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.

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    $\begingroup$ The question makes no sense. You can't compare these two in terms of stability. $\endgroup$ – Ivan Neretin Jul 21 at 12:07
  • $\begingroup$ Why not? I mean, na cation does have a stable electronic configuration? $\endgroup$ – stonecraft bros Jul 21 at 12:09
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    $\begingroup$ Na+ does not react with Cl-. NaCl(s) is collection of alternating Na+ and Cl- ions, each surreounded by 6 opposite ions. $\endgroup$ – Poutnik Jul 21 at 12:09
  • $\begingroup$ 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 $\endgroup$ – stonecraft bros Jul 21 at 12:10
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    $\begingroup$ @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. $\endgroup$ – Poutnik Jul 21 at 12:27

I think you are mixing things up in your question and it would be helpful to specify what the question in your textbook was.

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.

Also: Sodium ions do in fact react with other (uncharged) atoms, it's just a question of a strong enough reducing agent under the right environment. Commercially, sodium was produced in the Deville process at 1100 °C with carbon as the reducing agent:

$\ce{Na2CO3 + 2 C -> 2 Na + 3 CO}$

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  • $\begingroup$ I think I understand it better now. Thanks! Just have to read up on Born-Haber cycle and stuff, and will update you $\endgroup$ – stonecraft bros Jul 21 at 15:50

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