Place the following molecules and ions in order from smallest
to ...

Question

Place the following molecules and ions in order from smallest
to largest bond order: N22+, He2+, Cl2 H2-, O22-.

Accepted Answer

hey everyone in this example, we need to consider the given compounds and arrange them in order of decreasing bond order. So we should recall that to calculate bond order. We would take one half times our number of electrons in our bonding molecular orbital and then subtracted from our number of electrons in our anti bonding molecular orbital. So beginning with our first compound were given B two plus. We want to calculate the total number of valence electrons for this ion. So we would find beryllium on our periodic tables in group two A on the periodic table corresponding to two valence electrons because we were called at the group number and valence electrons are related. So we would have two atoms of berlian. So we would have two times and let's just make this clear Two times two electrons. We also want to recognize that we have this plus one caddy in charge, which we recall means we lose one electron. So that would be two times two which would give us four electrons. And then we would say -1 electron based on that Plus one candy in charge, which would give us a total of three electrons for our molecular orbital configuration. And so what we would have is because beryllium is in the second period or across the second period of our periodic tables. We would begin at the sigma to s bonding molecular orbital for configuration where we would fill in a total of two electrons moving forward in our molecular orbital configuration. We would have the sigma anti bonding two s molecular orbital where we would fill in only for one electron since we have three electrons total for this Or for the valence electrons in B two plus or plus one. Sorry. And so now we can calculate bond order By taking 1/2 and then we're going to count the number of electrons in the bonding molecular orbital. That's the molecular orbital without the asterix. So that would be this portion of our molecular orbital configuration where we would have a total of two electrons here. So we have two electrons in the bonding molecular orbital for B E plus one. And then we would subtract the number of electrons in the anti bonding molecular orbital, which is our molecular orbital with the asterisk here where we would count a total of just one electron. So that's one electron in the anti bonding Molecular orbital. And so what we would get is 2 -1, which would give us one. And so we would have a bond order equal to a value of one half times one, which is one half. And so what we're going to do is continue to follow these same steps to determine which of these Or what would be our arrangement for decreasing bond order based on our given molecules. So, our next given molecule is the helium di atomic molecules HE two. We want to begin by calculating its total valence electrons. So we would find helium in group eight a or noble gas group on the periodic table, with the atomic number two corresponding to its two valence electrons. And because we have two atoms we would have two times two electrons giving us a total of four electrons. So for our molecular orbital configuration we would recognize that helium is across the first period of our periodic table. So we would begin at the sigma one s bonding molecular orbital where we would fill in the first two electrons and then moving forward in our configuration, we have two more electrons to fill in. So we would go to the sigma anti bonding one S molecular orbital. We would fill in our next two electrons. And so now we can calculate the bond order for a G two. So we would have one half multiplied by our electrons and our bonding molecular orbital which is our molecular orbital without the astro here where we have two electrons filled in and then minus our number of electrons in our anti bonding molecular orbital. Where we have the asterisk where we filled in two as our exponents that's corresponding to two electrons in our anti bonding molecular orbital. So we would have to minus two which is zero and then one half times zero, which would give us a bond order for 82 equal to zero. So we have a smaller bond orbital so far for the helium di atomic molecule and now we want to go ahead and consider F two And so again calculating total valence electrons. We would recognize that flooring on our periodic tables is in group seven a. So we would have Two of our flooring Adams Times seven electrons which would give us a total of 14 electrons total. So writing out the molecular orbital configuration for F two. We would begin with the bonding sigma to us molecular orbital because florian is across the second period of the product table where we would fill in our first two electrons moving up in energy, we would have the anti bonding sigma two s molecular orbital where we would fill in our next two electrons. We then move up in energy to our sigma two p bonding molecular orbital filling in two more electrons. Next we have our pi two p bonding molecular orbital where we would fill in for a total of four electrons because we would have to orbital's here. The first one being P X. And then the second one being pY or PZ is the first one, and the second orbital would be P Y. We then have our higher up in energy our pie asterix anti bonding to p molecular orbital where we would fill in four more electrons to complete our 14 valence electrons for the F two molecule. So now we can go ahead and calculate the bond order of F two. So we would have one half times the electrons in the bonding molecular orbital's which are our molecular orbital's from our molecular orbital configuration without the asterisk. So we would count too, We would count four 56, 7, 8. So eight electrons in our Bonding molecular orbital's of F two minus our electrons in the anti bonding molecular orbital which were from our configuration. The orbital's with the asterix. We would count too. And then 4567 or rather six. Sorry. So we have six electrons in our anti bonding molecular orbital hair because we have plus