acid base - Hydrogen halide acidity

Why is it that $pK_{\ce{a-HF}} \lt pK_{\ce{a-HCl}}\lt pK_{\ce{a-HBr}}\lt pK_{\ce{a-HI}}$, although the electronegativity decreases? The more electronegative the atom accompanying hydrogen, the lower the energy of the $\sigma^\ast$ bond. The lower the energy of the $\sigma^\ast$ bond, the easier it is for nucleophiles to attach to it (i.e. hydrogen is more easily removed from the compound, thus making it more acidic).

Obviously, something is wrong with this reasoning. I could guess it is because hydrogen binds to different $s$ orbitals with the different halogens (for $\ce{F, Cl, Br, I}$ resp. $2sp^{3}, 3sp^{3}, 4sp^{3}, 5sp^{3}$).

Answer

Yes, you are on the right track, let's look at the situation in more detail. In the haloacid equilibrium $$\ce{HX <=> H+ + X-}$$ anything that stabilizes HX will push the equilibrium to the left and make the $\mathrm{p}K_\mathrm{a}$ more positive. Anything that stabilizes the proton and the halogen anion will push the equilibrium to the right and make the $\mathrm{p}K_\mathrm{a}$ less positive. Here's a table that summarizes some of the haloacid data:

$$\begin{array}{cccc} \text{X} & \text{H-X bond strength / kJ mol}^{-1} & \text{Electronegativity of X} & \mathrm{p}K_\mathrm{a}(\ce{HX}) \\ \hline \ce{F} & 565 & 4.0 & 3.1 \\ \ce{Cl} & 427 & 3.0 & -7.0 \\ \ce{Br} & 363 & 2.8 & -9.0 \\ \ce{I} & 295 & 2.5 & -11.0 \end{array}$$

There are 3 main factors that influence this equilibrium: bond strength, electronegativity and polarizabilty.

In the "Bond Strength" column we see that the $\ce{HX}$ bond strength decreases as we move down the column. This is because, as you noted, overlap between the hydrogen $\mathrm{1s}$ orbital and the halogen orbital is most effective with the fluorine $\mathrm{2sp^3}$ orbital (in fact, overlap is so good in the fluorine case that the bond has a significant covalent nature) and the effectiveness of the overlap decreases as we move down the column. Therefore, $\ce{HX}$ bond strengths should make the $\mathrm{p}K_\mathrm{a}$ decrease (less positive, more negative) as we move down the halogen column.

Your analysis of electronegativity is correct. The more electronegative the halogen, the more stable the halide anion. This factor should cause the $\mathrm{p}K_\mathrm{a}$ to increase (more positive) as we move down the halogen column.

The larger the atom, the more polarizable it is. The more polarizable an atom the more stable is its corresponding anion because we have our charge spread out over a larger volume which is a stabilizing feature. This factor should cause the $\mathrm{p}K_\mathrm{a}$ to decrease (more negative) as we move down the halogen column.

So we have two factors, bond strength and polarizability, that work together to push the $\mathrm{p}K_\mathrm{a}$ lower as we move down the halogen column; and one factor, electronegativity, that works in the opposite direction. Based on the observed $\mathrm{p}K_\mathrm{a}$'s, the first two factors win out over the electronegativity.

BTW, I like your $\sigma$, $\sigma^*$ reasoning. Here's how I think of it in those terms. As you bring two atomic orbitals closer together, they will split and form two new molecular orbitals. The degree of splitting is dependent upon the overlap of the two atomic orbitals. Overlap is a measure of electron density between the atoms. An electronegative atom will share its electron less readily decreasing the splitting. If there is a strong bond, then the electrons are well shared and the splitting will be larger.