(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 silver nitrate. State true or false.
(A) is $AlB{r_3}$ :
A. True
B. False

Hint: The detection of cations(basic radicals) and anions(acidic radicals) in a salt or in mixture is known as qualitative analysis. There are various methods by which we can detect the cation or anion in the mixture by noting its behavior with certain reagents.

Complete Step by step answer: we have been given the salt aluminium bromide and we have to find out whether it is the compound (A) and behaves in the similar way as stated above when reacted with the reagents as mentioned. Let us therefore detect the cation and anion in the salt, which is $A{l^{3 + }},B{r^ - }$.
If we assume that the compound (A) is $AlB{r_3}$ , then it reacts with NaOH to give a white ppt:
$AlB{r_3} + 3NaOH \to Al{(OH)_3} + 3NaBr$

$Al{(OH)_3}$ thus formed gives gelatinous white ppt. and therefore is compound (B). we know that white ppt. of $Al{(OH)_3}$ is soluble in NaOH. Since $A{l^{3 + }}$ is a group –III ion, it gets precipitated as their hydroxides in presence of group reagent $N{H_4}Cl/N{H_4}OH$. On dissolving in NaOH,

$Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O$

Then according to the question compound (C) is $NaAl{O_2}$. ${H_2}S$ is a colourless gas with rotten egg smell. $NaAl{O_2}$ does not give any ppt. with hydrogen sulphide but on boiling with $N{H_4}Cl$ it gives white ppt. of aluminium hydroxide thus formed which is (B). The compound has bromide anion which reacts with silver nitrate to give yellow ppt.
$AlB{r_3} + 3AgN{O_3} \to 3AgBr + Al{(N{O_3})_3}$
The silver bromide thus formed gives yellow residue.

Hence we can conclude that our assumption is correct and the statement is true and the correct option is A.

Note: Group –III radicals are precipitated as hydroxides and the $N{H_4}Cl$ suppresses the ionization of ammonium hydroxide so that only the group –III radicals are precipitated because of their low solubility product. Group-III radicals contain $F{e^{3 + }},A{l^{3 + }},C{r^{3 + }}$ .