Relation between ionisation enthalpy and nature of bonds in boron
At first, let me give a picture of what ‘relatively high’ actually means. I present the sum of the first three ionisation enthalpies of some elements typically found as metal(III) ions:


Boron $6887.4~\mathrm{\frac{kJ}{mol}}$
Aluminium $5139~\mathrm{\frac{kJ}{mol}}$
Scandium $4256.7~\mathrm{\frac{kJ}{mol}}$
Iron $5281.4~\mathrm{\frac{kJ}{mol}}$
Cobalt $5640.4~\mathrm{\frac{kJ}{mol}}$
Ytterbium $4195.2~\mathrm{\frac{kJ}{mol}}$


We see that boron’s ionisation energy sum is $1.2~\mathrm{\frac{MJ}{mol}}$ larger than that of the next largest in my sample. Also keep in mind that boron is rather small per se so a $\ce{B^3+}$ ion must be even smaller.

But what does that actually mean? To generate an ionic compound from its elements, one usually considers an overall formation enthalpy composed of bringing both elements into the gas phase, turning them into ions and letting the crystal lattice form — this last step liberates most energy and is usually the driving force for salt creation. In boron’s case, the cations of such a salt would be $41~\mathrm{pm}$ — exceptionally small; way smaller than $\ce{Al^3+}$ ($68~\mathrm{pm}$) which is already tiny and smaller than $\ce{Be^2+}$ whose $59~\mathrm{pm}$ are typically cited as the smallest ionic radius known ($\ce{H+}$ doesn’t count).[1]

The structure of $\ce{BeF2}$ can already be considered as the greatest possible compromise in keeping the cations and — more importantly — the anions apart from themselves; hence its spacious structure. These problems would be multiplied if an even smaller cation were used here: It would be extremely hard to generate a crystal lattice that would allow for sufficient separation of the anions from each other since the cation is so tiny (and we need lots of anions: three negative charges per boron!).

So all things considered, ionic boron salts would have a very unstable crystal structure and would hence be hard to generate, especially since the lattice enthalpy cannot counteract the large ionisation enthalpy required for boron adequately. Luckily, covalent compounds are always a possibility and boron’s atomic orbitals are well shaped and well placed in energy to allow covalent bonding.

Note that boron’s electronegativity of $2.0$ on the Pauling scale also shows how badly boron would form ionic bonds: The tiny sizes of ions would draw electrons in in an even stronger way, potentially reducing boron before it would reach the ionic state.



The Polish chemist Kazimierz Fajans simplified this in his set of Fajans’ rules:


A compound will be covalent, if:


there is a high positive charge;
the anion is large; and/or
the cation is small.

A compound will be ionic, if:


there is a low positive charge;
the anion is small; and/or
the cation is large.



Note that in ambiguous cases such as $\ce{AlF3}$, (small cation, high positive charge but small anion) the rules fail to predict a mainly ionic compound, but correctly make us assume a rather high covalent character of said ionic bonds.



[1] Even beryllium forms rather covalent compounds most of the time. Due to the electronegativity difference of $2.5$ on the Pauling scale, we may just be able to consider solid $\ce{BeF2}$ as an ionic compound. All of these are borderline cases in one way or the other, though.