Question

(A) (colourless solution) gives white ppt (B) with NaOH solution but ppt dissolves in excess of NaOH forming (C). (C) does not give ppt with $H_{2}S$ but on boiling with $N{H}_{4}Cl$, white ppt (B) appears. (A) also gives yellow ppt with $AgN{O}_3$.

State True or False.

(A) is $AlB{r}_3$:

• A. True
• B. False

Answer 1

The correct option is A True

Aluminum bromide (A) reacts with sodium hydroxide to give a white precipitate of aluminum hydroxide.
$AlB{r}_3+3NaOH\to \underset{whiteppt}{Al(OH{)}_3\downarrow }+3NaBr$
The precipitate dissolves in excess of NaOH to form sodium meta aluminate which is (C).
$Al(OH{)}_3+NaOH\to \underset{sodiummetaaluminate}{Na\left[Al{O}_2\right]}+2H_{2}O$
Sodium meta aluminate does not give precipitate with hydrogen sulphide.
When sodium meta aluminate is boiled with ammonium chloride, the precipitate of aluminum hydroxide is formed.
Aluminum bromide reacts with silver nitrate to form yellow precipitate of silver bromide.
$AlB{r}_3+3AgN{O}_3\to \underset{yellowprecipitate}{3AgBr}+Al(N{O}_3{)}_3$