Why is it that Boron does not form Diborane with water rather than $\ce{B2O3}$?


1 Answer 1


I assume the hypothetical reaction you are asking about is:

$\ce{3H2O + 2B -> B2H6 + 3/2O2}$

under standard conditions.

The standard enthalpy of formation of water is −285.8 kJ/mol, whereas that of diborane is 35.4 kJ/mol; this reaction would therefore be very endothermic and hence unfavourable. You're breaking up strong O-H bonds and B-B bonds to form B-H bonds (including two 3c2e bonds, which are very weak) and weak O-O bonds. In contrast the formation of boron trioxide by reaction of boron and water is very exothermic, and has been considered in potential applications of boron as a rocket fuel.

  • $\begingroup$ Whether 3c-2e bonds are "very weak" depends on what you're comparing them to. A 3c-2e bond is weaker than two separate bonds joining the atoms, but stronger than one ordinary pairwise bond would be. $\endgroup$ Dec 25, 2020 at 13:51

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