Question

Aluminate ion in aqueous solution actually exists as:
A. \[{\left[ {Al{{\left( {OH} \right)}_6}} \right]^{3 - }}\]
B. \[{\left[ {Al{{\left( {OH} \right)}_4}{{\left( {{H_2}O} \right)}_2}} \right]^ - }\]
C. \[\left[ {Al{{\left( {OH} \right)}_3}{{\left( {{H_2}O} \right)}_3}} \right]\]
D. \[{\left[ {Al{{\left( {OH} \right)}_5}{H_2}O} \right]^{2 - }}\]

Answer 1

Hint: Aluminate ion is an ion which is composed of more than one element i.e. multiple elements. So, the aluminate ion is a polyatomic ion. It forms a complex compound so can be found in complex structures.

Step by step answer: Aluminate ion contains one atom of aluminium and two atoms of oxygen. We know that aluminium is a metal, so aluminium will be electropositive by nature. That is why in aluminate ions, aluminium donates 3 electrons in order to obtain the inert state or say a complete valence shell and gets a charge of \[ + 3\]. Whereas the oxygen atom gains two electrons to complete its valence shell and obtains a charge of \[ - 2\].
\[A{l^{ + 3}} + {\rm{ 2}}{O^{2 - }} \to {\left( {Al{O_2}} \right)^{1 - }}\]
It is clear that since the aluminate ion is composed of one aluminium and two oxygen atoms, the overall charge on the aluminate ion will be $4O{H^ - }$ $ - 1$. So, we can conclude that aluminate ion in isolated form will be as $Al{O_2}$ $AlO_2^ - $. When the aluminate ion gets into the aqueous solution, it gets surrounded by the four water molecules i.e. ${H_2}O$ molecules. The two oxygen atoms present in the aluminate ion react with the two molecules of water to form $4O{H^ - }$ ions i.e. hydroxyl group. Now, the remaining two molecules of water form the coordination compound with the four hydroxyl groups and thus the aluminate ion complex in aqueous solution is obtained.

Hence, option (B) \[{\left[ {Al{{\left( {OH} \right)}_4}{{\left( {{H_2}O} \right)}_2}} \right]^ - }\], is the correct option.

Note: Aluminate ion is a polyatomic ion composed of two atoms of oxygen and one atom of aluminium with formula $AlO_2^ - $-. In aqueous solution the aluminate ion forms $4$ hydroxyl groups which later on react with two molecules of water and form complexes with coordination bonds. So, the aluminate ion in aqueous solution exists as a coordination complex with formula\[{\left[ {Al{{\left( {OH} \right)}_4}{{\left( {{H_2}O} \right)}_2}} \right]^ - }\].