Question

Conjugate base of \[\left[ {{\rm{Al}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}}
\right)}_{\rm{3}}}{{\left( {{\rm{OH}}} \right)}_{\rm{3}}}} \right]\] is:
(A) \[{\left[ {{\rm{Al}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_{\rm{3}}}{{\left( {{\rm{OH}}}
\right)}_2}} \right]^ - }\]
(B) \[{\left[ {{\rm{Al}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_{\rm{3}}}{{\left( {{\rm{OH}}}
\right)}_2}{\rm{O}}} \right]^ - }\]
(C) \[{\left[ {{\rm{Al}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_{\rm{3}}}{{\left( {{\rm{OH}}}
\right)}_{\rm{3}}}} \right]^ - }\]
(D) \[{\left[ {{\rm{Al}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_2}{{\left( {{\rm{OH}}}
\right)}_4}} \right]^ - }\]

Answer 1

Hint: As we know that, according to the Bronsted-Lowry acid base concept, a base has a tendency to attract protons and an acid has a tendency to release a proton. There are other theories for acid and base such as Arrhenius theory of acid and base.

Step by step answer: Now, let’s discuss the Bronsted-Lowry acid base concept in detail.
According to the Bronsted-Lowry acid base concept, a base forms conjugate acid by accepting a proton from an acid and an acid forms conjugate base by releasing a proton to a base. It is also important that if the base is strong then its conjugate acid will be weak and if the acid is weak then its conjugate base will be strong.
In this question, water molecules are there and hydroxyl groups as well. According to Arrhenius theory of acid and base water when accepts proton forms hydronium ion which is also conjugate acid of water and when it releases proton forms hydroxyl ion which is also conjugate base of water.
Now, it is clear from above explanation that when water molecule of the given question releases a proton forms conjugate base, so one water molecule of water will be changed into hydroxyl ion as-
\[
\left[ {{\rm{Al}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}{\rm{O}}{{\left( {{\rm{OH}}} \right)}_{\rm{3}}}} \right]\overset{ -H+}{\rightarrow} \left[ {{\rm{Al}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_{\rm{2}}}^ - {\rm{OH}}{{\left( {{\rm{OH}}} \right)}_{\rm{3}}}} \right]\\
\left[ {{\rm{Al}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_{\rm{2}}}^ - {\rm{OH}}{{\left( {{\rm{OH}}} \right)}_{\rm{3}}}} \right] \equiv {\left[ {{\rm{Al}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_{\rm{2}}}{{\left( {{\rm{OH}}} \right)}_{\rm{4}}}} \right]^ - }
\]
Therefore, the correct option is option(D).

 Note: There is also another acid base concept which was given by Lewis. So, according to Lewis concept a Lewis base has a tendency to donate a lone pair of electrons and a Lewis acid has a tendency to accept a lone pair of electrons.