The properties of H2S and H2Se are similar, although the selenide is more acidic with pKa = 3.89, and the second pKa = 11.0 at 25°C.
If this is true, why? Se would be less electronegative and thus HS- would be more stable than HSe-. Also, the more electron withdrawing effect of S
would tend to make H2S more acidic, right?
The only other thing I can think of is that Se is slightly larger than S leading to higher stability of HSe-. I suppose atomic radius trumps
inductive effects by a large margin.
Now that I think about it I think I answered my own question. This is very much related to the reason why HI is much more acidic than HF.
Should I even post this? Sure. I may not be right.vulture - 20-12-2010 at 13:36
Maybe the volatility of H2S is also a factor? Escape from solution would drive the equilibrium to the left. madscientist - 20-12-2010 at 13:52
Here's the explicit answer. Excerpt from "Modern Physical Organic Chemistry" by Anslyn & Dougherty.
The acidities of halogen acids follow the trend: HI > HBr > HCl > HF. A similar trend is found for group VIA acids:
H<sub>2</sub>Te > H<sub>2</sub>Se > H<sub>2</sub>S > H<sub>2</sub>O. The major factor that
contributes to this trend is bond strengths. The bond strength decreases in the following order: H-F > H-Cl > H-Br > H-I and H-OH > H-SH
> H-SeH > H-TeH. Although these are trends in homolytic bond strengths, not heterolytic bond strengths, the trends certainly contribute to the
increasing acidity with lower bond strength. However, when other factors are at play, homolytic bond strengths do not dominate trends in acidities.
For example, when one moves away from a homologous series of structures all in the same row of the periodic table, bond strength trends do not
correlate with acidities. R<sub>2</sub>N-H bond strengths (R = alkyl) are weaker than both R<sub>3</sub>C-H and RO-H bonds,
yet N-H bonds are more acidic than C-H bonds.
If we remove an electron from a hydrocarbon we weaken the bonds. Therefore, radical cations of weak acids can become very acidic. Radical cations
(R-H*+) can lose a proton by heterolytic cleavage to give a neutral radical (R* + H+), or undergo homolytic cleavage to give a carbocation by loss of
a hydrogen atom (R+ + H*). In polar solvents the loss of a proton is favored due to the favorable solvation energy. For example, in DMSO the
pK<sub>a</sub> of the methyl group in toluene is 43, but the pK<sub>a</sub> of the radical cation of toluene is -20. This
makes an amazing difference of 63 in pK<sub>a</sub> values.
