Question

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

• A. $\left[Al\right(H_{2}O{)}_2(OH{)}_4{]}^{-}$
• B. $\left[Al\right(H_{2}O{)}_3(OH{)}_3{]}^{+}$
• C. $\left[Al\right(H_{2}O{)}_4(OH{)}_2{]}^{+}$
• D. $\left[Al\right(H_{2}O{)}_6(OH{)}_3{]}^{-}$

Answer 1

The correct option is A $\left[Al\right(H_{2}O{)}_2(OH{)}_4{]}^{-}$

1. $\left[Al\right(H_{2}O{)}_4(OH{)}_2{]}^{+}$

2. $Al(OH{)}_3$ dissolves in $NaOH$ solution to give $Al(OH{)}^{-4}$ ion which is supposed to have the octahedral complex species $\left[Al\right(OH{)}_4(H_{2}O{)}_2{]}^{-}$ in aqueous solution.

3. $Al(OH{)}_3+NaOH(aq)⟶[Al\left(OH{)}_4\right(H_{2}O{)}_2{]}^{-}\left(aq\right)+NaO{H}^{+}\left(aq\right)$