# Molecular Orbitals and Triplets

Discussion in 'MCAT Study Question Q&A' started by SuperSaiyan3, 07.02.09.

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1. ### SuperSaiyan3

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I have this question and the answer explanation which does not seem to make any sense to me. Can anybody explain please? You would be my hero.

"Given the order in which orbitals are filled, which one of these would be a triplet in its ground state?"

- H2, O2, N2, F2

The answer to this question is found in the passage:You are given the order of orbital filling in paragraph three and are told in the last paragraph what a triplet state is. Remember that when there are orbitals of equal energy, each one will be half-filled before any of the orbitals is com-
pletely filled, and all the half-filled orbitals will have electrons with the same spin. Looking at the answer choices for the question, you can see that they are all diatomic; thus, there will be two electrons in a hydrogen molecule, fourteen in nitrogen, sixteen in oxygen and eighteen in
fluorine. The passage does not describe how to fill the n = 1 shell, but from knowing that the n = 2 shell has the sigma 2s bonding orbitals and one sigma 2s antibonding orbital, there will be sigma bonding and antibonding orbitals in the n = 1 shell, and these will also hold four electrons. Therefore, molecular hydrogen will fill half the n = 1 shell with the total of two electrons that it has and there will be zero electrons in its second shell. Working in a similar way you will have ten electrons in the second shell for molecular nitrogen, twelve for oxygen, and fourteen for fluorine. When you fill in the molecular orbitals, you will find that nitrogen just fills the pi 2x and 2y bonding orbitals. Since there are no unpaired electrons, molecular nitrogen cant be a triplet. Molecular oxygen has two extra electrons, which go into the pi 2x and 2y antibonding orbitals. In other words it has unpaired electrons with the same spin. If youre not sure, check the final choice: In fluorine, these pi 2x and 2y antibonding orbitals are filled with the two extra electrons, so fluorine is not a triplet. Thus, it turns out that the oxygen molecule is a triplet in its ground state, so choice B is the correct answer.

The Triple State is considered you have 2 unpaired electrons in separate orbitals.

The bolded is what I do not understand. I understand that when you fill up 1S and 2S, you'll have 4 electrons filled into the orbitals. But the 2p orbitals (x, y, and z) can only fill up 6 more, so I don't understand WHAT they mean when they say that you can JUST FILL IN 2x and 2y bonding orbitals of nitrogen.

Thank you!

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buuuuuuuump

4. ### minutemen11 7+ Year Member

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this is a kaplan FL #1 question, i just took it a week ago! when i looked at this question i had no idea what the heck the were asking, i chose O2 because I knew that O3 is stable and moved on... and i dont really get the explanation either

5. ### SuperSaiyan3

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ya this is the kaplan FL #1 question...

i thought the passages in that PS section was ridiculously hard.

DOES ANYBODY UNDERSTAND THIS THING!?!?!

6. ### neil100anti-dummies 2+ Year Member

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Yes but it is based on a passage. It would be easier to help you if you actually post the passage or the part that is relevant to the question.

The o2 has unpaired electrons in the 2x and 2y antibonding orbital so it is a triplet. This is what it means to be in the triplet state. It has three possible ways to align itself in those antibonding orbitals. Again post what the passage says about a triplet state.

7. ### SuperSaiyan3

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yessir here it is!

It might have come out a bit messed up from the copy and paste...

Every atomic orbital contains plus and minus regions,
defined by the value of the quantum mechanical function
for electron density. When orbitals from different atoms
overlap to form bonds, an equal number of new molecular
orbitals results. These are of two types: &#963; or &#960; bonding
orbitals, formed by overlap between orbital regions with
the same sign, and antibonding &#963;* or &#960;* orbitals, formed
by overlap between regions with opposite signs. Bonding
orbitals have lower energy than their component atomic
orbitals, and antibonding orbitals have higher energy. The
electron pairs reside in the lower-energy bonding orbitals;
the higher-energy, less stable orbitals remain empty when
the molecule is in its ground state.
A benzene ring has six unhybridized pz orbitals (one
from each carbon atom), which together from six molecu-
lar &#960; orbitals, each one delocalized over the entire ring. Of
the possible &#960; orbital structures for benzene, the one with
the lowest energy has the plus region of all six p orbital
functions on one side of the ring. The six electrons occu-
pying the orbitals fill the three most stable molecular
orbitals, leaving the other three empty.
Molecular orbitals are filled from the lowest to the
highest energy level. The number of bonds between atoms
is determined by the number of filled bonding orbitals
minus the number of filled antibonding orbitals; each anti-
bonding orbital cancels out a filled bonding orbital. For a
diatomic molecule, orbitals in the n = 2 energy level are
filled as follows: &#963;2s
, &#963;*2s
, &#963;2pz
, &#960;2px
and &#960;2py
(equal in
energy), &#960;*2px
and &#960;*2py
(equal in energy), &#963;*2pz
. (The
designation of the three p orbitals as px, py, and pz
are inter-
changeable.)
Absorption of a photon can raise an electron to a higher-
energy molecular orbital. The excited electron does not
immediately change its spin, which is opposite to that of the
electron with which it was previously paired. This singlet
state is relatively unstable: the molecule may interact with
Alternatively, there may be a change in spin direction some-
where in the system; the molecule then enters the so-called
triplet state, which generally has lower energy. The
molecule now cannot return quickly to its ground state,
since the excited electron no longer has a partner of opposite
spin with which to pair. It also cannot return to the singlet
state, because the singlet has greater energy. Consequently,
the triplet state, which has two unpaired electrons in sepa-
rate orbitals, is long-lived by atomic standards, with a life-
time that may be ten seconds or more. During this period,
the molecule is highly reactive.

