For the table that follows, write which orbital goes with the quantum numbers. Don't worry about x, y, z subscripts. If the quantum numbers are not allowed, write 'not allowed.' n l ml Orbital 2 1 -1 2p (example) 1 0 0 3 -3 2 3 2 -2 2 0 -1 0 0 0 4 2 1 5 3 0

Video transcript

Hey everyone we're asked to identify the orbital that corresponds to the set of quantum numbers in the table below the X. Y. Z. Sub scripts are not required if the set of quantum numbers is deemed impossible. Right? None First let's go ahead and define these symbols. So for N as we've learned this is going to be the energy level in orbital's And this value could be any positive integer starting from one now R. L. Which is our angular momentum quantum number. This tells us the sub shell or the sub level. And as we've learned, our angular momentum quantum number has to be at least one less than our end and it ranges between the values of zero up to N -1. Now for ml this is going to be our magnetic quantum number and this represents an orbital for the sub shell and this value is going to range between negative L. Two positive L. So now let's go ahead and answer our question. Starting with our first one we have an end of five with an L. Of two and an M. L. Of negative two. As we've learned an L. Of zero means rather S sub shell. An L. Of one means we're at our p sub shell an L. Of two means we're at our d sub shell and an L. Of three means we're at our f sub shell. So looking at our first one we have our 5D orbital since N is our energy level. And L tells us that we're at our d orbital since we have an L. Of two. Now looking at our second one, we have an end of three and an L. Of two. This tells us that we're at our three D orbital. Next looking at our third one, we have an end of two and an L. Of two. Now the answer for this is going to be none since it is impossible. The reason why it's impossible is because our L. Has to be at least one less than our end. And since our N N. R. L are both two, this means that this is impossible. So this will be none. Next looking at our fourth one, we have an end of zero and an L. Of one. Again, this is going to be none because R L has to be at least one less than our end. Next looking at our fifth one, we have an end of three and an L. Of one. So that means we're at our three P orbital. Next looking at our 6th 1, we have an end of two and an L. Of one. This means we're at our two P orbital and lastly were at an end of one and an L. Of zero. This means that we're at R one S orbital and this is going to be our final answers. Now, I hope that made sense. And let us know if you have any questions