tylerngo
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Why are lithides not known?
In the last few decades, many alkalides - anions of alkali metals - have been synthesised. The most famous is undoubtedly that of sodium:
[Na(2.2.2-cryptand)]+Na−, but the alkalides K−, Rb−, and Cs− are all known. However, Li−
is not.
James Dye writes in a 1984 review1 that for the product of the reaction
M(s)+N(s)+L(s)⟶[ML]+N−(s)
to be thermodynamically stabilised (M
, N are metals, and L
is the macrocyclic ligand), several criteria must be met:
Small lattice energies for L(s)
, M(s), and N(s)
so that the enthalpies of sublimation will not be too large.
Low ionization energy of M
.
High electron affinity of N
.
Large complexation energy of M+
by L
.
Large lattice energy of [ML+]N−
(which depends mainly on the interionic separation).
From these given factors, I suppose the only possible explanation is the larger heat of sublimation of lithium. However, lithium is still known to
form electrides in solution - compounds of the form [ML+]e−
. In order for these to be formed, the sublimation energy of Li
still has to be overcome.
Is there a thermodynamic reason why lithides, Li−
, have not yet been made?
Reference
Dye, J. L. Electrides, negatively charged metal ions, and related phenomena. Prog. Inorg. Chem. 1984, 32, 327–441. DOI:
10.1002/9780470166338.ch4.
Ta!
Brumer
[Edited on 11-10-2018 by tylerngo]
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Texium
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Thread Moved 11-10-2018 at 06:12 |
Assured Fish
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This exact question was asked at chemistry stack exchange.
https://chemistry.stackexchange.com/questions/73686/why-are-...
Twas also answered there as well.
Quote: |
Not too complicated of an answer - don't make a mountain out of a mole-hill. This is simple nuclear stability. You can just google "Nuclear Band of
Stability" and it explains the necessary balance between protons, neutrons and electrons. Li-1 would have 3 protons, 4 neutrons, and 4 electrons.
Crowded around such a small nucleus, the electrons would experience too much repulsion which would cause at least one to be ejected from the cloud.
Additionally, 3 measly protons does not suffice for another electron, at a greater distance, experiencing electron repulsion, to experience an even
greater force by effective nuclear charge.
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