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Pan AListthe values of the tour quantum numbers (n,€,mt,ma) tor each of the electrons in the ground slale dneanEssay answers are limited t0 about 500 words (3...

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Pan AListthe values of the tour quantum numbers (n,€,mt,ma) tor each of the electrons in the ground slale dneanEssay answers are limited t0 about 500 words (3800 characters maximum, including %p3ce ) 3800 Character(s) remanngSubmaBeeVmi

Pan A Listthe values of the tour quantum numbers (n,€,mt,ma) tor each of the electrons in the ground slale dnean Essay answers are limited t0 about 500 words (3800 characters maximum, including %p3ce ) 3800 Character(s) remanng Subma BeeVmi



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Give a possible set of values of the four quantum numbers for all the electrons in a boron atom and a nitrogen atom if each is in the ground state.

Let's start by writing the electron configuration for born. So it's gonna be one of squared to us squared. It's not one electron in the piece 02 p one. All right, now, going off that. Let's make a little chart. So over here you saw that we had one s squared, so there's gonna be two electrons in the oneness. We also had to us squared, so there's gonna be two electrons and then to us, and right here we've got to p one, which means there's one electron, the to P. So we just need to have one of those over here. Now, the quantum numbers are in go. It is about and the magnetic spin. So for the oneness, the in corresponds to this number here. So that's gonna be a one l l a zero corresponds to the S and Ellis. One corresponds to the p. Okay, So since we're in the s, I was gonna be zero here now in its abo Let's go ahead and write out what the quantum numbers could be, by the way. So images an integer, um el is anything from zero to n minus one imsa bell is anything from negative 02 positive l including zero and the MSR The spin is either negative 1/2 for a positive 1/2 spin after spin down Alright, So back to the 1st 1 for looking at IMSA Bill if l If instead all can be negative l a positive l But l zero intimate obviously has to be zero and the spin and we'll just go ahead and shoes spin up for now it could also be spin down either one It doesn't really matter Here we just know that they have to be opposites Mons So doing the same thing you've got basically the exact same quarter numbers except for the spin here needs to be spend down You can't have two electrons in the s orbital that air essential that are both spent up That would cause too much repulsion All right, living on to the to us So in his two or still ness So that means I was Nero. Impossible has to be zero. We'll give this one's been up The other electron, the to us is gonna be identical aside from the spin. Okay, so last one we've got is the to p It is to again l this time we're talking about the piece that I was gonna be one it is about can be either negative one positive one or zero So we'll just say negative one. And you can go and pick Either spin up or spend down on just pick spin up for now. Okay? And that's all we're looking at really? Right now. So let's we want to nitrogen now we're gonna do something pretty similar because that we're gonna art dealer trump invigoration again. So we've got one a square to a square to pee to the three for nitrogen So let's make our chart again. We've got to in the oneness to in the two s and three and the to be to pee 232 Okay. All right. So the inning and is just one Ellis here because we're in the S and development has to be zero and we'll give this one positive 1/2. So that means the next one in the s orbital the oneness is gonna be spend down opposite spent to us. We're now in the ankles too, But we're still in l equals here. Because of this, we'll give one of them spent up and the other one's been down our in the to P and it is still too. But Ellis one so invisible can be negative 10 or positive ones. I'll start with negative one and just go in order. We'll give this one spin up. Now I'll give this 10 Because, remember to electrons cannot have the same set of, um, quantum numbers. I'm gonna give it spent up again. 21 We're still in the peat and positive one. We'll give it, spend it begin. So here you can see that we went through all the options of the insa built to make sure that none of these air the same and right here I chose been up for all of them. That doesn't really matter. It could have been spin up or spend down, because if you remember what that to pee or more looks like it's like this. So in reality, there's six electrons. But we were just talking about one in each of these, so you could have chosen spent enforcement down. It doesn't matter at that point, and that's all for Introgen

Okay so since we have an N. Equals one and it is an SRL value is zero and your EMC Bell would be zero. And then you would have plus or minus one half for your spin for N equals two. We have an S. And a piece. That means you have a zero and a one for your L. Values your possible EMC Bell would be zero for L. Equals zero. And then a negative 10 or a positive one for L equals one plus or minus one half. For your spin For N. equals three. We've got an S. A. P. And a. D. So that means you can have 01 or two for your L. Values. That means um stability zero for L equals zero negative 10 plus one for L. Equals one. And then your options for the L. Equals to be negative to negative 10 plus 14 plus two. Still with your plus or minus one half. Then for your N equals four. You've got an S. And a piece. That means you'll have a zero and a one for your possible L. Values this would be zero or negative 10 positive one. And then a plus or minus one half for your N. Equals five. We only have an S. So these are both zero and then you have a plus or minus one half. Going down to sulfur. We have an N. Equals one of an S. So these are both zero. So the plus or minus one half. Then for your N equals two. We have an S. And a piece. That means you have a zero and a one for your L. Values. Then for your N equals three we have an S. And P. So this is going to match My n equals two. Then moving on to scan the um we have an N equals one with Justin S. So these are both zeros with the plus or minus one half. Then for N equals two we have an S. And P. Then for my end equals three. We have A S, A P and A D. C. This would be 01 and two from my possible L. Values. So we would fill this in accordingly. Then for my N equals four I only have an S. These are both zeros with a plus or minus one half.