8. ### SuperSaiyan3

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As I've said here, I do not understand how you'd be JUST able to fill in the 2P(x) and 2P orbitals of nitrogen.

Correct me if i'm wrong, but:
- each orbital can only hold 2 electrons of opposite spins, and every single orbital must be filled once before filling it with the electron of the opposite spin.

I think I am substantially confused with the concept of "antibonding orbitals." Do you have electrons filling up these orbitals as well? From what the answer to the passage said, it seemed like you filled up the bonding orbitals of 1s and 2s (because you used up 4 electrons to fill these babies up) without filling up the antibonding orbitals of 1s* and 2s*.

So when you have 10 electrons left to fill JUST 2Px and 2Py, how would you even be able to do this without filling up first 2Pz? and what about the antibonding orbitals?

Sorry for the longggggggggggggg post but I think everybody could benefit from this if somebody had the COHONES to take on this monster.

Thanks!

9. ### neil100anti-dummies 2+ Year Member

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Okay here goes it!

This is molecular orbital theory which you do not need to be familiar with hence it is given in the passage.

Molecular orbital theory is a way to think about the bonds in compounds in a more fundamental way than the VSEPR or other theory. It explains where the electrons are in the orbitals in the bonds and how many can be present in a given state.

so, onto O2, F2, and N2.

Here is the plot that is given. Note you have to be given this. Because there is this phenomenon of orbital mixing that occurs because you get close to each other in energy of the orbitals but it isn't a problem once you reach O2 and beyond.

2pz*

2px* 2py*

2px 2py

2pz

2s*

2s

1s*

1s

For O2 total electrons is 16. so you fill up bonding orbitals first because electrons are involved in bonding. The antibonding is there because of the e-e repulsion between the electrons. You can see this if you took a look at the sigma bond orbital image. There is a region of bonding where there is high electron density and there is antibonding where there isn't high electron density. This is how you get sigma and pi bonding.

So, when you reach to the p orbital

you have filled 2pz, 2px 2py, and parallel spin for 2px* and 2py* orbitals. note the hund's rule still applies of course. so, both are unpaired.

for the other two n2 and f2 you don't find an unpaired electron in seperate orbital as the definition of the passage says hence it ISN'T in the triplet state.

for n2 you have total 14 e. so you will have 2pz, 2px, 2py filled up. others are empty.
for f2 you have total 18 e. so you will have 2pz, 2x, 2py, and 2px* 2py* fill up and only 2pz* is empty. again no unpaired electrons in separate orbitals.
same reasoning with h2 of course since it has total 2 e.

Hence the answer according to the definition in the passage is O2. And indeed triplet state is o2's ground state.

How you like dem cohones??!!!

10. ### SuperSaiyan3

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Neil, I love your cohones. I finally understand it now. I never saw that point in the passage where they indicated that certain orbitals were EQUAL IN ENERGY, meaning that the Hund's rule applied to those 2 orbitals before applying to the orbitals above that energy level. I just assumed that P's x y and z were all the same because that's what I knew.

You are a terrific teacher. Thanks for your guidance buddy. I hope goodwill is headed your way.

SS3

11. ### s2thindi 2+ Year Member

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Is there anyway you could help answer this question (from the same passage):

Which of the following four depictions of molecular
&#960; orbitals represents the highest energy state for a 6-
carbon polyene molecule? (The signs given are the
signs for the mathematical functions defining the p
orbitals on one side of the molecule.)
A.      
B. + + +   
C. + +   + +
D. +  +  + 

I mean I kind of get it... but I'd like to understand it better.

12. ### eric51 5+ Year Member

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The answer is D. Pi orbitals that have the same sign that are next to each other are considered bonding, because the part of the orbitals that are in phase can overlap and interfere constructively. If orbitals that are next to each other have opposite signs, they interfere destructively and are anti-bonding. Having more bonding relationships in an orbital will make it lower energy, so the orbitals with the most anti-bonding relationships will be the highest energy.

13. ### s2thindi 2+ Year Member

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Could you just explain how one would know the ++ and -- mean they are bonding orbitals (constructive vs destructive)? I didn't see that in the passage...

14. ### eric51 5+ Year Member

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I think that might be something you just have to know. I don't see anything about it in the passage either.