Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: LaRealidad on May 28, 2010, 03:17:46 AM
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Hi,
I have a question here that asks me to predict if He2+ exists using Molecular Orbital theory?
Here's an example of predicting the existence of a molecule using MO Theory:
H2 exists as the 2 electrons of the H-H bond occupy the lower energy bonding molecular orbital (aka sigma bonding molecular orbital), where they are attracted to both positively charged nuclei . It is this electrostatic attraction that gives a covelant bond it's strength.
Another is the lack of existence of He2 because each He atom comes with 2 electrons. He2 would have 4 electrons, the first 2 filling the lower energy molecular orbital and the other 2 filling the higher energy anti-bonding molecular orbital(aka sigma* bonding molecular orbital). The 2 electrons in the higher energy orbital would cancel out the effect of the lower energy orbital therefore He2 cannot exist.
Any ideas on the existence of He2+?
Thanks,
LaRealidad
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What is the bond order of He2? of He2+ ?
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Any ideas on the existence of He2+?
Hint: where does the charge comes from?
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@sjb: (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fmath%2F9%2Fe%2Fd%2F9ed22a61a976240fa3cc5fed5b6dfd12.png&hash=8e56d59b77fd62c301bab2d6b0e1adfe23268eab)
Therefore for He2
BO= 2-2/2 = 0
and for He2+
BO= 2-1/2 =0.5
@Borek: I know that there are 3 electrons in this cation, and the charge comes the removal of one electron. I'm guessing that this means that the 3rd electron resides in the higher energy antibonding molecular orbital, is that correct?
Is there a threshold BO above which molecules can exist? Is this threshold just zero? Because if that's the case then I would say that He2+ will exist.
Thanks
LR
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s the removal of one electron. I'm guessing that this means that the 3rd electron resides in the higher energy antibonding molecular orbital, is that correct?
Yes, during ionization electrons of the higher energy are removed first (they need less energy to be removed from the atom/molecule). Assuming energy of bonding electrons and antibonding electrons is identical (which is about right, although not exactly) but different in sign, is there any energy gain when He2+ is created?
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There is a loss of energy when He2+ is created from He2 which is favorable, but isn't it more favorable to just have He on it's own?
LR
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I've finally found my solution manual lol It says:
"He2+ has three electrons. Two are in a bonding orbital and one is in an antibonding orbital. Because there are more electrons in the bonding orbital than in the antibonding orbital He2+ does exist."
From this I assume that:
I have to analyse each molecule on it's own, i.e. I should be able to analyse He2+ without comparing it to He2. (Any thoughts Borek? I'm curious as to what your thought process was in answering this and why you asked me that last question about energy.)
That "threshold Bonding Order" I was trying to figure out must be zero, as "more electrons in the bonding orbital than in the antibonding orbital" is what makes a molecule feasible.
LR
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There is a loss of energy when He2+ is created from He2 which is favorable, but isn't it more favorable to just have He on it's own?
Depends on teh situation. In most cases isolated He are the most stable thing, but in the positively charged He plasma He2+ is more stable than mixture of He and He+. Unless my memory fails.
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I have to analyse each molecule on it's own, i.e. I should be able to analyse He2+ without comparing it to He2. (Any thoughts Borek? I'm curious as to what your thought process was in answering this and why you asked me that last question about energy.)
Think - why do molecules exist? Think about energy of a molecule as compared with energy of separated atoms. Bond order is nothing else but a simplified attempt at estimating energy.
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Thanks, you're answer makes total sense. Correct me if I'm wrong but you're going back to basic chem principles saying that what ever is most stable (lowest energy) will be produced.
Appreciate it.
LR