# Spontaneous reaction between chlorine and a hypothetical atom

Suppose, my hypothetical element is $$\ce X$$, whose first ionization energy is $$\pu{200kJ/mol}$$. An atom of this element will react with an atom of chlorine (assume that I've already dissociated a chlorine atom from a $$\ce{Cl2}$$ molecule). The first electron affinity of chlorine is approximately $$\pu{349kJ/mol}$$.

We take those two atoms inside a non-reactive box in a controlled room. Inside the box is vacuum. I intend to make the following reaction happen:

$$\ce{X(g) + Cl(g)-> XCl(s)}$$

Now, I'm curious as to the course of events.

Hypothesis 1:

One hypothesis is that the reaction will not proceed even if you get the atoms really close together. For the reaction to proceed, you must provide the required ionization energy to $$\ce X$$ first. Then the removed electron from X will be acquired by Cl promptly. Energy will be released in this process. The energy released in this process will be greater than the ionization energy that was provided to $$\ce X$$ previously. In the final step, $$\ce {X^+}$$ and $$\ce {Cl^-}$$ will get stuck together due to the electrostatic force between them. In conclusion, the reaction will process in the following steps (in order):

$$\ce{X(g) -> X+ + e-}\tag{1}$$

$$\ce{Cl(g) + e- -> Cl^{-}(g)}\tag{2}$$

$$\ce{X+(g) +Cl^{-}(g) -> XCl(s)}\tag{3}$$

In this hypothesis, the reaction isn't fully spontaneous since heat/energy needs to be added in the beginning for the reaction to proceed.

Hypothesis 2:

Another hypothesis is that once you take the two atoms close together, electron transfer will take place i.e. the reaction will proceed even if you don't add any heat. In other words, in this hypothesis, heat/energy does not need to be added for the reaction to proceed: the reaction is fully spontaneous. Ionization energy will not need to be provided to $$\ce X$$ as the reaction is energetically favourable. The reaction will take place in the following steps:

$$\ce{X(g) -> X+ + e-}\tag{4}$$

$$\ce{Cl(g) + e- -> Cl^{-}(g)}\tag{5}$$

$$\ce{X+(g) +Cl^{-}(g) -> XCl(s)}\tag{6}$$

The commonality between hypothesis 1 and 2 is that there will be a net release of energy after $$(1)$$ & $$(2)$$ and $$(4)$$ & $$(5)$$. The net release of energy after $$(1)$$ & $$(2)$$ will be equal to the net release of energy after $$(4)$$ & $$(5)$$. $$(3)$$ and $$(6)$$ will be the same in both hypotheses. However, unlike hypothesis 1, $$(4)$$ and $$(5)$$ take place simultaneously while in hypothesis 1, $$(1)$$ takes place first and $$(2)$$ takes place second.

In this hypothesis, the reaction is fully spontaneous since heat/energy doesn not need to be added in the beginning for the reaction to proceed.

My question:

1. Which hypothesis is correct?
• In context of this question, one should distinguish thermodynamical and kinetical spontaneity. It can be thermodynamically spontaneous even if there is needed such a high activation energy the process is never observed. Like some isotopes are thermodynamically radioactive with thermodynamically spontaneous decays, but kinetically stable. Apr 14, 2022 at 8:30
• Both hypotheses are conditionally correct, as each describes a different possible scenario. I assume hypothesis 2 would still have some activation energy, but much lower than X ionization energy. Apr 14, 2022 at 8:35
• @Mithoron "That's it! No magical ionisation, no flying electrons, no ions."- I'm not confident about the ionization and flying electrons part, but I'm positive about the formation of ions: in the formation of NaCl, Na+ and Cl- ions are definitely formed and they get stuck together.
– user119245
Apr 14, 2022 at 13:13
• @Mithoron Sir, AmritanshSinghal used the Born-Haber cycle in his answer.
– user119245
Apr 14, 2022 at 23:38
• At contact of both atoms - no, it need not, as it is a different system than 2 separate atoms. Thete will be some activation energy, but lower than X ionization energy. Similarly as a planet could eventually escape a star easier, if some other star would incidentally come close to the first star. Apr 21, 2022 at 5:42

This thought experiment needs more thinking. Both scenarios are essentially Born-Haber cycles not possible mechanisms. Forming a positive ion always requires energy so that makes equations 4,5,6 not relevant to the second concerted mechanism. In the case of the concerted mechanism the X and Cl atoms must have sufficient KE to overcome electron repulsions as they approach and to continue to distort the atomic orbitals to form a molecular orbital. Ionic bonding requires a lattice energy to complete the cycle and two atoms will not suffice. Finally there must be a mechanism to remove the bond energy from the incipient bond. Radiation is a possibility.

The first proposal is not mechanistically reasonable. It requires external energy sources and the energy loss from the electron affinity is just lost and does not contribute to the overall reaction and the same applies to the final bond formation. Considering an initiated reaction such as a free radical chlorination where final lattice heat provides energy to maintain the reaction is probable but such a mechanism would not necessarily involve free chloride ions.

### There is a flaw in hypothesis 1 - getting the atoms "really close" will provide activation energy.

At sufficiently close distance the combined electron Pauli repulsion and nuclear core electrostatic repulsion will add significant energy. For example, $$2$$ protons $$7$$ angstroms apart have about $$200$$ kJ mol$$^{-1}$$ of electrostatic repulsion between them.

Also, $$200$$ kJ mol$$^{-1}$$ ionization energy is the absolute maximum activation energy required. As the two potential wells - the $$\ce{X}$$ nucleus and the $$\ce{Cl}$$ nucleus - approach, eventually the wells are close enough that an electron can tunnel from a $$\ce{X}$$ to a $$\ce{X}$$ to form $$\ce{X+}$$ and $$\ce{Cl-}$$ ions.

Alternatively, other bound electronic states across both atoms (essential a covalent bond) will form at much lower energies than $$200$$ kJ mol$$^{-1}$$ above the separated states.

Before that even, at much larger distances very weak dispersion interactions will draw neutral atoms towards one each over. The long-range potential is attractive and the first minimum is likely close enough to transfer an electron. The reaction is very likely to have no activation energy.

#### So diffusion will dominate to create dimers. Depending on the temperature I would expect it to look like hypothesis 2.

These dimers will essentially be entirely ionic - pulled as close together as possible until the Pauli repulsion of the electrons stabilises the electrostatic attractions.

As the ions are more thermodynamically stable, the balance of the depth of the attractive potential against entropy will determine the size of aggregate species - from free ions (high temperature plasma) to dimers (NaCl gas) to larger oligomers or condensation/sublimation.

• If the ionization energy of X would've been greater than the electron affinity of Cl, would the provision of ionization energy then for the reaction to proceed be non-negotiable? [Assume that the two atoms are not in contact and are a measurable distance apart]
– user119245
Apr 27, 2022 at 4:10
• If the atoms are very (infinitely) far apart and the ionization of X was greater (as is the case with all real elements), then the thermodynamically stable configuration is 2 neutral atoms. They have to be close enough for tunneling, or for some activation energy to be added as I mention - then the situation is only stabilised by the electrostatic attraction between the two charged ions. Apr 27, 2022 at 9:10