- Home
- Standard 11
- Chemistry
p-Block Elements - I
normal
The dissolution of $Al(OH)_3$ by a solution of $NaOH$ results in the formation of
A
$[Al(H_2O)_4(OH)_2]^+$
B
$[Al(H_2O)_3(0H)_3]$
C
$[Al(H_2O)_2(OH)_4]^-$
D
$[Al(H_2O)_6(OH)_3]$
Solution
$Al(OH)_3$ dissolves in $NaOH$ solution to give $Al(OH)_4^-$ ion which is supposed to have the octahedral complex species $[Al(OH)_4(H_2O)_2]^-$ in aqueous solution.
$Al{\left( {OH} \right)_3} + NaOH\left( {aq} \right) \to {\left[ {Al{{\left( {OH} \right)}_4}{{\left( {{H_2}O} \right)}_2}} \right]^ – }\left( {aq} \right) + N{a^ + }\left( {aq} \right)$
Standard 11
Chemistry
