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    FutureCSI's Avatar
    FutureCSI Posts: 34, Reputation: 1
    Junior Member
     
    #1

    Sep 9, 2010, 06:29 AM
    I need help with an organic Chem Problem. Please Help
    I am a college student and have my first exam coming up. I have the review sheet that is practically all done by me, but there is this one problem I cannot seam to understand and not even how to begin.

    If you can offer your help it is :


    * The hybridization of this oxygen atom 11 is _________. (please see picture above)
    * The hybridization of this oxygen atom 12 is ________.
    * The hybridization of this carbon atom 13 is ______.
    * The Hybridization of this carbon atom 14 is __________.

    Thank you so much, any help is greatly appreciated
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Sep 9, 2010, 07:10 AM

    Uh... your pictures are not showing up...

    I think that it's better to upload on a free image hosting site and post the link here. With the recent changes in the forum skin, there are things like the manage attachments, which I think is not working.
    DrBob1's Avatar
    DrBob1 Posts: 425, Reputation: 86
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    #3

    Sep 9, 2010, 08:29 AM
    If the structures won't post, just name the compound(s) and indicate the atom in question.
    FutureCSI's Avatar
    FutureCSI Posts: 34, Reputation: 1
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    #4

    Sep 9, 2010, 07:51 PM

    FutureCSI's Avatar
    FutureCSI Posts: 34, Reputation: 1
    Junior Member
     
    #5

    Sep 9, 2010, 07:51 PM

    Sorry guys I took a little long but it was hard to try and get this picture working. Thank you once again
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #6

    Sep 10, 2010, 09:19 AM

    Ok. I'm not sure about other elements, but I know that for carbon. In each case, an electron from the 2s orbital gets energy and goes into one suborbital of the 2p orbital.

    This is how I remember:

    (a) SP1 hybridisation is when you have 1 s and 1 whole p, meaning that there are two distinct groups of electrons. Since p is bigger than s, it can form more bonds, more specifically 3 bonds. Ethyne contains 2 carbon atoms in this hybridisation state.

    (b) SP2 hybridisation is when you have 1 s and 2 p (ie, the 3 p sub orbitals separated into 2; for example px and pypz), forming a total of three distinct groups of electrons. The 2s, the 2px and 2pypz can form single, single and double bonds respectively. Ethene is the example here.

    (c) SP3 hybridisation is when you have 1 s and 3 p (ie, the 3 p sub orbitals separated into 3; for example px, py and pz), forming a total of 4 distinct groups of electrons. The 2s, the 2px, 2py and 2pz can form single, single, single and single bonds respectively. Methane is the example here.

    I think Dr.Bob can confirm for oxygen, or correct me where necessary.

    I think it's time for you now to give the question a try and post what you think is the correct answer. We'll tell you whether you're right, and correct you if necessary. :)
    DrBob1's Avatar
    DrBob1 Posts: 425, Reputation: 86
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    #7

    Sep 10, 2010, 12:40 PM
    FutureCSI:

    Whenever these second row elements form a single bond they are sp3 hybridized. Accordingly, double bonds are sp2 and triple bonds are sp hybrids.

    To expand on Unknown008's answer, all of these atoms will have electrons in FOUR orbitals: 4 sp3 orbitals, 3 sp2 hybrid orbitals and 1 unchanged p orbital, or 2 sp hybrid orbitals and 2 unchanged p orbitals. This is why we find carbon, for instance, always forming FOUR bonds.

    Single bonds are formed by overlap of an spX hybrid with another s or spX hybrid. Double bonds by overlap of an sp2 with another sp2 (or sp) to form the sigma bond and with overlap of two p orbitals to form the pi bond. An sp and two p orbitals from both atoms form the sigma and pi bonds comprising a triple bond.

    So: Oxygen atom 11 will be sp2 hybridized (double bond)
    Oxygen aotm 12 will be sp3 hybridized (single bond)
    Carbon atom 13 will be sp3 hybridized (single bond)
    Carbon atom 14 will be sp hybirdized (triple bond)
    (and therefore, so is the N atom)

    I hope this answers your questions.

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