Answer:
\[BC{{l}_{3}}+3{{H}_{2}}O\xrightarrow{{}}B{{(OH)}_{3}}+3HCl\]
\[B{{(OH)}_{3}}+2{{H}_{2}}O\xrightarrow{{}}{{[B{{(OH)}_{4}}]}^{-}}{{H}_{3}}{{O}^{+}}\]
\[\ln {{[B{{(OH)}_{4}}]}^{-}}\],
boron is in \[s{{p}^{3}}\]-hybrid state.
\[AlC{{l}_{3}}+9{{H}_{2}}O\to
[Al{{({{H}_{2}}O)}_{6}}]{{(OH)}_{3}}+3HCl\]
In\[{{[Al{{({{H}_{2}}O)}_{6}}]}^{3+}}\],
aluminium is in \[s{{p}^{3}}{{d}^{2}}\]-hybrid state.
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