Apply molecular orbital theory to predict if each molecule or ion...

Question

Apply molecular orbital theory to predict if each molecule or ion
exists in a relatively stable form.
a. C22+
b. Li2
c. Be22+
d. Li22-

Accepted Answer

hey everyone in this example, we need to determine whether the given molecule or ion is going to exist or not according to molecular orbital theory. So what we should first recognize in part a is that our Adam helium is located in group two a of our periodic table. And in this helium molecule we have two atoms of helium. So we would recall that our group number corresponds to two valence electrons and we're going to multiply this by our two atoms of helium. To give us a total amount of four electrons from each of our helium atoms. We're going to use this information to go ahead and fill in a molecular orbital diagram where we would begin at the lowest sub level where we have our bonding orbital's at sigma s. We also want to go ahead and fill in our anti bonding molecular orbital which is represented by sigma astrid s. So because we know we have four electrons total, we want to start off at the lowest energy level, which is our s sublevel and we would go ahead and fill in our first two electrons here honoring our principal in Huntsville. And then we have two more electrons contributed from our other helium atom which we will fill in our anti bonding molecular orbital. I'm sorry. That should be a upward arrow here Now. In order to determine whether our H E two molecule exists or not. We're going to recall that. We can use bond order to determine that. And we would recall that bond order is found by taking one half multiplied by the difference between our bonding electrons minus our non bonding electrons and so based on example a we would say that our bond order is equal to one half multiplied by our two bonding electrons which we find in our sigma s bonding elect electrical orbital. Or sorry, molecular orbital. And this is subtracted from the two electrons that we find in our anti bonding molecular orbital. And so this gives us a value equal to one half times zero, which gives us zero. And so because we have a value of zero for our bond order, we would say therefore HD two does not exist so we can go ahead and box this in as our first answer here for part A and move on to part B. So in part B We want to recall that beryllium is also in group two a of our periodic tables corresponding to two valence electrons. However, we should recognize that we have a two plus charge. And so this means that overall we still want to multiply by our two atoms of our beryllium to give us four electrons total. But because we have that plus charge, we're going to therefore subtract two electrons to give us a total of two electrons total. And so we're going to use this total to draw in our molecular orbital. So again, starting with our sigma s molecular bonding molecular orbital, we would go ahead and fill in these two electrons and this would actually complete our molecular orbital diagram for B E 22 plus. So now we would go ahead and find our bond order where we take one half multiplied by the number of bonding electrons, which in this case is two electrons subtracted from the number of anti bonding electrons, which according to our diagram we have none filled in. So that would be minus zero and this is going to give us a value of one half times two. And so we're going to get a value of one. And so therefore because we have our bond order of one hour given ion B E 22 plus does exist. And so this would be our answer for part B of this example. So moving on to part C, we have the atom B 22 plus. So we would look for boron on our periodic tables and see that it's in group three A Corresponding to three valence electrons, which we will then multiply by the two atoms in our molecule here to give us a total of six electrons total. However, we do have another two plus caddy in charge. And so therefore we would go ahead and subtract two electrons because a positive charge we recall means we lose two electrons and this would leave us with a total of four electrons total. And so we would go ahead and draw in our molecular orbital diagram first. Starting out at the lowest energy level here where we have the sigma s bonding molecular orbital. We would then fill in our sigma asterix as molecular orbital. And so what we should have is first our bonding molecular orbital filled in according to the hunt's rule an alfa principle, leaving us with two electrons left to fill into our anti bonding molecular orbital where we would fill in appropriately. And so now that we have our four electrons from our B 22 plus ion filled in our molecular orbital diagram, we can go ahead and calculate our bond order to give us one half, multiplied by the difference between our bonding electrons in which we filled in two bonding electrons here. So we would have two electrons minus our number of anti bonding electrons. Where again we have two electrons giving us two minus two which is zero and then one half times two which is going to give us a bond order of zero and so therefore we would say that the B 22 plus Catalan does not exist. And so this would be our answer for part C. Of this example. And next we have part D where we have B2 as our molecule. So we again want to recall that boron is in group three A. Of our periodic table corresponding to three valence electrons and we have two atoms. So we're going to multiply by two atoms to give us a total of six electrons. Now this will stay as six electrons total because thankfully this is a neutral molecule. We don't have any charges. And so we would go ahead and draw in our molecular orbital diagram. Starting off with the lowest energy level for our bonding molecular orbital sigma s And then above that we would fill in our sigma anti bonding molecular orbital. So following Hunt's rule and all spouse principle, we would recall that We fill in our first two electrons here and I'm just going to make more room actually because we have a total of six electrons. So we would fill in our first two electrons in the bonding molecular orbital. Then we would go ahead and fill in our Next two electrons in the anti bonding molecular orbital. And so far we fill the total of our of four of our six total electrons. So after our anti bonding molecular orbital sigma gastric esc, we have the pie p orbital which is our bonding molecular orbital and we should recall that R P orbital contains a total of two slots where we would fill in according to our Hunts rule Our last two electrons. And so this gives us a total of six electrons filled in our molecular orbital diagram and now we can go ahead and find our bond order, which we recall is found from taking one half multiplied by the difference between our bonding electrons in which we can count to here in our sigma s molecular bonding orbital and then another two electrons in our pie bonding molecular orbital. So that would give us a total of four bonding electrons subtracted from the number of anti bonding electrons in our diagram where we have near sigma asterix s our two anti bonding electrons. So we would subtract two of our anti bonding electrons and what we would get is one half times two, which is going to give us another bond order equal to a value of one. And so therefore we would say that our given molecule B two does exist. And so for our final answer, this would be our final answer to complete this example. So I hope that everything that we reviewed was clear. But if you have any questions, please leave them down below and I will see everyone in the next practice video.

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