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Formal Charge and Molecular Configuration

Discussion in 'MCAT Study Question Q&A' started by supafield, May 12, 2008.

  1. supafield

    supafield Dream Big 7+ Year Member

    May 18, 2006
    I'm a little confused about this..

    I was doing a question in the TPR orgo that had me rank CO, CO2 and CO3^2- bond length...

    It was intended to have you figure out which molecule had the most s character in its hybridized orbitals...

    I got it right just because I knew carbon monoxides chemical configuration..... my question is what if I hadn't known it...

    How would I be expected to know the C would triple bond to O and take on an extra electron?

    All things taught in normal prep books would have you believe that it would try to assume the configuration with the lowest Formal charge, and if CO double bonded with 2 lone pairs on Oxygen and one on carbon.... both atoms would have a formal charge of 0

    In reality Carbon has a formal charge of -1 and Oxygen +1
    Given what I know about electronegativity I would expect Oxygen to hog the electrons??

    So is there any way to figure this out with normal MCAT knowledge or would I just have to had known CO behaves the way it does.... and if so are there other molecules the MCAT would expect me to know behave differently... ie: cyanide? Or would the structure be drawn in most cases if it didn't follow expected trends?
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  3. BloodySurgeon

    BloodySurgeon Moderator Emeritus Lifetime Donor Classifieds Approved 10+ Year Member

    Nov 19, 2006
    hSDN Member
    hSDN Alumni
    Carbon has 4 valence electrons, Oxygen has 6. together there are 10 valence electrons. To obey the octet and only use up 10 valence electrons, carbon monoxide has a triple bond and a lone pair of electrons on each atom.

    Now both oxygen and carbon both share 8 electrons (octet).

    Obeying the octet is more important than having 0 formal charge. Hope this helps.
  4. supafield

    supafield Dream Big 7+ Year Member

    May 18, 2006
    It sure does, thank you... I got caught up in calculating these formal charges that I forgot rule #1..... thanks again
  5. BerkReviewTeach

    BerkReviewTeach Company Rep & Bad Singer Exhibitor 10+ Year Member

    May 25, 2007
    SDN Exhibitor
    When it comes to resonance stability, the Octet rule ranks first. This means that forming the third bond between C and O is necessary for stability, even at the expense of the formation of formal charges.

    Not sure what you mean by normal, but we address the structure in our materials on two occassions (once with hemoglobin and the other with a resonance/IR spectroscopy question).

    You could have deduced the structure by knowing that CO binds hemoglobin more tightly than O2 and that it binds in a linear fashion. That can only be true if the carbon binds the Fe using an sp-hybride, which occurs because C carries a negative charge. There's a chance that some people could be exposed to the structure of CO via hemoglobin.

    Other anomolies in structure include B2H6 (dimer of BH3), SO2 (being bent), the various allotropes (sulfur, carbon, etc...), and as you mentioned CN-.
  6. TJames

    TJames 2+ Year Member

    Feb 18, 2008
    As another poster said - the most important thing is to give each atom eight electrons.

    I wanted to add that the charges on each atom are only FORMAL charges, not the actual charges that end up on each atom.

    In CO, even though C has a - formal charge and O has a + formal charge, the ACTUAL charge on each atom is just about neutral. CO has a very small dipole moment for an asymmetric compound.

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