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How come hybridisation of orbital doesn't violate quantization of energy of electrons

  1. Aug 1, 2012 #1
    Couldn't decide where to post... Chemistry or quantum mechanics... But posted here cause I wanted to know a physicist's view...

    We know that the electrons in the atom have discrete energy,I mean not just any energy... An electron can't have the energy between 2s and 2p orbitals... But after hybridization i.e. sp hybridization the electron will have energy which is the average of s and p orbitals... How? Is this 'legal' to have energy between the orbitals we know?
     
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  3. Aug 1, 2012 #2
    Re: how come hybridisation of orbital doesn't violate quantization of energy of elect

    Hello!

    The answer is yes, it's perfectly legal.

    Quantization of energy in quantum mechanics is the consequence of the fact that (usually) only for some discrete set of energies the Schroedinger equation, which connects the wave function and the potentials acting on the particle it describes, can be solved.

    When an atom is close to another such set of energies (and the wave functions related to each of these) changes with respect to the set of allowed energies of an isolated atom. This is because the electromagnetic potential of the close atom changes the potentials in the Schroedinger equation.

    This is approximately what happens, I hope this helps.

    Ilm
     
  4. Aug 1, 2012 #3

    fzero

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    Re: how come hybridisation of orbital doesn't violate quantization of energy of elect

    The energy spectrum of a system depends on what components make up the system. For example, the hydrogenic atom is a fairly simple system involving a single nucleus and a single electron. It's relatively easy to solve Schrodinger's equation to find the energy spectrum.

    If we now add another electron to the system, for instance to describe the helium atom, then we are making a significant change with respect to the hydrogenic atom. The original electron now sees the repulsive potential due to the 2nd electron, in addition to the original nucleus. So the energy spectrum will not be the same. In fact, we can no longer solve for the energy spectrum exactly, as we could for the hydrogenic atom. However, we have approximation techniques that allow us to solve for the system to an excellent degree of agreement with measurements. Many of these techniques involve starting with the solutions to the hydrogenic atom and then studying the effects of adding the 2nd electron. So the picture that emerges can be phrased in terms of the orbitals familiar from the hydrogen atom, with some differences that can be understood once you go into the details.

    A similar situation emerges when you consider molecules instead of atoms. We still have a picture of orbitals, but their details, in particular their energies, are modified by the presence of the additional interactions present because we're dealing with many bodies. The energy levels of the orbitals and the particular combinations of base orbitals that result in molecular bonds are different from the simple hydrogen atom, but the orbital picture is still useful as a framework for our models.

    So, yes, it is legal to have energy between the orbitals we know. This is precisely because the molecular system is more complicated than the hydrogen atom and the additional interactions lead to a different energy spectrum.
     
  5. Aug 1, 2012 #4
    Re: how come hybridisation of orbital doesn't violate quantization of energy of elect

    Thank you very much Ilm & fzero for your good answer and elaborate explanation... :)

    So in molecules the energy spectrum changes from the atom...
    But in an atom an electron can't have any energy between the energy levels,right?(unless we change anything.I mean add or remove any electrons)
     
  6. Aug 2, 2012 #5
    Re: how come hybridisation of orbital doesn't violate quantization of energy of elect

    Almost, but it actually depends on your definition of energy.

    In QM a physical systems can be in a superposition of states.

    For example one electron can be in different positions at the same time, say [itex]\vec{r}_1[/itex] and [itex]\vec{r}_2[/itex]. When you measure position you can fin it in [itex]\vec{r}_1[/itex] or in [itex]\vec{r}_2[/itex] but you can't know where until you actually do your measure.
    If you do the mean value of the position of such an electron, you will find a result between [itex]\vec{r}_1[/itex] and [itex]\vec{r}_2[/itex], say [itex]\langle\vec{r}\rangle[/itex], but you will never find the electron in [itex]\langle\vec{r}\rangle[/itex].

    The same thing can sometimes happens for energy levels of an atom. The atom can be in [itex]E_1[/itex] and [itex]E_2>E_1[/itex] but then it has mean energy [itex]E_1 < \langle E\rangle < E_2[/itex].
    So if by energy you mean the value(s) you can measure, then you cannot (by definition) have an energy between two consecutive levels.
    But if with energy you mean the mean value of energy, then it can be between the minimum and maximum value you can measure.

