|Jun2-08, 09:54 PM||#1|
So acidity increases across a period because the electronegativity of the conjugate base increases as move from left to right. This means it can better stabilize the negative charge. Something else that contributes to this trend is the fact that as move from left to right across a period, the bond between H and the other atom becomes more polar, allowing the bond to be broken easier (ex. HF is more polar than NH3). Is that correct?
However, isn't HF a weaker acid than HI because HF is in fact a stronger bond. Doesn't this contradict the fact that HF is more polar and the bond is easier to break. Which one of the statements is right? I'm aware that the fact that I- is bigger allows the electrons to be delocalized better in the periphery and this is probably the major contributing factor to the acidity down a group but can someone please explain the electronegativity bit?
|Jun2-08, 11:40 PM||#2|
HF is more polar, but has a stronger bond - i think you're reasoning is correct, but let me rephrase.
Because 'I' has so many electrons already, the electrons in the bond with 'H' are "sheilded" from the nucleus. i.e. the outermost electrons are not only attracted to the nucleus, but also repelled by coulomb forces with the other electrons - decreases the bond strength. This is a fairly complicated issue and its hard to incorporate it in the general trends of the periodic table.
Electro negativity increases - in general - from left to right, and top down; but not universally. As soon as you form large nuclei, the electro negativity decreases again.
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