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Help me walk through the logic here:

1) HF is a weaker acid than HCl.

2) This is as far as I can understand. How does sigma star affect the acidities of the binary halogen acids?

If a covalent bond was formed, you would expect a decreasing acidity down the periodic table of the halogens, because the energy of the σ∗ would increase forming a less attractive LUMO (lowest unoccupied molecular orbital), which would result in less hydrogen atoms being pulled of the halide (i.e. lower acidity, higher pKa).

Here are some questions about the above excerpt:

1) Covalent bond formed between what?

2) Why would the sigma star orbital be populated if the hydrogen were removed and a bond formed in place of it? Or is a bond going to be formed with the hydrogen still present?



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1 Answer 1


Some initial comments on the above:

1) Bonding in the $\ce{HX}$ series is a varying mix of ionic and covalent bonding. $\ce{HF}$ has the largest covalent bonding component and $\ce{HI}$ the smallest. This trend goes against what electronegativities alone would suggest, but is consistent with what overlap integrals (e.g. overlap between a $\ce{1s}$ orbital and a $\ce{2sp^3}$ orbital is better than overlap between a $\ce{1s}$ orbital and a $\ce{5sp^3}$ orbital) would predict. Therefor, the latter effect appears to predominate.

2) An MO diagram depicts only the covalent portion of the bonding situation. Therefor it is unlikely to accurately predict acid strength over the entire $\ce{HX}$ series where the (covalent\ionic) contribution to bonding is varying.

3) The MO diagram should probably have hybridized the $\ce{Cl}$ AO's prior to mixing them with the $\ce{H}$ AO. That is, I'd guess that the covalent portion of the $\ce{HX}$ bond is between a $\ce{H}$ $\ce{1s}$ orbital and a $\ce{Cl}$ $\ce{3sp^3}$ orbital, not a $\ce{Cl}$ $\ce{3p}$ orbital. This could be assessed by looking at the photoelectron spectrum of $\ce{HCl}$ and analyzing it (e.g. how many chlorine lone pair signals?) in terms of hybridization.

4) $\ce{\sigma^{\ast}}$ orbitals don't really need to be brought in to the discussion since they are unoccupied throughout the $\ce{HX}$ dissociation process.

5) Increased $\ce{HX}$ acidity correlates inversely with $\ce{HX}$ bond strength. $\ce{HF}$ with the best overlap and strongest bond is the weakest acid; $\ce{HI}$ with the poorest overlap and weakest bond is the strongest acid.

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Response to OP update

I'm not sure I understand just what % ionic means in the Table you posted. When bonding is discussed, usually 3 types of bonding are presented (see this link for example) 1) covalent bonding where electrons are roughly equally shared, 2) polar bonding where we have a covalent bond with unequal electron sharing and 3) ionic bonding where electrons are not shared. For $\ce{HX}$ you can draw 2 resonance structures, one covalent and one ionic that both contribute to the true picture of the covalent bond in $\ce{HX}$. Is the % ionic in the Table referring to the polarization in the $\ce{HX}$ covalent bond, or is referring to a truly ionic bond? How does the increasing covalency suggested by the Table as you move down the $\ce{HX}$ series square with decreased bond strength down the series? Since bromobenzene and methyl bromide have dipole moments, the equation used to generate the % ionicity in the Table could also generate a percent ionicity for these compounds, but they are certainly not "ionic" compounds. I suspect the Table is measuring the polarization of a covalent bond and not referring to the ionic character found in a salt.

  • $\begingroup$ Thanks! Why would overlap be better with F and not I? What do the overlap integrals mean, physically? $\endgroup$
    – Dissenter
    Jun 19, 2014 at 14:56
  • $\begingroup$ When we say "overlap" that means we're talking covalent, not ionic, bonding. The electron density in iodine's $\ce{5 sp^3}$ orbital is much more diffuse (spreadout over a larger volume) than it would be in fluorine's much smaller $\ce{2 sp^3}$ orbital. Therefor, overlap (covalent bonding) should be much stronger (e.g. the overlap integral should be larger) between hydrogen with its $\ce{1s}$ orbital and fluorine. Think of overlaying a small circle with another circle of similar size and one of much larger size. Say there is one electron uniformly distributed in each of the 3 circles, $\endgroup$
    – ron
    Jun 19, 2014 at 15:07
  • $\begingroup$ which situation will lead to the greatest absolute electron density in common between the overlapping circles? $\endgroup$
    – ron
    Jun 19, 2014 at 15:09
  • $\begingroup$ The former example. This is very interesting. My prof only rationalizes binary halide acid strengths based on the stabilities of their conjugate bases. I'll have to bring this up to him. He doesn't like quantum mechanics but this concept of overlap originated with Pauling if I remember correctly (whom he adores). $\endgroup$
    – Dissenter
    Jun 19, 2014 at 15:10
  • $\begingroup$ This also reminds me of another question I asked some time ago - why is HBr considered polar. Someone brought up bromobenzene and reaction rates. Your explanation could be another justification. $\endgroup$
    – Dissenter
    Jun 19, 2014 at 15:12

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