I've been reading about the one dimension particle in a box model, and from some websites, I found that it can be applied in a conjugated system. However, the system is for a single particle in a box, so I don't quite get how they assume all the electrons can be taken to be one particle. Anyone has any idea why?
Applications of Particle in a Box model (Chemistry)
Announcements  Posted on  

Complete this short survey for a chance to win an iPad mini!  22092014  
Got a question about Student Finance? Ask the experts this week on TSR!  14092014 


(Original post by Lifeisnice)
I've been reading about the one dimension particle in a box model, and from some websites, I found that it can be applied in a conjugated system. However, the system is for a single particle in a box, so I don't quite get how they assume all the electrons can be taken to be one particle. Anyone has any idea why? 
Yeah, but the application assumes a single particle in a box which causes the conjugated system; the conjugated system is caused by many electrons 2 of each occupying each orbital.

Yes, it's an approximation. You can treat the electron individually and ignore things like electronelectron correlation. It's not a great model for such systems but it is simple.

(Original post by EierVonSatan)
Yes, it's an approximation. You can treat the electron individually and ignore things like electronelectron correlation. It's not a great model for such systems but it is simple.
However, how do you interpret the wave function solution? From what I understand, the square of the wave function has something to do with the probabilities of finding the electron there, but I can't quite grasp how to interpret the graphs properly.
Also, isn't the particle in a box model assuming that each individual electron acts by itself, and is spread over the WHOLE molecule, and not localized in a specific orbital? Electrons are supposed to occupy orbitals in pairs. 
(Original post by Lifeisnice)
However, how do you interpret the wave function solution? From what I understand, the square of the wave function has something to do with the probabilities of finding the electron there, but I can't quite grasp how to interpret the graphs properly.
Yes the square of the wavefunction is related to the probability of the particle being within a defined area. By definition the integral of the square of the wave function over all space is equal to unity (effectively the particle must be somewhere).
Also, isn't the particle in a box model assuming that each individual electron acts by itself, and is spread over the WHOLE molecule, and not localized in a specific orbital? Electrons are supposed to occupy orbitals in pairs. 
There is a very small part of me that loves quantum. I'm still semidetermined to get to the end of MQM properly  in which case I might sit an extra exam and inch a little closer to degree safety.
(well, I say inch  I of course mean 254 000 000 angstroms ) 
(Original post by EierVonSatan)
A wavefunction tells you everything about a physical system, when you solve a system (by applying a given operator onto a wavefunction) it spits out a set of states in which the particle can now behave.
Yes the square of the wavefunction is related to the probability of the particle being within a defined area. By definition the integral of the square of the wave function over all space is equal to unity (effectively the particle must be somewhere).
The assumption here relies on the electron being trapped in a long thin molecule and pretty much nothing else. things like nuclearelectron, electronelectron interactions are ignored. Atomic orbitals are probability density solutions of the Schrodinger equation for a hydrogen atom (a much more complex model).
What about the quantum numbers? I saw some graphs of wavefunctions of n=1 to n=4 being plotted of a conjugated system. What do these quantum numbers correspond to? 
(Original post by Lifeisnice)
Oh so basically, you're assuming each electron contributes to every single bond!
What about the quantum numbers? I saw some graphs of wavefunctions of n=1 to n=4 being plotted of a conjugated system. What do these quantum numbers correspond to?
Spoiler:ShowThe idea is you approximate a conjugate bond as a box with n particles "electron contributes to every single bond" [you can "get rid" of all non bonding electrons [I don't know if you have come across MO theory yet]. If we take the bonding electrons into account and ignore any electrostaic interaction between them then you can model them as n particles in a box. The wave function should not be viewwed as how the particles travel, where you will find the particles, etc. It is well defined such that for some wavefunction [or limits a and b if you know your particle is bound in some way. I.e. we take the absolute value of psi (the wavefunction  as it will in some cases be complex), square the wavefunction and do an appropiate definite integral with the contraint that it equals 1  the wavefunction psi mustsatisify this. This is extremely crude though and even approimations using more sophisticated methods don't really give us an accurate idea.
I couldn't possibly comment on the graphs without seeing them
Lot probs wrong  been too long 
From http://bouman.chem.georgetown.edu/S02/lect13/lect13.htm and http://74.125.153.132/search?q=cache...&ct=clnk&gl=sg
The second site says, "To a rough approximation, electrons in a conjugated hydrocarbon can be modeled as a particle in a onedimensional box." 
(Original post by Lifeisnice)
Oh so basically, you're assuming each electron contributes to every single bond!
What about the quantum numbers? I saw some graphs of wavefunctions of n=1 to n=4 being plotted of a conjugated system. What do these quantum numbers correspond to?
And in terms of what's really going on with bonding, often the concepts can be more complicated than 2 electrons forming one bond between 2 atoms. You can have many atoms contributing partially to bonds. 
(Original post by Kyle_SC)
n=1 to n=4 will correspond to the states of the particle in a box (and the molecular orbital). This quantum number is just the n in .
And in terms of what's really going on with bonding, often the concepts can be more complicated than 2 electrons forming one bond between 2 atoms. You can have many atoms contributing partially to bonds.
Don't each electron from each other go into specific molecular orbitals, rather than all of them contributing to the bonds? 
(Original post by Lifeisnice)
so which molecular orbital would n=3 represent then?
Don't each electron from each other go into specific molecular orbitals, rather than all of them contributing to the bonds? 
(Original post by cpchem)
Yes, they go into specific MOs. Remember that the MO model of bonding is different to the valence bond model. 
(Original post by Lifeisnice)
If so, how do you justify using the particles in a box method? The molecular orbitals are in specific places, not along the entire length of the molecule. Also, how do you interpret those graphs of n=1 to n=4.
The interpretation of the graphs is that the MO wavefunctions look similar to the particle in the box wavefunctions, where L is the length of the molecule and n = 1 to x/2 where x is the number of electrons in the conjugated system. 
(Original post by Kyle_SC)
The MOs aren't in specific places, they're spread over the entire molecule (and beyond, what we usually draw is an isosurface at which point the integral has some fixed value). Take a look at benzene's MOs. The MO approach can explain some things which valence bond theory (i.e. electrons sit in bonds) cannot, like conjugation and aromaticity. It also has limitations (e.g. it can't do dissociation).
The interpretation of the graphs is that the MO wavefunctions look similar to the particle in the box wavefunctions, where L is the length of the molecule and n = 1 to x/2 where x is the number of electrons in the conjugated system.
So basically this particle in a box model assumes that each electron acts independently and contributes to EVERY observed bond in the system? That's the only way I can think of how a (single) particle in a box can model the molecule.
Also, the graphs of say n=1 and n=4 look quite different, they have different nodes. How do these correspond to specific instances of the molecule? 
Try looking at the lectures 'Particles in Boxes' and 'Spectroscopy in Boxes' here. That's a reasonable introduction.

