a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure 9.46, how many MOs would you expect for the HF molecule?

Video transcript

Hello everyone. So in this video we're given this big energy level diagram that's for the atomic and molecular orbital's in N. O. Molecule. So scrolling down for our question, it says that the same models being followed. Count the molecular orbital involved for R I B. R molecule. And the hint is to use only the bands of how make orbital's of iodine and bromine atom. So scrolling down to give us a little bit more space. The first step is to write the condensed or noble gas electric configuration of iodine and grooming. So for iodine the closest noble gas is K. R. And then continuing our electron configuration we have four D 10, 5 S two and five P five. Then for our bromine atom the closest noble gas is a R continuing our electric configuration. It is for us to three d 10 and four p 5. So for our paradigm the highest end value is five. So we have two and 52 plus five. And a total of seven fans electrons for our paradigm Then for bringing the highest end value will be four. So we have two and five again to post five is seven. So burning also has seven valence electrons. and those two numbers up we get a total of electrons for our molecule. Alright, so the second step is to go ahead and draw the molecular orbital diagram. So of course we have this big energy arrow going up in the center will be about R I P R molecule. Over to the left can be about our iodine adam then to the right will be our roaming. Okay, so we started off with our sigma to the right. We have our four s to the left. We have our five s on top will be our sigma start and that's like a diamond shape Frankie. Next set we have our signal right above that. We have our pipe to the right. Let's go ahead and actually try to make this meter. We have our four p. And then on top we have our five p above that we'll have our pie start. And lastly we have our sigma start and again they have to be connected. All right, so let's start off with our iodine atom. Of course we have the seven veils electrons. Let's go ahead and use Hunt's rule growing from the bottom to the top. All right. So we have 12 and seven. And we know that browning has also seven valence electrons will do this on the right. So we have 123456 and seven. Lastly for our molecule. Right over here in the center we have a total of 14 valence electrons. Let's go ahead, fill that out. So we have 123456789, 10, 11, 12, 13 and 14. All right now that we filled it out. The question is asking us how many molecular orbital is that the molecule is using up? So let's go and count that. We have 123456 and seven. So the final answer for this question is seven molecular articles used, and this right here is going to be my final answer for this question. Hopefully this all helped.