# Paramagnetic and diamagnetic species

Discussion in 'MCAT Study Question Q&A' started by growingpains, Apr 14, 2012.

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1. ### growingpains

57
0
Dec 22, 2011
I'm having trouble understanding why molecular fluorine and hydrogen are diamagnetic. In F2, after the fluorine atoms form a single bond with each other, each atom has 6 valence electrons left over, Are 2 of these electrons in the 2s orbital and the other 4 in the 3 2p orbitals? If so, wouldn't the fluorine have an unpaired electron in the 2p orbitals?

3. ### pfaction 7+ Year Member

2,195
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Mar 14, 2010
WV
MDApps:
Diamagnetic: everything is paired.
Paramagnetic: one electron THINK RADICAL is left out.

So F2 = complete 16
BrO is paramagnetic: 7+6 = 13
Fluorine by itself is paramagnetic.

4. ### growingpains

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Dec 22, 2011
Ah, so the bonding electrons are included in both fluorines. Thanks.

5. ### folktale

333
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Jun 30, 2010
California
Can you explain how you got 13 for BrO?

6. ### growingpains

57
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Dec 22, 2011
Br has 7 valence e-, O has 6. 7+6=13

7. ### folktale

333
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Jun 30, 2010
California
Then why would F2 be 16, if both atoms have 7 valence e-

8. ### chiddler 5+ Year Member

Apr 6, 2010
should be 14 i think. fluorine by itself has 5 electrons in the p shell so it is paramagnetic. then you make it F2 so that electron #5 is shared with the other F and the other F shares electron #5 with the first F. Now they both have 6 electrons in their p shell and there is no unfilled orbitals. They are diamagnetic.

9. ### folktale

333
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Jun 30, 2010
California
I think of it as F2 having a total of 18 electrons b/c each F atom will have 9 e- each. Then if you add that up, that's 18. There are 3 atomic orbitals in the p-orbital and 18e- are completely paired up w/ each other.

EDIT: and even if you just count valence e-, it'd be 7 + 7 = 14. And an F2 molecule will have all its valence e- paired F-F. So it is diamagnetic.

Last edited: Apr 14, 2012
10. ### chiddler 5+ Year Member

Apr 6, 2010
yeah that sounds right.

oops @ my numbers.

11. ### folktale

333
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Jun 30, 2010
California
This is a great topic. Let's keep going with other examples.

12. ### growingpains

57
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Dec 22, 2011
Alright let me take another stab at this. CN- has a triple bond and both C and N have 8 valence e- by sharing the triple-bond e-. Thus, with 8 valence e- each, the compound is diamagnetic?

13. ### folktale

333
1
Jun 30, 2010
California
See this is what I'm confused about.

I'm fine with single atoms. Like Li would be paramagnetic b/c it has one unpaired electron in the 2s orbital, and Be would be a diamagnetic because it has the 2s orbital completely filled. But I am somewhat still confused on molecules like CN- or any other ones. Anyone wanna take a stab at this?

14. ### hellocubed

313
1
Jul 9, 2011

I'm not quite sure how you came onto the figure 8

C has 4 valence electrons, nitrogen has 5. They should have 9 valence electrons together. If they had 8 valence electron each they would both be noble gases, and the bond would not be possible.

9e-, and hence should be diamagnetic...

15. ### folktale

333
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Jun 30, 2010
California
But there's an extra e- because CN- is negatively charged, so overall it would be 10e- correct? and how is 9e- going to result to all electrons paired? Isn't a 10e- going to make CN- diamagnetic? 9e- seems like it would have one unpaired electron, only filling two of the 3 2d orbitals, leaving the last one with an unpaired e-.

16. ### hellocubed

313
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Jul 9, 2011
Oh wow I screwed up everything on that response. haha

10 e-, diamagnetic

17. ### growingpains

57
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Dec 22, 2011
When we're looking at paramagnetism/diamagnetism of compounds, aren't we looking at the paramagnetism/diamagnetism of each of the atoms in the compound first? (e.g. If a compound contains all diamagnetic atoms, the compound is diamagnetic, but if the compound contains at least one paramagnetic atom, the compound is paramagnetic).

In CN-, both C and N have full octets (8 e-) via the triple bond. So would they be both diamagnetic, making CN- diamagnetic? Is this the right approach to paramagnetism/diamagnetism of compounds?

