What does this even mean? (energy levels of orbitals)

[pfaction]

I got this during a Princeton Review passage earlier today. I kind of understand it, but then I got this:

Bond order is defined by the expression 1/2(# of bonding electrons – # of anti-bonding electrons)
A bond order of zero indicates an unstable molecule or ion. A higher bond order means a shorter bond length and a greater bond dissociation energy. If a molecule has unpaired electrons, then it is paramagnetic, as opposed to diamagnetic.

An ion of O2 2+ would be predicted by this theory to have a bond order of:

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So, O2(2+) = 8*2 -2 = 14 electrons.
So, going up: if each orbital is 2 electrons...I got 4? But the answer is no, bond order is 3.

Therefore (filling orbitals from bottom to top), the s1s, s*1s,s2s, s*2s, p2p, and s2p orbitals would be filled.

How does this even work?! What is going on!!
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[MedPR]

It's molecular orbital theory and I don't think you have to know it for MCAT. You probably very briefly touched on it during gen chem 1. O2 2+ would have 10 electrons, not 14. O2 has 12, 12-2 = 10.

So for every bonding orbital you have an antiboding orbital. start at 2s and fill that first (2 electrons). Then put 2 electrons in the 2s* (*=antibonding orbital). You have 6 electrons left to account for. Next is the 2p orbital (there's 2px, 2py, and 2pz) After you fill all of those, you have 0 electrons left. So you have two in 2s, two in 2px, two in 2py, and two in 2pz. That means you have a total of 8 bonding electrons. You also have two in 2s*, so you have 2 anti-bonding electrons.

1/2(8-2) = 1/2(6) = 3.

Edit: If the answer isn't 3, then ignore me. I'm not exactly sure I remember this either.

 

[pfaction]

Hmm, I'll go over what you wrote. Here is what TPR says:

Explanation:
C. Oxygen (1s2 2s2 2p4) has eight electrons, so O2 2+ has 2 × 8 – 2 = 14 electrons. Therefore (filling orbitals from bottom to top), the s1s, s*1s,s2s, s*2s, p2p, and s2porbitals would be filled. The two filled s* orbitals represent 2 × 2 = 4 antibonding electrons, so there are 14 – 4 = 10 bonding electrons. According to the equation given in the passage, then, the bond order is (1/2)(10 – 4) = 3.

EDIT: After reviewing what you said and what they said, I still have no idea what's going on.
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[Dasypus]

Check this out:



As the OP stated, there are 14 electrons available in the system. You fill from the bottom up, and keep things unpaired at first when there are two equivalent energy-level orbitals available. I numbered where each electron goes in my image.

Now, how many of those electrons are in orbitals for which the matching anti-bonding (*) orbital is unfilled? Numbers 9 through 14, or 6 electrons. Bond order is that number of un-cancelled-out electrons divided by two, which gives the answer: 3.

Edit: thanks!

 

[MedPR]

Check this out:

http://i.imgur.com/z7Ddr.png

As the OP stated, there are 14 electrons available in the system. You fill from the bottom up, and keep things unpaired at first when there are two equivalent energy-level orbitals available. I numbered where each electron goes in my image.

Now, how many of those electrons are in orbitals for which the matching anti-bonding (*) orbital is unfilled? Numbers 9 through 14, or 6 electrons. Bond order is that number of un-cancelled-out electrons divided by two, which gives the answer: 3.

Edit: can you tell me how to make that image visible in-post?

use [i.m.g]whateverimageyouwant.jpg[/i.m.g] remove the periods from img.

 

[pfaction]

Wow, this is awesome. Thanks! I didn't understand the pi orbitals got 2 of them.