11th Class Chemistry The p-Block Elements-I

  • question_answer 77) When \[BC{{l}_{3}}\] is treated with water, it hydrolyses and forms\[{{[B{{(OH)}_{4}}]}^{-}}\] only whereas \[AlC{{l}_{3}}\] in acidified aqueous solution forms \[{{[Al{{({{H}_{2}}O)}_{6}}]}^{3+}}\] ion. Explain what is the hybridisation of boron and aluminium in these species?

    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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