Question

In group III, the basic radicals are precipitated as their ____.
(A) Hydroxide
(B) Carbonate
(C) Sulphate
(D) All of the above

Answer 1

Salt contains two radicals- acidic and basic. Precipitation reaction is used to determine these radicals. In group III, $N{H_4}OH$ is used in presence of $N{H_4}Cl$ as a regent in order to determine the basic radical.

Complete Solution :
-In salt analysis, the salt contains two radicals- acidic and basic. The basic part is the cationic part and the acidic radicals are the anions.
-Analysis of basic radicals involves following steps:
(i) We have to prepare aqueous solution or original solution.
(ii) Then Separate the basic radicals present in different groups.
(iii) Finally Analyze the precipitate obtained for each group. When substance is changed into the original solution, its constituents are ionized and each basic radicals a specific route of tests is used.
-Now, let us discuss the group III cations.
The group III cations include the ${\rm{A}}{{\rm{l}}^{{\rm{3 + }}}}{\rm{,F}}{{\rm{e}}^{{\rm{3 + }}}}{\rm{ and C}}{{\rm{r}}^{{\rm{3 + }}}}$.

-Now, we add the reagent used that is $N{H_4}OH$ is used in presence of $N{H_4}Cl$. Let us see the reactions now.
$A{l^{3 + }} + N{H_4}OH \to Al{(OH)_3}$
Here, the Aluminium hydroxide gives a white precipitate, thus confirming its presence.
$F{e^{3 + }} + N{H_4}OH \to Fe{(OH)_3}$
Here, the$Fe{(OH)_3}$ gives a red precipitate, thus confirming its presence.
$C{r^{3 + }} + N{H_4}OH \to Cr{(OH)_3}$
Here, the Chromium hydroxide gives a green color, thus confirming its presence.
Clearly, we can see that the group III basic radical forms hydroxide and confirms its presence.
So, the correct answer is “Option A”.

Note: ${K_{sp}}$ is called the solubility product. Thus, the ${K_{sp}}$ values of $Al{(OH)_3}$ and the other same group hydroxides are low. $N{H_4}Cl$ produces a common ion effect with $N{H_4}OH$ and thus decrease the ionization of $N{H_4}OH$, thus producing lower concentrations of $O{H^ - }$ ions.