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 \[{{H}_{3}}B{{O}_{3}}\]
Correct Answer: B
Solution :
\[{{B}_{2}}{{H}_{6}}+{{O}_{2}}\to {{B}_{2}}{{O}_{3}}\] Boric anhydride \[{{B}_{2}}{{H}_{6}}+{{H}_{2}}\to {{H}_{3}}B{{O}_{3}}\] Boric acid.
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