A) By donating \[{{H}^{+}}\]
B) By accepting \[O{{H}^{-}}\]
C) By donating lone pair of electrons
D) By accepting lone pair of electrons.
Correct Answer: B
Solution :
[b] \[B{{(OH)}_{3}}+2{{H}_{2}}O\rightleftharpoons {{[B{{(OH)}_{4}}]}^{-}}+{{H}_{3}}{{O}^{+}}\]
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