• # question_answer Diborane $({{B}_{2}}{{H}_{6}})$reacts independently with ${{O}_{2}}$and ${{H}_{2}}O$ produce, respectively: A) ${{H}_{3}}B{{O}_{3}}$ and ${{B}_{2}}{{O}_{3}}$ B) ${{B}_{2}}{{O}_{3}}$ and ${{H}_{3}}B{{O}_{3}}$ C) ${{B}_{2}}{{O}_{3}}$ and ${{[B{{H}_{4}}]}^{\,-}}$ D) $HB{{O}_{2}}$ and ${{H}_{3}}B{{O}_{3}}$

${{B}_{2}}{{H}_{6}}+{{O}_{2}}\to {{B}_{2}}{{O}_{3}}$ Boric anhydride ${{B}_{2}}{{H}_{6}}+{{H}_{2}}\to {{H}_{3}}B{{O}_{3}}$ Boric acid.