The precipitate of $\mathrm{Al}(\mathrm{OH})_{3}$ dissolves in $\mathrm{NaOH}$ solution. It is due to the…

The precipitate of $\mathrm{Al}(\mathrm{OH})_{3}$ dissolves in $\mathrm{NaOH}$ solution. It is due to the formation of
  1. $\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}ight)_{4}(\mathrm{OH})_{2}^{+}$
  2. $\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}ight)_{3}(\mathrm{OH})_{3}$
  3. $\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}ight)_{2}(\mathrm{OH})_{4}^{-}$
  4. $\mathrm{Al}(\mathrm{OH})_{6}^{3-}$

Solution

Dissolution of $\mathrm{Al}(\mathrm{OH})_{3}$ by a solution of $\mathrm{NaOH}$ produces a complex compound shown as below:
$\mathrm{Al}(\mathrm{OH})_{3}+\mathrm{NaOH} \longrightarrow\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}ight)_{2}(\mathrm{OH})_{4}ight]^{-}$ /

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more P-BLOCK ELEMENTS GROUP 13 & 14 questions on Aicharya