six here circled. And so what we want to do is first take 8 -6. That would leave us with two and then we would have one half times two which would give us a bond order for F2 equal to a value of just one. And so now what we have to do is continue on with our next molecule here, which is our 02 plus one cat ion molecules. So we have 02 with a plus one charge. We want to count its total valence electrons. So, finding oxygen on our periodic table, we see that it's in group six A corresponding to six valence electrons. So we would have two times six electrons which would give us 12. However, because we have this plus one charge, this means that we would lose one electron so we'll circle that in red For -1 electron based on that plus one caddy in charge, which would leave us with a total of now 11 electrons for our configuration. And so for our molecular orbital configuration we're going to begin because oxygen is across the second period of our product table at the sigma two s bonding molecular orbital where we'll fill in our first two electrons. Moving up an energy. We have our anti bonding sigma two s molecular orbital filled in with our next two electrons. We then have our Sigma two P bonding molecular orbital filled in with two electrons. Then we have higher up and energy are Pi two p bonding molecular orbital filled in with four electrons. And then we would move up an energy to our pie anti bonding to p molecular orbital where we would only fill in one electron to get to our 11 electrons total for our configuration of 02 plus Or for our molecular orbital configuration. And so now we want to go ahead and calculate the bond order for 02 plus. So we would have one half multiplied by the electrons in the bonding molecular orbital which again were are orbital's our molecular orbital that did not have asterix. So we would have here two electrons plus two more four and then we would have four plus four which is eight. So this would give us eight electrons minus our electrons in our anti bonding molecular orbital which were the ones with the abstract so we would count 12 and then three. So we have three electrons here in our anti bonding molecular orbital. We have eight minus three which leaves us with five and then one half times five which would give us a bond order of five halves. And so now we just have one more molecule to draw out according to the prompt, that was our S two molecule. So we're going to calculate its total valence electrons. So we have us to we find sulfur on our periodic table in group six. A and so we would have two of our silver molecules here, multiplied by six electrons to give us a total of 12 electrons total. And so to write out its configuration, we would begin because sulfur is across the third period of our periodic tables. We would begin at the sigma three s bonding molecular orbital where we would fill in our first two electrons. Moving up an energy, we have our sigma anti bonding three s molecular orbital filled in with two electrons. Further up in energy we have our sigma two p bonding or sorry, sigma three p bonding molecular orbital filled in with two electrons. We then have our pie three p bonding molecular orbital filled in with four electrons. So we just have a total right now of 10 electrons filled in. We need two more for our 12 total electrons. So we would move up in energy to our pie anti bonding three p molecular orbital where we would only fill in two electrons here and now we can go ahead and calculate our bond order for the S two molecule. And so we would have one half times the electrons in the bonding molecular orbital's which were the ones without the asterix. So we would count 12, 34. We would have four plus four which is eight. So this is eight electrons in our bonding molecular orbital, minus our electrons in the anti bonding molecular orbital which were the one in the or with the asterix in our configuration. So we would have 12 and then 34, so minus four electrons in our anti bonding molecular orbital. So 8 -4 would leave us with four. We would have one half times four which would give us a bond order equal to a value of two. And so now that we have all of our bond orders, we can determine our ranking. So going back up we have for our B. E two or plus one carry on a bond order of one half. And comparing that to all of our other bond orders, we can say that the lowest bond order was from our oh two plus one cat eye on where we had a bond order of five half. So this was our lowest bond order. And then for our highest bond order we got a value of two for our sorry that's backwards there. So here this would be five house would give us our highest bond order. And then for our lowest bond order We have a value of zero Which came from our H. E two molecule. So this would be our lowest bond order. And we want to list again according to the prompt, our bond order in order of decreasing bond order. So because we know that 02 plus one carry on was a. Or yielded the highest bond order. We can begin that first in our ranking. So we have 02, make sure everything is visible. Oh two plus one has a bond order. That was greater than the next highest bond order. Which came from our S. Two molecule here. So that would have the second highest bond order. And then our third highest bond order came from our f. two molecule which had a bond order equal to one, so that's third in ranking. And then next we have from our first calculation for B. E. Plus one as a cattle and we had a bond order of one half. So we would have that next up in our ranking. And then for our lowest bond order as we stated which gave us a value of zero. We have our H. E two molecule which had a bond order of zero. And so for our final answer, we have this ranking here which is an order of decreasing bond order. And so what's hided in yellow here. Our ranking is our final answer. To complete this example. I hope that everything I explained was clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.

Place the following molecules and ions in order from smallest to largest bond order: N22+, He2+, Cl2 H2-, O22-.

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