How do we differentiate between \[F{e^{3 + }}\] and \[C{r^{3 + }}\] in group \[III\]?
A. By taking excess of \[N{H_4}OH\]
B. By increasing \[N{H_4}^ + \]ion concentration
C. By decreasing \[O{H^ - }\] ion concentration
D. Both \[(b)\]and \[(c)\]

Hint: The group reagent for group \[III\] is \[N{H_4}OH\] in presence of \[N{H_4}Cl\], which leads to formation of their respective hydroxides\[ - {\text{ }}Fe{\left( {OH} \right)_3}\]and \[Cr{\left( {OH} \right)_3}\]. Now minimum quantity of \[dil.\,HCl\] is added and then excess of \[NaOH\] is added boil and filter. The filtrate contains sodium aluminate which separates \[A{l^{3 + }}\] ion another member of this group, and residue is then tested for \[F{e^{3 + }}\] and \[C{r^{3 + }}\].

Complete Step-by-Step Explanation:
In this question \[F{e^{3 + }}\]and \[C{r^{3 + }}\]are given, an aqueous mixture of \[C{r^{3 + }},\,A{l^{3 + }},\,F{e^{3 + }}\]and \[N{i^{2 + }}\] is first treated with a mixture of \[NaOH\] and \[NaOCl\] solutions.
This is cause to the iron and the nickel cations to the precipitation out as the hydroxide salts while the chromium and the aluminium cations can remains in solution,

We can test for \[F{e^{3 + }}\], then we have: Dissolve a part of residue in \[dil.\,HCl\] and solution is divided into two parts:- To first part add potassium ferrocyanide, a Prussian blue colour ppt. of ferric-ferrocyanide is obtained as Prussian blue:
\[FeC{l_3}{\text{ }} + \;3{K_4}Fe{\left( {CN} \right)_6}{\text{ }} - - - - - - - - - > {\text{ }}F{e_4}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_3}\; + {\text{ }}12{\text{ }}KCl\] Now, test for \[C{r^{3 + }}\], that represents to a portion of residue containing \[Fe{\left( {OH} \right)_3}\] and \[Cr{\left( {OH} \right)_3}\]are heated with \[NaOH\] and bromine water, \[Cr{\left( {OH} \right)_3}\]dissolves forming sodium chromate, and \[Fe{\left( {OH} \right)_3}\] remains as residue, when acetic acid and lead acetate is added to above solution, a yellow ppt. of lead chromate is obtained: \[N{a_2}Cr{O_4}\; + {\text{ }}Pb{\left( {C{H_3}COO} \right)_2}{\text{ }} - - - - - - - - - > {\text{ }}PbCr{O_4} + {\text{ }}2C{H_3}COONa\]
We noticed that \[N{H_4}\] ions are increased to suppress the release of \[O{H^ - }\]ions,
Hence, the solubility product of \[Fe{\left( {OH} \right)_3}\] then the colour of precipitate is different.
Then, \[{k_{sp}}\] value of \[Fe{\left( {OH} \right)_3}\]is less that of \[Cr{\left( {OH} \right)_3}\]. Therefore, the concentration of \[O{H^ - }\]ions in the solution can be limited by either increasing \[N{H_4}\]ion concentration or by decreasing \[O{H^ - }\]ion concentration.

Thus, the correct option is (D): Both \[(b)\]and \[(c)\]

Note: As we know that the group \[III\]radicals contains the elements aluminium, ferrous, and ferric ions. The group reagent is: Salt aqueous solution should be introduced to Ammonium Chloride in the presence of excess Ammonium Hydroxide. However, if ferrous ions (a light green salt) are present, concentrated nitric acid should also be added in order to change the ferrous ions into ferric ions.