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Molecular Orbital Theory Question about bonding/antibonding orbitals

by mayer
Tags: molecular, orbital, orbitals, theory
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mayer
#1
May20-14, 11:10 PM
P: 9
Hi I spotted on an MCAT book I am studying off of that the shading of the lobes(from a figure of p and sp3 orbitals) in an orbital represents the direction of spin for the electron and that in order for the electron density to overlap, the electrons must have the same spin.
Firstly, I thought the shading represents the phase of the orbital, as in +/-. Is spin another way of saying phase? If this is indeed the case, this seems to conflict with the fact that electrons in one orbital must have opposing spins.

Just a little curious, because this one portion of the book completely threw me off.
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dextercioby
#2
May21-14, 12:44 AM
Sci Advisor
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P: 11,952
You may wanna name the book.
mayer
#3
May21-14, 12:48 AM
P: 9
its from berkeley review, ochem part 1, chapter 1, page 11 at the bottom of the page. Not sure if it varies by edition.

DrDu
#4
May21-14, 01:19 AM
Sci Advisor
P: 3,633
Molecular Orbital Theory Question about bonding/antibonding orbitals

If that is what the book sais, it is complete nonsense. Your explanation of the phase is the correct one.
abitslow
#5
May21-14, 04:38 PM
P: 140
No. Electron spin is either + or - and is one of the quantum numbers which characterize the orbital. The full set of quantum numbers (analogous to the atomic Quantum Numbers) 'completely' characterize the orbital (not literally 'completely', since nuclear charge (among other things) isn't part of them). No two electrons may have the same quantum numbers (in the same location) and since spin is the easiest (lowest energy) QN to change, it works quite well to pretend that there is this 'thing' called an orbital which can be occupied by two electrons (as long as they have opposite spin.) Bonding and Antibonding Molecular Orbitals are made up of a LCAO (linear combination of atomic orbitals) and have a 'phase'. The phase can be thought of as the sign of the wave-function ψ. You may recall that ψ is the electron probability density (the probability of a electron being at a particular point in space). But ψψ = (-ψ)(-ψ), so the phase of the wave isn't relevant to where the electron is. You don't want to go into the quantum mechanics to try to understand the reasons why, I think. When you compile a set of LCAO, the 'housekeeping' requires you to have equal numbers of both +ψ and -ψ, so that your MOs are made up of the overlap of +ψₐ with +ψᵦ, +ψₐ with -ψᵦ, -ψₐ with +ψᵦ, and -ψₐ with -ψᵦ. You can consider each 'lobe' of an atomic orbital to have either + or - phase. The simplest example of how this effects bonding is by using two atomic orbitals...say atom A has an s orbital, which has no nodes and is everywhere positive (this is only true for a 1s orbital, but is a great approximation for 2s, etc.) You probably know that the 'shape' of the orbital is a sphere, right? Ok, now lets say atom B has a p orbital. You (hopefully) know that there are 3 p orbitals, px, py, and pz and each is 'shaped' like a dumbell (or figure 8). Well, thats correct, but it is also true that one lobe has + phase and the other -. You'll have to believe me on this (although I simplify the situation). So picture O ⟷ 8 moving together. If you accept that +ψₐ with +ψᵦ results in bonding and +ψₐ with -ψᵦ results in antibonding, then you may be able to see that as the two orbitals overlap, the amount of ++space is exactly the same as the amount of +- space? This results in non-bonding (a non-bonding MO). Same process but now lets change the p orbital to this ∞, where the left lobe is +. As you move them together O ⟷ ∞ the result is a lot of ++ overlap and no +- overlap (unless you shove the p orbital into the O so far that the nucleus is inside the other atom's radius). This is bonding. Antibonding is just as easy to picture, just flip the p orbital end-over-end and approach the O orbital with a negative lobe.
mayer
#6
May21-14, 04:42 PM
P: 9
0.0, ill start reading
mayer
#7
May21-14, 04:50 PM
P: 9
Went a little ahead of the point of the question but it has helped clear out my confusion tremendously, Thanks all of you.


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