I am imagining that the occasional oxygen might be floating close enough to a certain calcium atom and take the two electrons, which will ionize them and they will now 'glue'. Maybe the same thing happens in a hydrogen and chlorine, and a covalent bond is smoothly and calmly formed.

I expect these cases to be the exception, because if atoms have speeds of hundreds of meters per second, violent collisions would be the norm for their encounter. The question is: Is it the case that electron transfer occurs in collisions, but the kinetic energy overcomes the electrostatics and the atoms fly apart, and then they eventually find other partners to bond as they wander around? Even in covalent bonds? (It was the latter part that sparked the question, because in covalent 'no complete electron transfer occurs', but I couldn't picture atoms nicely slowing down to share electrons). Or is it always (collision=bonding-and-becoming-molecule-in-one-shot)? Or both?


2 Answers 2


Your intuition is spot on. If we have two atoms approaching each other with a large kinetic energy then they will have too much energy to form a stable molecule. Their electrons will interact as they approach, but the two atoms will simply whizz past each other and head off into the distance.

In many cases the reaction is more like:

$$\mathrm{AB + CD \to AC + DB}$$

or variations thereof, and in this case the reaction products can carry away excess energy as kinetic energy so the problem is avoided. However when you have a reaction like:

$$\mathrm{A + B \to AB}$$

The $\mathrm{AB}$ molecule doesn't have any way to dissipate the energy.

But in practice reactions like this do happen. For example, if you start with a gas of hydrogen atoms at room temperature you very quickly end up with $\mathrm{H_2}$ molecules. Typically the reaction is possible because the reaction products can shed their excess energy by colliding with other molecules. The process would be something like:

$$\mathrm{H + H \to H_2^*}$$

where the $*$ on the $\mathrm{H_2^*}$ indicates it is in a highly excited state that would typically rapidly fall apart into two atoms again:

$$\mathrm{H_2^* \to H + H}$$

However if the $\mathrm{H_2^*}$ can collide with a hydrogen atom the extra energy can be transferred to kinetic energy:

$$\mathrm{H_2^* + H \to H_2 + H + \text{kinetic energy}}$$

where now the $\mathrm{H_2}$ molecule and the $\mathrm{H}$ atom on the right hand side have a large kinetic energy. This large kinetic energy is what we call heat, so the end result is that the energy involved in the chemical reaction makes the gas hot.


To complete John's answer, chemical reactions happen in matter which is at a certain temperature. All matter follows roughly the black body radiation curve which has an energy distribution for the atoms/molecules that comprise it , given its temperature.

black body low temp

The temperature is connected in statistical thermodynamics with the average kinetic and vibrational/rotational energies in the sample. One can see that at room temperature (~300K), the radiations emitted by the interacting atoms/molecules is at small fractions of electron volts. The kinetic energy of atoms and molecules is low enough to allow capture by the available channels. The tails of the distribution with high energy collisions have very small probability, falling exponentially.


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