The atomic number of Na is 11 (2, 8, 1). Na loses an electron to achieve octet stability but why can't it just accept an electron to complete duplet stability?
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4$\begingroup$ Why, of course it can. The problem is, there are not enough free electrons for everyone. $\endgroup$– Ivan NeretinFeb 9, 2017 at 8:05
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$\begingroup$ @Ivan Neretin Would you elaborate "there are not enough free electrons for everyone. "? $\endgroup$– MockingbirdFeb 9, 2017 at 9:26
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$\begingroup$ I mean, all $\ce{Na}$ can't convert to $\ce{Na-}$. But I guess that's not what the question was about, anyway. $\endgroup$– Ivan NeretinFeb 9, 2017 at 9:34
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3$\begingroup$ Possible duplicate of Is it possible to have a diatomic molecule of sodium in gaseous state? $\endgroup$– bobthechemistFeb 9, 2017 at 13:12
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1$\begingroup$ @bobthechemist The questions are completely different even though they might have a similar answer. It is absolutely not the same asking "Why doesn't it accept an electron?" and "Does a dimer exist?" $\endgroup$– Martin - マーチン ♦Feb 9, 2017 at 14:39
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2 Answers
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Well, sodium does have non-zero electron affinity, so it surely can acquire another electron, provided it can get one for free. That's what happens when a lone sodium atom meets a lone electron somewhere in deep space. But here on Earth, with all that condensed matter around, we don't have many free electrons (*). They all belong to some other atoms, and if you want to get an electron (that is, to get reduced), you have to oxidize something else.
(*) Then again, some might recall metals in which the electrons are kinda free, but wait: they are not really free. They are bound by the collective field of metal cations, and you need to apply certain energy to take one out.
It turns out that $\ce{Na}$ and other alkali metals can actually be reduced to form alkalides, but those are exotic compounds. They are not particularly stable, and get oxidized with pretty much anything. It is much easier for $\ce{Na}$ to form a cation, thus achieving the stable octet, and call it a day.
There is a whole different dimension to the problem: why wouldn't two sodium atoms share their electrons, establish a bond, and form a diatomic molecule? Again, they would, but only in a vacuum! When you have many such molecules, the sodium atoms would share their other empty orbitals, find out they don't have enough electrons to fill them all, and just resort to being a metal.
answered Feb 9, 2017 at 10:40
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4$\begingroup$ Yes, I believe the species $\ce{Na2}$ (and other such diatomic alkali metals) have been observed in the gas phase, as is predicted by qualitative MO diagrams. $\endgroup$ Feb 9, 2017 at 11:31
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$\begingroup$ Published a little after all the above: Hyperfine structure of alkali-metal diatomic molecules J. Aldegunde, J. M. Hutson arxiv.org/abs/1708.05734 $\endgroup$– iSeekerAug 22, 2020 at 18:39
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Sodium not only forms species with two electrons in the $3s$ subshell ($\ce{Na^-,Na2}$). It is also known to form a dimer of the $\ce{Na^-}$ion, even though the individual atoms therein possess closed-shell electronic structures and would not be expected to form a covalent bonds to offset the repulsion between the negative charges.
The dimeric $\ce{Na_2^{2-}}$ anion occurs in barium azacryptand sodide [1]. The crystal structure, pictured as shown below from [1], features the large, blue atoms which are the paired sodium atoms in this dianion (red atoms are barium, the small green atoms are from the cryptand). The large atomic size, corresponding to a highly diffuse electron cloud, reduces the like-charge repulsion while improving polarizability for a VdW attraction.
Reference
1. Mikhail Y. Redko, Rui H. Huang, James E. Jackson, James F. Harrison, and James L. Dye, "Barium Azacryptand Sodide, the First Alkalide with an Alkaline Earth Cation, Also Contains a Novel Dimer, (Na2)2-", J. Am. Chem. Soc. 2003, 125, 8, 2259–2263. https://doi.org/10.1021/ja027241m
answered Jul 28, 2021 at 0:15