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

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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.

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