A) \[\operatorname{B}{{\left( OH \right)}_{3}}\]
B) \[\operatorname{PO}{{\left( OH \right)}_{3}}\]
C) \[\operatorname{SO}{{\left( OH \right)}_{3}}\]
D) \[{{\operatorname{SO}}_{2}}{{\left( OH \right)}_{2}}\]
Correct Answer: A
Solution :
\[\operatorname{B}{{\left( OH \right)}_{3}}\] Boric acid \[\operatorname{B}{{\left( OH \right)}_{3}}\]behaves as weak monobasic acid. It does not act as proton donor i.e protonic acid but behaves as lewis acid. \[\operatorname{H} -OH + B {{\left( OH \right)}_{3}} \to {{\left[ B {{\left( OH \right)}_{4}} \right]}^{-}} + {{H}^{+}}\]
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