So this problem we're going to be working with scan Diem and what we want to do is to write out the quantum numbers for electrons within the Skandia madam. So I'm going to do is give you all of the possible quantum numbers available to each of the electrons within the atom. And you can write those individual configurations out yourself. So the first thing I want to do is to write out the electron configuration for scanned IAM, which is going to be, um, equivalent to the electron configuration for neon and skinny is also going to have a filled three s orbital filled three p orbital. We have one electron and it's three d two electrons, and it's for us. So reason I didn't wanted to write this out because I wanted to emphasize that skin Diem has three electrons or has in the unequal three energy level. Skinny um, has electrons and s orbital's pure metals and D orbital's. So if you write those possibilities out, we already know that for quantum numbers, the end in this case is always going to be equal to um, and it's always gonna be equal to three, because again, we're dealing with the an equal three energy level, so we rent out the possible l quantum numbers for each of these orbital's. We know that l more or less corresponds to the sub orbital's themselves. So for S Orbital's Ellis, always zero for P. Orbital's L. A is always one Fergie Orbital's Ellis Always, too. We can then write out the ML values so the ML values are always going to be plus or minus the the range of negative lt positive l over s Orbital's. There's only one possibility at zero for p orbital's. There's three possible sub shells or three degenerate orbital's, and we represent those using the manager's negative 10 and one and then for D. It's the same thing. We go from negative L through positive l and arrange. So we go from negative to two. And, as is always the case, um, for these electron configurations are for these quantum numbers. The spin on these electrons could neither be plus or minus 1/2. That's going to be true for all of these orbital's, except for D. C. Since there's only one electron within the D orbital, you're actually only going tohave one possible spin there. Um, so these are all of the possible combinations. So the different ways that you can write these out, obviously, or for the three s orbital's, you're only re electron configurations are going to be an equal three. L equals your OML equals zero, and then a spin of either 1/2 or a spin of negative 1/2. And you can take these values. We had them listed here and make the individual configurations for the other sub orbital's. So for the P orbital, for example, one of the electrons is going to be an equal three l of one Emily of negative one and a spin of plus 1/2 on DSO. From this table, you can extrapolate all of the possible combinations again. The only one that's going to deviate is going to be this three d one, in which case only one of the electrons there's only one electrons. Only one of these possible combinations is going to be true.

Okay, so we have to write down all this sets off the quantum numbers. So here we have the principal onto number. Here we have the orbit over one to number. Do you magnetic? I want a number. And this being quantum number, they're corresponds to the electrons in a completely feel for F so show. So first of all, a four f social means that if Greece IPO quantum number maximum is actually the principal hunter, number is always for even maximum. It's always four. Therefore, we can say that to the 14 sets always begins with before in the principle of quantum number. So we just need to calculate order three numbers. So that's speaking with the 1st 1 The I'm good luck. One to a number always is one size lesser than the principal quantum number. Therefore, we can see that all the sets off the angular want a number is going to be three. So all sides of the three simple begins before and all sets off. The orbital begins with three and the list goes on here. Now we have to look to the two magnetic quantum number. So these two are the only ones that change. So, um that brings suppose is going to be four in all. The orbital is going to be three for the 14. So that's right. Only the magnetic and is speaking right now. So we have to right now. 14 authorities. So here we have the Magan attic. So the magnetic has a property that he goes from minus l to l. So if we have 14 of this, we can say that can say that we have to be at least two off each number. So we have true three's then Chu Tu's than the ones Ben Zeros. Let's continue it here. And we have Ah, let's see. Minus one my no swung minus two minus two minus three. Ministry. So this needs to be all the 14 set numbers off the magnetic. So now let's do this. Same for the speed thespian quantum number. And they speak one to number. Has a property. They speak wantto number. Let's put you here. Oops. Sorry. The split I want a number has a property for the electron that he can only be less or minus. Yes. Therefore we have here for the is split until number values off what you have negative have was half then negative half and don't go along. Was chief narrative was negative and wasn't it again? And this is the least off all the 14 sets of numbers here we have Hell, no. All deals are Greece and all the bring suppose our force. What difference it boys for? For for for and then go on. That's all the forking numbers off the sub show. Thanks for watching.


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