First, I understand that, on $\ce{BH3}$, they eletronegativity difference between $\ce{B}$ and $\ce{H}$ is very small. Which mean that the $\ce{B-H}$ bond is less polarized. But, I couldn't find a relation of this, to the fact that he is a Molecular hydride. And, secondly, I couldn't understand why $\ce{AlH3}$ its "intermediary", and not molecular, as well...

  • $\begingroup$ (What do you mean by "relation"?) BH3 exists as the gaseous dimer molecule diborane. AlH3 is a solid that decomposes at 100°C. It's not molecular. $\endgroup$ – Karl Jun 28 '17 at 20:07
  • $\begingroup$ I see. But, i still don't understand why this happen, you know? $\endgroup$ – Bruno Jun 28 '17 at 20:27
  • $\begingroup$ "Intermediary" is meant in comparison to BH3 on one side and "real" metall hydrides on the other. The differences are explained in any advanced inorganic chemistry textbook. $\endgroup$ – Karl Jun 28 '17 at 20:42

Alane is not a molecule: https://en.wikipedia.org/wiki/Aluminium_hydride You can't isolate molecules easily because they polymerize.

Borane on the other hand is known to exist as a dimer: https://en.wikipedia.org/wiki/Diborane

The difference in bonding is likely due to "hypervalency" (or similar concept with different terminology) in aluminum.

  • $\begingroup$ Article you linked shows that alane monomer was isolated and how its polymers bonding looks, so I encourage you to expand answer. $\endgroup$ – Mithoron Jun 28 '17 at 20:32

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