(Original post by cpchem)
Try looking at the lectures 'Particles in Boxes' and 'Spectroscopy in Boxes' here. That's a reasonable introduction.
I read through it, and while I can understand the mathematics of solving the Time Independent Schroedinger Equation, I don't understand what the n in the eigenfunctions correspond to. It has something to do with energy being quantized, but what physical reality is n supposed to correspond to? 
(Original post by Lifeisnice)
Thanks for the link!
I read through it, and while I can understand the mathematics of solving the Time Independent Schroedinger Equation, I don't understand what the n in the eigenfunctions correspond to. It has something to do with energy being quantized, but what physical reality is n supposed to correspond to?
By only allowing discrete values of n (and not continuous ones) we get a quantum system. Energy levels have an n^{2} dependence and the wavefunctions nodes is affected by the value of n, number of nodes = where the wavefunction = 0 and so the particle can not be found there, here nodes = n  1. Again we can related the wavefunction back to the probability density. 
(Original post by EierVonSatan)
Yes, n is the quantum (~quantisisation) number for the system and is a variable (n = 1, 2, 3... (n is an element of the positive integers)) the physical realities i.e. the energy levels and wavefunctions.
By only allowing discrete values of n (and not continuous ones) we get a quantum system. Energy levels have an n^{2} dependence and the wavefunctions nodes is affected by the value of n, number of nodes = where the wavefunction = 0 and so the particle can not be found there, here nodes = n  1. Again we can related the wavefunction back to the probability density.
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following: