Why is a third body needed in the recombination of two hydrogen atoms?

Question

In the article Discuss. Faraday Soc. 1962, 33, 205, the authors say that in order to form molecular hydrogen from collision to atom we need a third body to remove the excess energy. That is we have the reaction $\ce{H + H + M -> H2 + M}$.

I suppose the reason is because if we have the excess this energy will break the bond. If it this the case why the energy can not be transferred to the electron and so would have $\ce{H + H -> H2^*}$, where $\ce{H2^*}$ denotes an excited state of $\ce{H2}$?

Answer 1

Yes there is validity to the finding that two atoms of hydrogen form molecular hydrogen at very slow rates. Review Gould, R.J., and Salpeter, Edwin E., Astrophysical Journal, Vol 138, pg. 393. It is publicly available at: http://articles.adsabs.harvard.edu/pdf/1963ApJ...138..393G

The exact answer to your question is found on page 138 of that work, under the heading “II. Mechanisms for the Formation of Molecular Hydrogen.”

In space, molecular hydrogen is destroyed (by cosmic radiation) at rates that exceed its formation without some type of surface catalyst. This conundrum motivated significant research to find the actual mechanisms of molecular hydrogen formation in natural settings. More recent work can be found by Gianfranco Vidali, et.al., Journal of Physics: Conference Series:6, 2005:
https://iopscience.iop.org/article/10.1088/1742-6596/6/1/003/pdf
This more recent work cites the Gould and Salpeter work. It appears to be still accepted as the explanation for the slow rate of formation of molecular hydrogen from atomic hydrogen.

According to this explanation, after ground state hydrogen atoms collide, the new hydrogen molecule will be in a triplet state, and it cannot decay into a singlet state because this is a forbidden transition. A cursory search on the PES's of triplet states of diatomic hydrogen show that indeed these are shallow curves which would lead to rapid dissociation.