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Ch.5 - Periodicity & Electronic Structure of Atoms

Chapter 5, Problem 82

What is the maximum number of electrons in an atom whose highest-energy electrons have the principal quantum number n = 5?

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Hello everyone in this video we're given four different orbital's and we're trying to see the maximum number of electrons that can be contained in H. So bro, number one or the first part were given the number one? S So we can see her right off the bat there. End value is equal to one and for L value, the L value that relates with the S. Sub shell is going to be Sarah. Alright, so the next quantum number is going to be our M L value and our ml value is depending on our L value and it can range from negative L. They're a positive elk We said that our our value is going to be zero which means it's going to be negative zero through positive zero but we know that zero doesn't really have a negative or positive value sir, Ml simply just zero then next let's see here. So we only have one orbital because L equals to zero and because we have one orbital we can only have two possible M S values. So the M s value is equal, so either negative half or positive half. So each one of these represent one electron and because we have one commodore Then it's only possible to have maximum two electrons. So let's see here that the electrons welcome is going to be too too. So that's the answer for the first part. Now, Next we're going to move on to the second part. And the second part is The quantum numbers that were given is going to be four. Now, right off the bat again you can see that our end value is going to be equal to four and for our L value, what correlates to the d sub shop It's going to be l equals to two. Now solving for the M. L. Number again that's ranging from negative L two, positive L. That's going to be negative two through positive too. Now drawing that out, I just want to go ahead and make it kind of visual because me myself, I am a visual learner. So let's see here we have they're going to negative one, zero one and two. And what I just drew here is just seeing the values for this range is running out or expanding this range. So all these numbers are going to fit in this range and we know that only two electrons can fit in to each orbital here. So we have let's see, we have 1234 and five. So do some quick math here five times 2 because five or our shells there are here Orbital's I apologize, we have five, orbital's here and then two electrons can fit into each orbital five times two equals to 10. And therefore the maximum number of electrons that can be filled in this orbital, It's going to be 10. So that's going to be our answer for the second part of this question. Now let's go ahead and scroll down a little bit. Alright, so number three were given four and again right off the bat you can see here That and it goes to four and the I'll buy you that relates to a F. Sub shell is equal to three. Now moving on to the M. L. Value again ranging from negative L. Two positive L. In our case we said LF 33 and that should be from negative three to positive three. And again drawing that out we have negative three negative two negative one zero for positive one positive too and positive three. So again let's see here we have one two three or fine six and 7 orbital's so seven times two is equal to 14. So the maximum number of electronic fit in this orbital's is going to be 14. And that is our answer for the third part of this question now last one let's go ahead and do that. I'll go ahead and scroll down. Alright the last one were given six gee. Alright so now our quantum number is going to be equal to six. And the L. Value that relates to a G. Sub shell is going to be equal to four. So now finding the M. L. Value Again ranging from negative L. two positive l. We said our L. E. goes to four. So the new range is going to be from negative for two positive Again drawing that out we have negative four negative three negative two. Yeah one zero positive one positive too Positive three and positive four. Let's go ahead and help those orbital's 12, three, four, 56, 7, 8, and nine, And again nine times 2 it goes to 18. So the maximum electrons can fit here. It's going to be 18, and that is going to be our last answer for the last part of this question.