    Ilm
     
  7. Aug 3, 2012 #6
    Re: how come hybridisation of orbital doesn't violate quantization of energy of elect

    Yeah... I understand what you are saying... But by measuring we will force the electron to have a certain place and energy,so from this view the energy levels are discrete... And if we want to say what's the energy before measuring we will get a value in between...

    However, one thing crossed my mind... As an electron could be anywhere,let's say a 1s e got to the 2s orbital,then how did it get the energy to go there? (Cause energy of 2s is higher than 1s)
     
  8. Aug 4, 2012 #7
    Re: how come hybridisation of orbital doesn't violate quantization of energy of elect

    Yes, measuring energy we cause the collapse of wave function in an energy eigenstate (we do not force the electron to have a certain place, only to have a certain energy).
    After the measurement the energy is well defined (all subsequent measure of energy we can do will have the same result) and is one of a certain discrete set of allowed energies.

    If we measure another observable, for example position, we will know exactly the position but energy is not defined (the electron will be in a superposition of different energy eigenstates). Nevertheless, as we know exactly the wave function after a measure of position (and spin if we are talking about an electron), we will know exactly the mean value of all the observables.
    For example we will know the mean energy, and this can be between two energy eigenvalues. This does not mean that we know what we will obtain in subsequent energy measurement, we only know what is the probability of measuring every given energy.

    You are right, for an atom to go from a superposition of 1s and 2s to a 2s it needs some energy. This energy is given to the atom during the interaction with our measurement apparatus.

    Ilm
     
    Last edited: Aug 4, 2012
  9. Aug 4, 2012 #8
    Re: how come hybridisation of orbital doesn't violate quantization of energy of elect

    Thank you very much for your answers Ilm... :smile:

    But I don't understand this... If we know where the electron is,then why can't we know its energy? Isn't energy quantized for orbitals? Or are there no certain limits for orbitals(I mean limit of distance from nucleus and a certain region of space)?
     
  10. Aug 5, 2012 #9
    Re: how come hybridisation of orbital doesn't violate quantization of energy of elect

    You're welcome :smile:

    You have some nice quantum intuition!
    Actually each orbital is not limited in space. If you have an hydrogen atom in his fundamental state the probability of finding the electron very far (with distance [itex]r \gg 10^{-10} m[/itex]) from nucleus is very small*, so small that is zero for all practical purposes, but is not zero.

    The mathematical reason for that is that in QM a system can't have two defined quantity at the same time if the relative observables does not commute. You surely knew about momentum and position.

    In atoms Hamiltonian (the energy operator) and position does not commute too. This is, loosely speaking, because an atomic Hamiltonian is rotationally invariant (spherical symmetry) while obviously coordinates do vary with rotations.

    Ilm

    * I'm too lazy to take my calculator now, but I will if you want some numbers.
     
  11. Aug 5, 2012 #10
    Re: how come hybridisation of orbital doesn't violate quantization of energy of elect

    wavefunction square is the probability to find an electron somewhere. so it relates the position. if you know the position precisely you will loose your precise answer of momentum. momentum relates the energy. in mathematical terms, momentum space is the fourier transform of wavefunction. thats why these two cannot be measured simultaneously.

    one better view is to compare the time domain and frequency domain representation of a signal. wavefunction is something we don't know. but we know the physical interpretation of time domain and frequency domain representation of a signal (wave). this analogy may come to your use to understand the uncertainty.
     
  12. Aug 5, 2012 #11
    Re: how come hybridisation of orbital doesn't violate quantization of energy of elect

    In the electron orbitals we do not exactly know the position of the electrons.
     
  13. Aug 5, 2012 #12
    Re: how come hybridisation of orbital doesn't violate quantization of energy of elect

    Thank you very much for your answers Ilm,Sagar And Kholdstare...

    Heisenberg's uncertainty principle explains why can't we know both position and momentum(energy) at the same time... Suppose we precisely measure the position of the e... But our act of measuring would change the momentum of the e(the more precise measurement means photons with lower wavelength,that implies higher energy photons and thus we are more and more uncertain abt momentum,cause the photon would change the momentum of e )... And if we want to measure energy precisely, we would need photons with low energy,but that would make us uncertain us about the e's position... (I think the low energy photons can't be detected that precisely)

    The thing that confused me was that I was thinking there was certain regions for orbitals... But it should be there are certain regions for orbitals where the probability of finding the e is large... And that's why we can't know its energy for sure even after knowing its position cause as Ilm said it would be in superposition of different energy states... Cause there is no certain region for orbitals...
    Am I right?
     
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