gained two electrons. And so this is going to be our final answer. Overall, I hope that this helped and until next time.