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Chemistry Question on Group 13 Elements

The dissolution of Al(OH)3 {Al(OH)_3} by a solution of NaOH results in the formation of

A

\ce[Al (H2O)4(OH)2]+\ce{[Al \, (H_2O)_4 (OH)_2]^{+}}

B

\ce[Al(H2O)3(OH)3]\ce{[Al (H_2O)_3 (OH)_3]}

C

\ce[Al(H2O)2(OH)4]−\ce{[Al (H_2O)_2 (OH)_4]^{-}}

D

[Al(H2O)6](OH)3]{[Al (H_2O)_6] (OH)_3]}

Answer

\ce[Al(H2O)2(OH)4]−\ce{[Al (H_2O)_2 (OH)_4]^{-}}

Explanation

Solution

Dissolution of Al(OH)3Al ( OH )_{3} by a solution of NaOHNaOH produces a complex compound shown as below: Al(OH)3+NaOH⟶[Al(H2O)2(OH)4]−Al ( OH )_{3}+ NaOH \longrightarrow\left[ Al \left( H _{2} O \right)_{2}( OH )_{4}\right]-