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So in our previous topic, we learn how to write the electron configuration for atoms, and so in this video will be doing something similar except for the case of islands. And so when we write the electron configuration for islands, you basically find the electron configuration for the atom first, and then, depending on the charge of the atom, we can add or remove electrons from the atomic orbital. And so when we do this, it's important to keep in mind that when you write the electron configuration for islands, thes tend to have noble gas configurations. And so that means that when you add electrons or move electrons, it will have the configuration as neon argon are kept on or things of that sort. And so that's one thing to keep in mind whenever you were writing the electronic duration four islands and so this is best demonstrated with a couple of examples. So let's say, for our first example, we want to figure out the electron configuration for a chlorine, and so first we begin with writing the electron configuration for the atom, so chlorine on DSO we can use the same strategy that we've used before so we can start from the top left corner and then work our way down towards chlorine. And so we know that we can fill the electrons in one s and we can fill electrons and to us. And so we get to us, too. And subsequently, two p six and then three. Us, too. And now, to get corn, we have three p five. And so that is theologically Ron conversion of the atom. And to get the electron configuration of Florida, that means that we need to add another electron. And so, instead of three b five, we now have 336 which will be your final answer to the first part. And so, as you notice, if we add an extra Ultron, this means that the chloride will actually have an electron configuration that is the same as Arugam, which makes a lot of sense because the reason why these ions form in the first place is to achieve a stable configuration. Our in other words, the noble gas configuration. And so now let's go on to another example. So let's say that you want to figure out the electron configuration for aluminum, three bus and so we'll be using the same strategy as before. So aluminum is looking in here, and so we want to write the electron configuration for the item. And then once we figure that out, we can determine the electron configuration for the. So again, we can fill up the electrons and wants to and to us too, as well as two p six. And there has to and Olodum has to be in three p one, and now we know that aluminum has a charge of plus three. So that means that we need to remove three electrons. And so that means we need to remove the electron and three B and three US too. And so this would be your electron configuration for aluminum, which again makes sense because if we lose the electron and aluminum and those two and the three and still orbital's, we get an electron configuration off neon, which is stable. And so this makes a lot of sense and moving on to our next problem. Let's say that we need to figure out the electron configuration for strontium, which makes a two plus Kyle. So strontium is located here, and so, like before we'll be using the same strategy, Um, specifically writing the electron configuration for the atom first and then figuring out what the electron configuration for the ironist. And so again we can feel electrons and want us to to us too. Two p six three us too. Three p six for us to remember that three d time comes after he for us to we can feel these orbital's with electrons and we have four p six. And finally we have shown damn which is five as to. And so now we know that to the strontium makes a two plus cat on and so we can remove electrons from the s orbital. So we end up with a final electron configuration of one s to to us to t v 63 us to three p six for us to three d 10 and four p six. And if we double check, this is the noble gas configuration specifically for a crypt on. And so let's make a lot of sense and moving on to our last example, we have go bald, which has a plus to charge. And so again, using this change using the same strategy as before we will right the electron configuration of the atom. And then from there we can determine the electron conversion for the ion. And so here we're going to use the same strategy so you can feel in the one that's too to us, too. Tp six, three us too. Three p six for us to on 1234567 So this is three D seven. And so now we can figure out the electron configuration for the on. And so we need to remove two electrons because this is a canton. And so that means it's in a 37. We need three d five, which will be your final answer to the problem.
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