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Carleton College
University of Kentucky
University of Toronto
03:09
Shahriar Khan
Suppose that a buffer contains equal amounts of a weak acid and its conjugate base. What happens to the relative amounts of the weak acid and conjugate base when a small amount of strong acid is added to the buffer? What happens when a small amount of strong base is added?
00:48
Eugene Schneider
If energy flows out of a chemical system and into the surroundings, what is the sign of Esystem?
01:49
Kevin Chimex
What is internal energy? Is internal energy a state function?
02:03
If the internal energy of the products of a reaction is higher than the internal energy of the reactants, what is the sign of E for the reaction? In which direction does energy flow?
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So for our next problem will be determining the electron configuration for a variety of different aluminum countdown's and then using this to determine the point at which the aluminum count on experiences largest change in organization energy and if the first part will be figuring out the electron inspiration for owner. So to do this, let's start with the electron configuration for the neutral aluminum. So aluminum is located here, and so we can use the periodic table to determine the electron configuration. So from them we can fill up. They want us to orbit all to us, too. Two P six three us too, and three p. One. And for the next few electron configurations will be gradually removing an electron from the highest energy level shelf and then continuing onward. And so if we look at a limit plus this means that will be removing one electron from the P orbital. And so we have one us too, to us too. Two p six and three s to. And so we removed the electron from the P orbital. So now we just have the filled three ask for both and for a little too. Plus, we are removing two electrons. So we will be removing an electron from the B atomic orbital as well as Theus, Atomic orbital. So now we have one is to to us too. Two p six and three s one. And for the next one, we have over three plus. And so, in this case, will be removing three electrons. And so now we will be removing all of the electrons and the energy level through. So we're left with one. Us too. Thio us too. And to peace. Yeah, finally for alumina, for bless. We will be removing an electron from the to P orbital. And so now we have one us too. Thio us two and tp five. And Senna. This means that we figure out the different ionization energies for each removal of an electron. And so this would be the first organization energy because we removed one electron and this would be the second because we removed an addition electron. And so now we can see that this is the third and this is the fourth Insana that we've determined the electron configuration for all of these different carry ons. We can now determine which one will experience the large change in organization energy. And so do this. We need to understand that when you have filled atomic Orbital's, this means that the Adam for Ion is in a stable configuration, and so if it is stable, that means that you need a large amount of energy to remove the electron. But if you have valence electrons in a partially filled atomic orbital than that means that it is much easier to remove these electrons in comparison to those in filled atomic orbital's. And so those probably won't have as much of a high ionization energy. And so with that, we can see that it is likely that the aluminum will not have the greatest change organization energy, because we're removing an electron from the atomic orbital and so we can rule this one out as well as the aluminum, mainly because you have an electron in the transorbital, and so that's pretty easy. Thio live, um, and saving across that out as well. And now, in terms of the other species, we know that, um, if you remove an electron from the three s to, this would be someone more challenging, mainly because you would disrupt the fully occupied. That's orbital, which is already stable, and so we can leave us alone for now. And we could say the same for a little three plus as well. Because thebe orbital is welcome and ask for one of four plus in comparison to the 1st and 3rd civilizations, we know that this is a partially filled atomic orbital. And so if you remove an electron after that, it would still be pretty high. But in terms of the difference between to be five and two before, this would not be very significant. And so we would cross this out as well. Do not we need to compare the change in ization energies for the three s two and the to be six. And if you create these theologian of three plus must be the point at which you experience largest change organization, energy or, in other words, the removal of the electron from the TV set. And the main reason why this is the case is because the two p six is very stable, and the two s two and to be six actually have a full octet and has the noble gas configuration of neon. And so That means that you are taking away an electron from a very stable configuration. And so if you do this, that means that you need a lot of energy to remove the electron. And so that means that we will probably see the largest spike and the reason why three us too, is probably not, um, in terms off the spike as high as boyfriend three s 2 to 3 s one is because the three s to does not have the noble gas configuration. And so even though it is stable because the atomic orbital is filled in comparison to the octet, it is not as stable. And so for be, our final answer would be the transition from the third ionization energy toothy fourth ionization energy.
Chemical Bonding
Molecular Geometry
Gases
06:52
12:58
