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Hybridized orbitals, more s character = shorter bond??

Discussion in 'MCAT Study Question Q&A' started by engineeredout, Jun 21, 2008.

  1. engineeredout

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    From EK Orgo: "The more s character a bond has, the more stable, the stronger, and the shorter it becomes".


    I thought double and triple bonds were shorter though, so why doesn't a greater p character cause a shorter bond length?
     
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  2. 161927

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    alkyne < alkene < alkane with respect to bond length and p character

    an sp hybridized carbon has more s character (50%) than an sp3 carbon (25%), and it has a shorter bond length
     
  3. ThePandaFactor

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    alkane = sp3 hybridized, one s and 3 p so only ~25% s character
    alkene = sp2 hybridized, one s 2 p so ~33% s character
    alkyne = sp hybridized, one s one p so ~50% s character
     
  4. engineeredout

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    Ah yeah thanks. Good ****. Been two years since orgo
     
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  5. wrathofgod64

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    I was also reading this and I had one question.

    It says "the more s character a bond has the more stable, the stronger, and the shorter it becomes."

    Here's my contradiction. Generally, shorter bonds, such as double and triple bonds, have more energy than single bonds. But the S orbital is a lower energy orbital than the P orbital, so bonds with more S character should have lower energy than bonds with more P character. I'm specifically referring to a Kaplan problem

    Which choice correctly lists the energy level of the C-H bonds, from lowest to highest?

    The answer is HCCH, C6H6, CH4

    Can someone help explain this? Thanks.
     
    #5 wrathofgod64, Jul 14, 2008
    Last edited: Jul 14, 2008
  6. Quantum Chem

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    Think of it like this: An electron in an S orbital feels more attraction to the nucleus than an electron in a P orbital. Bonds are basically attractions between electrons of the atoms and the nuclei of the atoms. So the more S character a bond has, the more stable in energy it is. I'm going to give an example that to help understand the concept. Look at this pic: http://oregonstate.edu/instruct/engr321/Exams/ExamsW02/MTOneImages/MT1a.jpg

    Notice that the X-axis is bond length and the Y-axis is the bond energy. There is an optimal length to get the lowest relative energy. Let's imagine that this optimal length is the bond between H-H, so the bond would be two S orbitals (not saying it is, just using this as an example). Now imagine that as you add P character to make it SP the bond the length shifts to the right, meaning that the energy increases, meaning that the bond is less stable. This is similar to moving an electron from a low energy state to a higher energy state, it takes energy. By adding P character you are slightly moving the bond electrons to higher energy states.

    So in your question the C-H bond in acetylene would be SP, the C-H bond in Benzene would be SP2, and the C-H bond in methane would be SP3. Now look back at the hypothetical example and notice that the SP3 bond would have the highest energy because its less negative and the out of these three the SP bond would have the lowest energy because its the most negative.

    I hope this analogy doesn't confuse anyone. Its easier to say in person than it is to type.
     
  7. SaintJude

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    like this explanation. note for later reference.
     
  8. MedPR

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    This is a good explanation, but in fewer words:

    Higher bond energy = harder to break = more stable bond = stronger bond = shorter bond. So bond energy is sp>sp2>sp3.
     
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  9. Buttafuoco

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    Too lazy to read the responses, but a double bond or triple bond is a sigma bond plus one or two pi bonds, respectively. The sigma bond is between the hybridized orbitals and has S character so it's strongest. The pi bonds between P orbitals are less strong, but the overall bond (or combination of bonds, if you wanna look at it that way) is stronger with stronger bond order.

    sp3 sigma bonds have the least S character too so they are weaker in general than sp2, sp, s-s bonds under hybridization theory.
     

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