Take sodium iodide for example. I understand both elements can obtain a full outer shell via the electron transfer. But doesn't the electron move from a lower-energy orbital ($3s$ in sodium) to a higher-energy one ($5p$ in iodine)? So how could that transfer be energetically favorable? Yet it is exothermic, with enthalpy of formation of $\Delta H_f = -287\ \text{kJ/mol}$.

I don't understand how having a full outer shell inherently lowers the energy of the system. I guess that means it's not just about the energy of the electron that is transferred -- it must be that the presence of that electron lowers the energies of the other electrons in the valence shell as well. Maybe this is partially about pairing of up/down spins, but wouldn't that only matter for filling one orbital, not the full shell? What's the deal?

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    $\begingroup$ 5p of some atom is higher in energy that 3s of the same atom. With different atoms, not necessarily so. $\endgroup$ Commented Oct 4, 2022 at 18:07
  • $\begingroup$ @IvanNeretin Oh okay -- so what if I ask the reverse question -- why would cesium chloride formation be exothermic when the electron is going from $6s$ in cesium (many protons, hence lower energy) to $3p$ in chlorine (fewer protons, higher energy)? $\endgroup$ Commented Oct 4, 2022 at 19:00

3 Answers 3


Generally, forming 2 isolated ions from 2 isolated atoms is not thermodynamically favoured, as ionization energies are higher than electron affinities.

But, ion formation is thermodynamically favoured, if the above is backed up by releasing of energy due ion pairing, ionic lattice formation or ion solvation.


I don't understand how having a full outer shell inherently lowers the energy of the system.

This depends on which exact systems are being compared. For your example, the ions with two full valence shells, Na+ and I–, need to be compared to the atoms Na and I where the valence shell in Na has one electron and the valence shell in I is one electron away from being full. Energy changes are about the total energy after an event minus the total energy before the event. You can't just think about the position of that one electron like you can in hydrogen-like atoms.

Some quantum chemistry considerations can be taught to general chemistry students with hand-waving arguments like "noble gas electron configurations are low energy configurations" as long as the reason for those arguments is also taught ("this is due to the Schrödinger model of the atom.") More advanced undergraduate students and graduate students might learn that approximate methods like Hartree-Fock for many-electron atoms are needed instead of finding actual solutions to the Schrödinger equation, like what's possible for hydrogen-like atoms with one electron.

General chemistry students should recognize that the process you're describing requires multiple steps other than the electron transfer. The electron transfer is described by the ionization energy of a sodium atom (496 kJ/mol) and the electron affinity of an iodine atom (-295 kJ/mol). Those two things happening together in the absence of other processes actually would not be energetically favorable. But the enthalpy of formation of sodium iodide also involves converting solid iodine to gaseous I2, breaking the bond in I2 to get individual I atoms, converting solid sodium to gaseous sodium to get individual Na atoms, then applying the ionization and electron affinity values to create two ions before finally considering the lattice energy of letting these gaseous ions come together to form solid NaI (–700 kJ/mol). The value you wrote of –287 kJ/mol for the enthalpy of formation of NaI should be the sum of all those values when Hess's Law is applied.


Essentially, metals will form ionic compounds with non-metals of a lower period because the energy released in the formation of the bond is vastly greater than the energy required to promote an electron.

Reactions which cause an increase in entropy (I.e. kinetic instability) or a decrease in enthalpy (I.e. Thermodynamic instability) will be favourable and will naturally occur.

In this case, only considering enthalpy is particularly relevant.

So a decrease in enthalpy will occur when the energy of the products is less than the energy of the reactants, and if this is the case, the reaction will be favourable and occur naturally.

It is important to note that as we’re talking about an ionic bond, e.g. NaI, only one electron from the sodium atom is given to the iodide atom.

So you’re correct, that promoting this electron requires energy (or is an endothermic process in itself) because energy levels are quantised, and so the electron must be promoted to a higher level to be a part of the iodine ion.

However, the energy released by the formation of the ionic bond is vastly greater than the energy required to promote the electron, and so the net change of the reaction will be that the energy released is greater than the energy supplied, and so the reaction is exothermic, leads to a decrease in enthalpy, and so is favourable.

The presence of the additional electron does not decrease the energy of the other valence electrons as they are quantised and do not change their energy level.

The concept of bond formation releasing energy can be conceptually difficult to grasp, so it is best to think of bond formation as the opposite of bond-breaking, which it inherently is. And since it will obviously require energy to be supplied in order to break a bond, by extension, forming a bond releases energy.


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