# Is Na$^+$ more stable than Na? [duplicate]

I had a question in my textbook:-

Na -> Na$$^+$$ + e$$^-$$

Cl + e$$^-$$ -> Cl$$^-$$

But Na$$^+$$ and Cl$$^-$$ have a stable electronic configuration. So, why do they react to form NaCl?

So, this is my question. Is Na$$^+$$ really more stable than Na?

Na$$^+$$ doesn't react with atoms, only with ions, like Cl$$^-$$. Moreover, Na$$^+$$ has a noble gas electronic configuration, and Na is highly reactive BECAUSE of the extra electron in its valence shell. So, in this respect, Na$$^+$$ should be more stable, right?

Edit: I looked this up on StackExchange, and got varying answers. That's why I'm confused.

• The question makes no sense. You can't compare these two in terms of stability. – Ivan Neretin Jul 21 at 12:07
• Why not? I mean, na cation does have a stable electronic configuration? – stonecraft bros Jul 21 at 12:09
• Na+ does not react with Cl-. NaCl(s) is collection of alternating Na+ and Cl- ions, each surreounded by 6 opposite ions. – Poutnik Jul 21 at 12:09
• So, is the question given in my book wrong? I suppose they were trying to ask why does NaCl form, when Na+ and Cl- already have stable electronic config – stonecraft bros Jul 21 at 12:10
• @stonecraft bros If you had 2 sets of balls charged positively and negatively, what would they do if you mixed them together ? NaCl as a molecule does not exist. Even in a gaseous or plasmatic phase, it is a ion pair. If dissolved in water, you get separated independent hydrated ions. If melted, then again independent "naked" ions. – Poutnik Jul 21 at 12:27

NaCl is a solid made up from Na$$^+$$ and Cl$$^-$$ ions, not from uncharged atoms as stated in your question, arranged in a periodic crystal lattice. So the redox equations for oxidation and reduction in your question are not relevant for the formation of NaCl from its ions e.g. from a supersaturated solution, where the formation of the solid releases lattice energy. Although the redox equations are relevant, when looking at the formation of NaCl$$_{(s)}$$ from its elements Na and Cl, which you rightly pointed out yield the favorable noble gas configuration for both. (See Born-Harber cycle for more information on the formation of salts from its elements and the related energies)
As stated in Nilay Ghosh's comment, a question (and answer) comparing the stability of Na vs Na$$^+$$ can be found here.
$$\ce{Na2CO3 + 2 C -> 2 Na + 3 CO}$$