Assign a set of four quantum numbers to each electron in oxygen.

Video transcript

Welcome back everyone. We need to assign N L, M L and M S values to all seven nitrogen electrons. We need to begin by writing out the electron configuration of nitrogen. So utilizing our periodic table here and I'll make it a bit bigger. We go through our one s two sub level recall that R. S sublevel can hold a maximum of two electrons or rather R. S orbital because it only consists of one orbital. Then we go through our two s two sub shell, so that's our next part of our configuration. And then we hit our P block where we see nitrogen and we can count a total of three units until we hit nitrogen, meaning that we would next have to pee because this is our P block and then we have an exponent of three because those three units to get to nitrogen represent our three electrons that we fill in our p orbital's. So note that as we stated, our s orbital holds a maximum of just two electrons in the single orbital. And our P sub level consists of three orbital's represented by these boxes which can hold a maximum of six electrons where we would have to paired up in each orbital with opposite spins. So with this outlined, we next need to assign our quantum numbers. So beginning with n. Recall that N represents our principal quantum number and I'm using shorthand here and sorry, just so everything is visible. This is and so far Recall that for N our values can range from positive one. So only positive integers all the way up to infinity. And this is given by the energy levels in our configuration. So for our first two electrons in our s sub level we have the principal quantum number being our first Level and So N is going to equal one. So we'll place a one here and this represents our first two electrons. So we have to one so far for our end values. Next we have our L values to give which represent our angular momentum quantum number. I'm not going to write that in since we are short on room but recognized that RL values are determined from the range of 012 and on and we can find it by taking N our principal quantum number minus the value one. So for our first two electrons we would have n being one here -1. And that would give us our first two l values as zero. And just so things are clear. We have N in pink And -1 to get our L values and we'll make l red. So let's keep going for the rest of our quantum values. We have next our end value being too for the two s sublevel. So we'll fill that in accordingly where we have two electrons filled there. So we have 22 so far and actually I'll change the color to purple and then notice that we have our next sublevel being R two P sublevel which also has a principal quantum number of two and we fill that in twice for our or rather three times for our three electrons that fill in that two p sub level. So we should have a total of five values of two for our principal quantum number representing the five electrons that we filled in our two S and two P sub levels. Now figuring out the L values, as we stated, that would be n minus one. So in this case we have also an L value equal to zero. And that is for these two purple principal quantum numbers of two. Because notice that we are still in our s sublevel and we want to recognize that for the S sublevel we should always have an L value of zero as we've shown for these electrons. Now when we get to two p sublevel, we are in our p sub shell meaning that we will now have a different value where whenever we have the p sub shell, R L value is going to be the value one. So just to take note our values for the s orbital's are going to equal zero and R L values for the p orbital's equal one. This is always true And this is why we have the range of N -1 which can range from zero all the way to infinity for L as we've shown for our s and P orbital's here for R L values. So now let's move on to our ml values, recall that these are given from the range of negative L two positive L. So based off of our L values, we have negative zero which is just going to be zero again negative zero again. So zero. Then for our third electron we have also another L value of zero. Sorry an ml value of zero and then we have another L value of zero corresponding to an ml value of zero. Now, when we hit this fifth electron here we have an L value of one. We know the range for Ml is negative to positive L. So we would begin with an ml value of negative one here and counting up the range to positive L. We have a L value of one meaning we would then go to zero and then we have our final electron with the L value of one. So we have an ml value of positive one plus one. Because as we stated again, the range is from negative L two positive L. So now moving on to our last quantum number set, which is our M. S values, we want to take note that our M. S values are special because our M. S values do not depend on our other quantum numbers. So the way that we want to think about these values is plus or minus one half with regard to the direction of our electron and how it's filled in the orbital. So either an upward spin or a downward spin. So just to make things extra clear, I'm going to note that our first two quantum numbers are for the s orbital as well as our following two quantum numbers. These are both us and then we have our p orbital's here with this in mind recognize that because we know that the s orbital only has a maximum of just one orbital that can hold two electrons max recall that for the magnetic spin quantum numbers, M sub S. When we have the same level orbital, the electrons will be filled in singly first before pairing the electrons doubly with the opposite spin. And so in other words for this first orbital we would have an Ml or sorry an M. S. Value for our upward electron arrow being positive one half. And then for the downward electron arrow we would have the next M. S value be negative one half. Then moving on to our two S sub shell, we have still the s orbital meaning that we will follow that same arrangement. Where we have first a positive M. S value for the upward electron filled in. And then for the next electron in our s sub shell, sorry, R two S sub shell to be specific, we have the downward spin being negative one half. For RMS value. Now take note for our p sub level we have a different case because we have a total of three orbital's now and so recognize that the electrons will be filled in singly in each orbital first before being paired. Doubly so in this case we would have all positive one half as R M. S. Values for the last three electrons. So we would have plus one half for each of these M. S. Values for our two P three electrons Now notice that we could have also began by filling in our electrons with a downward spin, meaning that we have two options for M. S. values. This would be option one and we can say or for option to our M. S values and I'm going to take away this periodic table so that things are a bit clearer. So we can say or for our second option of M. S values, we can just do the opposite where we filled in with a downward electron first. So we would have negative one half. Then we would have for the next line positive one half for an upward spin. And then following suit for our third electron in the two s sub shell, we would have a downward spin first, so negative one half. Then for the second electron we would have positive one half and use opposite colors. So let's make this one pink half and then Positive 1/2 for the upward arrow again for our fifth electron in the two piece sub shell we would have instead of filling them all with upward spins, we would imagine filling them with downward spins singly first. And so we would have all negative one half values instead for our M. S magnetic spin quantum number values. And so to complete this example, we can define that our final answers for each of our quantum numbers M n l m l M s. We have two options for M s M. S one and S two, or going to be all of the highlighted yellow values that I've outlined, which represent our seven electrons that make up our configuration of nitrogen. So each of our answers, highlighted in yellow, will correspond to choice C in the multiple choice as the correct answer. So I hope that this made sense. And let us know if you have any questions.