# 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. $2sp^{3}, 3sp^{3}, 4sp^{3}, 5sp^{3}$).

• There are a few things to keep in mind. The $pK_a$ values you typically see depend on approximations for thermodynamic activity that only hold for dilute solutions, and are solvent-specific (typically water). Concentrated solutions of $\ce{HF}$ are actually extremely acidic, stronger than concentrated $\ce{HCl}$. $\ce{HF}$ is quite a strange beast, and its reactivity is very complex and concentration-dependent. Finally, the consideration of the acid's LUMO energy is probably more relevant to reaction kinetics, while acidity (as measured by $pK_a$) is a thermodynamic parameter. – Greg E. Jun 18 '14 at 4:06
• Also, I'm not sure your reasoning about the LUMO energies is correct. See my discussion with ron in the comments to his answer. I could well be mistaken, but I think the LUMO energy of $\ce{HF}$ is higher than $\ce{HCl}$, for example. Keep in mind also that the extent of ionic character in a bond depends on a lot of factors, but generally you can expect that larger energy differences between interacting atomic orbitals (which tends to correlate with electronegativity difference) favors greater ionic character. Still, one also has to consider atom size, orbital overlap, polarizability, etc. – Greg E. Jun 18 '14 at 4:26
• @GregE. I think I finally see my mistake now. I was assuming that if the $\sigma$ bonds rises in energy, the $\sigma^{*}$ would also do, but the opposite is actually happening: when the σ bond rises in energy, the $\sigma^{*}$ lowers, making it more easy to remove the proton in our case. I looked at colby.edu/chemistry/PChem/notes/AOIE.pdf and it seems that the energy of the hydrogen $1s$ orbital does not differ that much from the energies of the $2sp^{3},3sp^{3},4sp^{3},5sp^{3}$ orbitals of our halogens, which if you come to think of it isn't very strange (continued) – Jori Jun 19 '14 at 13:45
• because, although the orbitals with a higher principal quantum number have higher energy, they are also more shielded of from the nucleus by the electrons from the lower shells. This seems to cancel each other out mostly for atoms in the same group, leaving the energies of the $3p, 4p, 5p$ orbitals a bit higher for $\ce{Cl, Br, I}$ (higher bond energy, lower bond strength, more acidic). There are of course other factors to consider, but since I'm not studying quantum mechanics, but organic chemistry, I'll leave it up to you guys to explain the remaining granularities. – Jori Jun 19 '14 at 14:16
• Last thing I noticed is that fluorine's $2s$ orbital has significantly lower energy than the corresponding $ns$ orbitals of $\ce{Cl, Br, I}$. Also note that the table I listed doesn't say anything about the $d$ orbitals, I don't know what effect these have on the hybridized orbital energy, but I think not much as they will be very high energy. – Jori Jun 19 '14 at 14:23

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.

• Oops, I forgot the lone pairs of the halogens, they will be hybridized into $sp^{3}$ of course :) – Jori Jun 14 '14 at 15:31
• What do antibonding orbitals have to do with the binary acids of group 7? I don't think any of the antibonding orbitals are populated in the binary acids ... – Dissenter Jun 14 '14 at 23:55
• HF has a significant covalent component to its bond. The covalent nature of the HX bond drops off rapidly going down the column. As the bond becomes less covalent (more ionic) there is less electron density between H and X causing the energy of $\ce{\sigma^{\ast}}$ to decrease. HX acidity depends on a number of factors as I mentioned above. No one factor alone can adequately explain the trend. – ron Jun 15 '14 at 15:13
• Are you saying that the HX series becomes more ionic as you go down the column? That doesn't make sense because the halogens get less electronegative as you go down the column. – Dissenter Jun 18 '14 at 2:12
• and also less capable of forming a covalent bond with a small 1s orbital – ron Jun 18 '14 at 2:15