$$\ce{AlF3 + NaF \to Na3[AlF6]}$$
My question is why don't we get $\ce{Na[AlF4]}$, since when we do the same thing with hydroxides instead of fluorides ($\ce{Al(OH)3}$), we end up with $\ce{Na[Al(OH)4]}$. I know that it is a 3rd period element so it can expand its octet, so boron being in the same group forms $\ce{NaBF4}$. I've also heard that aluminium is larger so it can accommodate more fluorine atoms around it. But why would it want to do that?
In fact sodium tetrafluoroaluminate can be formed[1]. However, it requires quenching vapor from a temperature of 1000°C, definitely a non-equilibrium process.
The formation of the hexafluoroaluminate under more common conditions correlates with two features of this compound. First, it can form a relatively stable solid lattice, which formerly appeared in nature as the mineral cryolite. (Mining of this mineral for use in aluminum smelting has made it rare, and now the flux is made synthetically for industrial use.) The stability of the lattice is evident from cryolite being essentially insoluble in plain water, a rarity for sodium salts.
Second, $\ce{Al-F}$ bonding is more amenable to forming an octahedral complex than $\ce{Al-OH}$ or $\ce{B-F}$. As explained here, when an element with $s$ and $p$ valence orbitals forms an "extended octet" species such as $\ce{SF6}$ or the formally isoelectronic $\ce{AlF6^{3-}}$, there are antibonding interactions between the ligands which must be minimized. Using a larger central atom (aluminum instead of boron) and ligands that hold the electrons more tightly (fluoride instead of hydroxide) does that minimization.
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