18. ### EnginrTheFuture 2+ Year Member

511
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Nov 22, 2011
http://scramlinged.com/resources/Notes+on+Molecular+Orbital+Theory.pdf

You have to count up the total electrons available (not just valence)... then you have to throw it in the MO diagram shown in the link above. I believe this includes antibonding orbital filling which is why people are getting thrown off attempting the individual fill out method. I do not think we are responsible for knowing how to predict paramagnetism/diamagnetism of molecules, just atoms. Although who knows.

I do not think you can just take "half the bonding electrons" or all the bonding electrons and analyze each atom individually. (not sure if thats what people were doing up there, I skimmed). One thing is for certain, if you add up all the electrons in the system and it's not an even number --> definitely paramagnetic. Other than that you have to fill out the orbital diagram or have the numbers of electrons needed memorized. This isn't like doing the usual 1s 2s 2p 3s filling!!! These are bonding orbitals, not the usual individual atomic orbitals. Not only that but there are two forms of bonding theory, the molecular orbital theory and valence bonding theory. This is all explained fairly well in the above link.

If total electrons add up to 2,4,6,8,12, 16, 20, or 22... it's diamagnetic BUT if it's a bond between oxygens or fluorines the orbital filling changes. This gets real nasty and this is why I think your better off sticking to individual atoms and ions.

I could be wrong and please let me know if this is the case... but from what I remember of g-chem years ago and some searching around, this is what I came up with.

Last edited: Apr 14, 2012
19. ### chiddler 5+ Year Member

Apr 6, 2010
for mcat, I don't think we have to know molecular orbital theory, do we?

after all, for something like cyanide, it would take too long to draw out the orbitals and see if it is indeed para or diamagnetic. plus i couldn't find it on the list.

20. ### johnwandering 7+ Year Member

430
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May 18, 2009
You are definitely right
It's not quite as simple as we are making it

O2 and N2 both have even number of valence electrons (Now that I think about it, all real compounds apparently do, or else they would be free radicals. BrO has a free radical, it is supposed to be BrO2)

But O2 is paramagnetic, even though the Oxygen atoms are DIAMAGNETIC.

weird~

21. ### folktale

333
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Jun 30, 2010
California
I still don't see how an Oxygen atom by itself is diamagnetic. it would have 4 e- in its 2p orbital, with one of the 2p orbitals filled and 2 2p orbitals with an unpaired e- each.

And to clarify, do we use the orbitals s p d f for elements/atoms like N and the LUMO/HOMO bonding orbitals for molecules like N2?

22. ### EnginrTheFuture 2+ Year Member

511
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Nov 22, 2011
Check out the link I posted a few up... it should explain everything you are curious about in a fairly easy-to-read 2 page pdf

cheers!

23. ### johnwandering 7+ Year Member

430
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May 18, 2009
Ah, so... It has really not much to do with being "even or odd."

24. ### EnginrTheFuture 2+ Year Member

511
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Nov 22, 2011
well if the total # of electrons in the molecule is odd... Boom, you know there is no way of pairing --> paramagnetic. If it's even number though, things get complicated.

25. ### folktale

333
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Jun 30, 2010
California
Alright I get it now. So my statement holds true. If an atom is in question, you can use the regular 1s2s2p orbitals, but if we're talking about a molecule, then MO diagrams are necessary. I'm just going to remembre the 1s, 1s*, 2s, 2s*, 2py-2pz, 2px, etc. Hopefully, the MCAT shouldn't be this complicated.

26. ### EnginrTheFuture 2+ Year Member

511
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Nov 22, 2011
That should do it for most cases, yup. If anything more complicated comes up... everyone is in the same boat to wtf-town

27. ### folktale

333
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Jun 30, 2010
California
Yep. If something comes up that starts dealing with the d-orbital comes up, I'm gonna mark that question and move on lol. Well, I'd guess first at least. But I'm thinking if they're gonna ask something in this topic, it would be like an atom instead of a molecule. That'd be a great discrete.

28. ### Dasypus

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Apr 25, 2011
Do we really have to know any of this stuff? It doesn't seem to be on the AAMC outline....

29. ### MrNeuro 5+ Year Member

604
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Mar 17, 2010
Danger Zone
we do need to know how to determine if an atom is diamagnetc/paramagnetic

we don't need to know mo theory...although it could show up in a passage where it is explained....(TPR had a passage on this)

30. ### pfaction 7+ Year Member

2,195
35
Mar 14, 2010
WV
MDApps:
Wait, how is O2 paramagnetic?

193
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Apr 25, 2011