• # question_answer 83) The dissolution of $Al{{\left( OH \right)}_{3}}$ by a solution of NaOH results in the formation of: A) ${{\left[ Al{{\left( {{H}_{2}}O \right)}_{4}}{{\left( OH \right)}_{2}} \right]}^{+}}$B) $\left[ Al{{\left( {{H}_{2}}O \right)}_{3}}{{\left( OH \right)}_{3}} \right]$C) ${{\left[ Al{{\left( {{H}_{2}}O \right)}_{2}}{{\left( OH \right)}_{4}} \right]}^{-}}$D) $\left[ Al{{\left( {{H}_{2}}O \right)}_{6}}{{\left( OH \right)}_{3}} \right]$

$Al{{\left( OH \right)}_{3}}$ dissolve in NaOH solution to give $Al\left( OH \right)_{4}^{-}$ ion which is supposed to have the octahedral complex species ${{\left[ Al{{\left( OH \right)}_{4}}{{\left( {{H}_{2}}O \right)}_{2}} \right]}^{-}}$ in aqueous solution $Al{{(OH)}_{3}}\,+\,NaOH(aq)\,\,\to \,{{[\underset{(aq)}{\mathop{Al}}\,{{(OH)}_{4}}{{({{H}_{2}}O)}_{2}}]}^{-}}\,+\,N{{a}^{+}}\,{{\,}_{(aq)}}$