# Why does boron form only covalent compounds? [duplicate]

I got two options,

1. The size of $$\ce{B^3+}$$ ions, is very small and the ion has high charge density. Thus, following Fajan's rule, boron forms covalent compounds.
2. The sum of the first three ionisation enthalpies of boron is too high, thereby restricting the formation of $$\ce{B^3+}$$ ions. Thus, boron is forced to form covalent bonds with other compounds.

Which one is a better explanation?

• Some borides can be considered metallic. – Mithoron Jun 12 at 18:02
• Another thing is what you consider a covalent compound... – Mithoron Jun 12 at 18:06
• – Nilay Ghosh Jun 13 at 2:33

## 2 Answers

Another explanation.

It costs something to remove one election from an neutral atom, whatever the nature of this atom. It costs more to remove a second electron from an atom that has already lost one electron. But it is feasible. As a consequence, removing a third electron form an atom that has already lost two electrons may be too much, if there is another possibility like covalences. This is the reason why Boron makes $$\ce{BF3}$$ or $$\ce{BCl3}$$. The real question is : Why does Boron chloride not make dimers like $$\ce{B2H6}$$ or Aluminum chloride, which is made of covalent $$\ce{Al2Cl6}$$ ?

To say that boron forms only covalent compounds is an oversimplification. It can be involved in metallic or even predominantly ionic bonding, especially when combined with elecyropositive elements like magnesium. Compounds such as magnesium diboride ($$\ce{MgB2}$$), discussed in the referenced answer, or calcium hexaboride ($$\ce{CaB6}$$) are best described as having covalent bonds between boron atoms, but mostly ionic bonds between the boron (which is assembled into an electron-deficient macromolecular structure) and the metal. Transition metals are in general less electropositive, and similar borides of these metals have more metallic bonding between boron and the metal.