Reaction between 2 hydrogen halides

Question

I know that hydrogen halides can react with each other to produce halogen halides.

As an example here is a reaction between $\ce{HCl}$ and $\ce{HF}$: $$\ce{HCl + HF -> H2 + ClF}$$

As for the mechanism it is like this (taking into account that chlorine is hypervalent but fluorine isn't): $$\ce{HCl + FH -> HCl\bond{...}F + H. -> H. + ClF + H. -> H2 + ClF}$$

I know that both of these are acids and that $\ce{HF}$ is weaker than $\ce{HCl}$. However with the halogen bond the negative charge on the fluorine (partial negative after the formation of the halogen bond) is more delocalized. This stabilizes the $\ce{H-Cl\bond{...}F}$ intermediate and makes the dissociation easier for the hydrogen and harder for the fluorine. And yes the hydrogen radical is very reactive but that is exactly why $\ce{H2}$ forms in this process.

But I could see some problems with this.

1. It would have to be in a vacuum for it to not be explosive(most reactions involving fluorine or fluorine compounds are explosive because of fluorine's high electronegativity (same reason reactions with $\ce{O2}$ burn things))

2. If it is in air $\ce{O2}$ could react very violently with both the chlorine and the fluorine giving you these products: $\ce{ClO}$, $\ce{ClO2}$, $\ce{ClO3}$, $\ce{ClO4}$, $\ce{Cl2O}$, $\ce{Cl2O2}$, $\ce{Cl2O3}$, $\ce{Cl2O4}$, $\ce{ClO6}$, $\ce{ClO7}$, $\ce{OF2}$, $\ce{O2F2}$, $\ce{ClO2F}$, $\ce{ClO3F}$

3. Chlorofluorocarbons could also form from reactions with carbon based compounds in the air like methane and $\ce{CO2}$.

And this is all because the halogens are very reactive.

Similar things would be true for other halogen fluorides.

So is there a more practical way to form halogen halides than reacting $\ce{HX}$ with $\ce{HY}$ where both $\ce{X}$ and $\ce{Y}$ are halogens?

Answer 1

The reaction you are proposing is very unlikely to happen under standard conditions. Taken average bond energies from UC Davis ChemWiki,^[1] the approximated heat of reaction would be $\Delta_\mathrm{R}H = 307~\mathrm{kJ\,mol^{-1}}$. $$\ce{HCl + HF <=> H2 + FCl}$$

The reaction mechanism you are proposing is also not quite feasible. If we would ignore the energetics of this reaction, and we speak hypothetical about this reaction, we would still have to consider an activation step. What we also need to consider is the high affinity of fluorine towards electrons and therefore have to exclude anything that would enhance the heterolytic bond cleavage of hydrogen fluoride. Hence this reaction can only take place in gas phase in an inert atmosphere.
A feasible starting point would be the thermolytical or photolytical cleavage of one of the bonds. $$\begin{align} \ce{HCl &->[\Delta T/ h\nu] H + Cl}\\ \ce{HF &->[\Delta T/ h\nu] H + F} \\ \end{align}$$

Then we would need to consider propagation of the reaction. $$\begin{align} \ce{HCl + H/Cl/F &->{} [ HCl.(H/Cl/F)]^{\ddagger} && -> H(H/Cl/F)/Cl(H/Cl/F) + H/Cl}\\ \ce{HF + H/Cl/F &->{} [ HF.(H/Cl/F)]^{\ddagger} && -> H(H/Cl/F)/F(H/Cl/F) + H/F}\\ \end{align}$$ And then finally recombination of the remaining free radicals.

We are basically looking at the reverse mechanism of the reaction of hydrogen and chlorine.^[2]

The practical approach is given in orthocresol's answer. Interhalgen compounds, i.e. halogen halides, are formed by reacting the pure halogen gasses. For the reaction of chlorine and fluorine we find the approximate enthalpy of reaction to be $\Delta_\mathrm{R}H = -113~\mathrm{kJ\,mol^{-1}}$. $$\ce{F2 + Cl2 <=> 2FCl}$$ Side products also include $\ce{ClF3}$ and $\ce{ClF5}$. The reaction itself is exothermic and ignoring entropy should proceed spontaneously. $$\begin{align} \ce{Cl2 &->[h\nu] 2Cl} &&\text{(start)}\\ \ce{Cl + F2 &-> ClF + F} &&\text{(propagation)}\\ \ce{F + Cl2 &-> ClF + Cl}\\ \ce{F + Cl &-> ClF} &&\text{(stop)}\\ \end{align}$$

All reactants are highly reactive and may decompose explosively when other reactants are involved. Even in diluted systems these compounds react with almot everything, see also Why does free chlorine in the stratosphere lose its ozone-depleting potential after about 100,000 reactions?

Conclusion

The formation of interhalogen compounds via their hydrogen halides is energetically not feasible. They can be synthesised from their pure compounds under moderate conditions.
The mechanisms for these reactions can only develop in a closed system as radical chains, which need an initiation step.

Appendix

1. Average bond energies (Bond dissociation energies; BDE) from UC Davis ChemWiki: $$\displaystyle\begin{array}{lr} &\text{BDE}/\mathrm{[kJ\,mol^{-1}]}\\\hline \ce{HCl} & 427\\ \ce{HF} & 565\\ \ce{H2} & 432\\ \ce{FCl} & 253\\\hline \ce{F2} & 154\\ \ce{Cl2} & 239\\\hline \end{array}$$
2. A. L. Marshall, J. Phys. Chem. 1925, 29 (7), 842–852.
   Carl M. Furgason and John W. Moore, J. Chem. Educ. 1943, 20 (1), 41.

Answer 2

I think the term you want is "interhalogen compounds", not halogen halides.

Personally, I have never heard of them being prepared via the reaction of the hydrogen halides. Most common synthetic routes involve direct reaction of the elements $\ce{X2 + Y2}$, or reaction of one element with a different interhalogen compound, such as $\ce{X2 + XY$_n$}$.

Chlorine monofluoride is not a great example because no matter what synthetic route you take, it is going to be kind of dangerous. Greenwood & Earnshaw's Chemistry of the Elements 2nd ed. (p 825) gives two routes:

$$\ce{Cl2 + F2 -> 2ClF}$$

(the product must be purified from the by-product $\ce{ClF3}$, as well as unused reactants)

$$\ce{Cl2 + ClF3 -> 3ClF}$$

(must be purified from excess $\ce{ClF3}$).

As for the exclusion of oxygen or organic compounds, that's not difficult at all, as long as you set up your experiment properly.