Dismiss Notice

Interview Feedback: Visit Interview Feedback to view and submit interview information.

Interviewing Masterclass: Free masterclass on interviewing from SDN and Medical College of Georgia

strong acids + atom size

Discussion in 'MCAT Study Question Q&A' started by yestomeds, Aug 3, 2015.

  1. yestomeds

    May 12, 2014
    Likes Received:
    Really sorry for the simple question, but i'm trying to read diff internet sources and my head is just... feeling jumbled up. Thought it might just be quicker for me to pose such simple question to you guys and maybe learn that way.

    This is an ex. of what I saw on the internet:
    HI > HBr > HCl > HF Size. When comparing atoms within the same group of the periodic table, the larger the atom the weaker the H-X bond and the easier it is for the conjugate base to accommodate negative charge (lower charge density)

    1) I get how big atoms make weaker bonds with the other atoms
    (wait, why is this? Because of all the orbitals in b/w the big atom and the 2nd atom?)

    2) What I don't get is how you speak of the conjugate base + accommodating a negative charge?
    3) Also, read somewhere that bigger atoms are better able to stabilize a -ve charge? How so? :'( :(
  2. theonlytycrane

    2+ Year Member

    Mar 23, 2014
    Likes Received:
    Medical Student
    I think this is a great question :) here's my take-

    1. A larger halogen atom has a longer bond with a Hydrogen atom than a smaller halogen atom. Longer bonds are weaker and more easily broken.

    2/3. Larger halogens are able to better accommodate a negative charge because the negative charge will be distributed over a larger area vs. smaller concentrated in a small area (Flourine). In organic chemistry I usually draw resonance structures to show how a negative charge will be delocalized among atoms in a molecule. Usually more resonance structures of a conjugate base -> more stability of the conjugate base -> more acidic starting molecule.

    If you haven't taken Orgo yet, just focus on the larger halogen size being able to better accommodate a negative charge due to charge distribution over a larger area.

    Additional link: http://www.chemguide.co.uk/inorganic/group7/acidityhx.html
  3. 7331poas

    2+ Year Member

    Jun 16, 2015
    Likes Received:
    Medical Student
    For 1.

    Bond length is more a product of electronegativity, not really atom size. (Although they do certainly correlate)

    For 2/3.

    The answer is put shorter; the charge is spread over a larger area. There is less charge density. (lowering the potential energy)

    Charge density is the smoking gun here.
  4. BerkReviewTeach

    BerkReviewTeach Company Rep & Bad Singer
    Vendor 10+ Year Member

    May 25, 2007
    Likes Received:
    This can be true when comparing elements in the same row, but the OP was working with elements in the same column of the period table. In that case, size is the primary factor.

    The HI bond is made from a 1s of H and an sp3 hybrid orbital of 5s with 5p orbitals on I. Because of the fifth shell being involved, the I atom is large and thus cannot get as close to H as a smaller atom (such as F using the 2nd shell). Consequently, HI forms a longer bond with less overlap than HF. So the atomic radius is in fact the primary factor is an H-I bond being longer than an H-F bond. This is also the explanation for why the sidechain pKa of cysteine is notably lower than the sidechain pKa for serine.

    But you make a valid point when considering HF vs. H2O vs. NH3.
    Stop hovering to collapse... Click to collapse... Hover to expand... Click to expand...

Share This Page