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Hybridization, Antibonding Orbitals and All that Jazz

  1. Aug 7, 2017 #1
    This going to be a long post !

    Orbital Dilemma:

    i) Why there is opposite phase orbital for s-orbital? [ Couldn't find any representation on the web except for a pic in Ian Fleming's Molecular Orbitals and Organic Chemical Reactions,pg3)
    upload_2017-8-7_20-3-58.png
    *I assert no right over the image,image is referenced under fair-use policy

    ii) As shown in the image is it possible that or what is probability that in a given sample of Hydrogen atleast n hydrogens are out of phase and therefore doesn't react with other chemical species(can be hydrogen itself or any other chemical)

    iii) Why the phenomenon of phase arises? Lets say two opposite phase orbitals makes anti-bonding orbital , the probability that electron is present is negligible or zero , the mathematical argument behind this also seems fine!
    But why ?(what's physical intuition behind it), does electron field a kind of repulsion ? (Suppose I fire electron at such orbital with sufficiently low KE/negligible electron, will electron feel a repulsion there/what is preventin its presence there?)

    iv) Can phase of the wavefunction be reverted? (in some scenario)

    b)Ghosts of Hybrids

    i)Lets start with carbonyl atom , the mayor of the world of organic reactions , I am new to the world organic reactions .In Clayden , the author says that a Nucelophile (CN) gives its electron to Electrophile CO's antibonding low energy LUMO π* ...<details of Mechanism skipped>
    The thing which confuses me: The side-on combination of p-orbitals gives rise to π orbitals , as per the diagram of Carbonyl molecule both the left/unhybridized p orbital of sp2 hybridized atoms of C and O atom are in phase and side-on combo yields a π orbital....makes sense, but if both of the unhybridzed p-orbitals are in phase then how does the π* comes into the picture?

    .......Pausing for now as when I started writing questions I was 'in zone' and now due to certain circumstances I'm 'out of the zone'
     
  2. jcsd
  3. Aug 7, 2017 #2

    jim mcnamara

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    Staff: Mentor

    Are you trying to ascribe what you ask to the effect of an electron's position? Absence or presence?
     
  4. Aug 8, 2017 #3
    I am guessing that Adjax was unable to understand the concept of bonding and antibonding orbitals. I think I should provide the answer for the first question only, and that may help Adjax to think about the other questions.


    Here's an intuitive explanation:
    The origin of the bonding and antibonding orbitals arise from the "phase" of the orbitals. Remember, + and - of a wavefunction? Wavefunction of 1s-orbitals are either fully + or fully - (other s-orbitals are slightly different but from the bonding point of view, it is pretty much the same thing). So then we have two possible combination of these wavefunction: + & + (or - & - but that's the same thing under the definition of wavefunction), and + & - (or - & +, which is the same thing). For a + & + combination, the wavefunction can spread throughout the two atoms and the electrons can move freely within. This is why we call it "bonding orbital". However, for a + & - combination, the resulting wavefunction conflicts and cannot combine at the interface of the two wavefunctions. Electron on both sides will not be able to go to the other side. We call this "anti-bonding". For example like the figure you provided, the antibonding orbitals do not look like they are forming a bond. This is why for a hydrogen case like you written above, we have a bonding and anti-bonding orbital.
     
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