Few drops of $HN{{O}_{3}}$ are added to the II group before proceeding to the III group to:
(a)- convert $F{{e}^{2+}}\text{ to }F{{e}^{3+}}$
(b)- convert $F{{e}^{3+}}\text{ to }F{{e}^{2+}}$
(c)- ppt. III group
(d)- None of these.

Answer Verified Verified
Hint: We have the cations divided into groups for detection. To proceed to III group concentrated nitric acid is added to increase the oxidation number of the ion. Ferric ions can completely precipitate out.

Complete answer:
Quantitative analysis of cation is the technique that is used to detect the cation of the compound by stepwise addition of specific reagents and solutions.
The cations are classified because each group of cations is precipitated at every step. Each cation behaves to a common test of reagent that is different from another cation.

There are five groups in which the cations are precipitated:
(i)- Group I cations- These are silver ion ($A{{g}^{+}}$ ), mercury ion ($H{{g}^{2+}}$ ) and lead ion ($P{{b}^{2+}}$ ). These are insoluble chlorides and precipitate out at the first step.
(ii)- Group II cations- These are mercury ion ($H{{g}^{2+}}$ ), lead ion ($P{{b}^{2+}}$ ), copper ion ($C{{u}^{2+}}$ ), bismuth ion ($B{{i}^{3+}}$ ), cadmium ion ($C{{d}^{2+}}$ ), arsenic ion ($A{{s}^{3+}}$ ), tin ion ($S{{n}^{4+}}$ ) and antimony ion ($S{{b}^{3+}}$ ). In acidic medium they form in soluble sulfides.
(iii)- Group III cations- These are aluminium ion ($A{{l}^{3+}}$ ), ferric ion ($F{{e}^{3+}}$ ), cobalt ion ($C{{o}^{2+}}$ ), nickel ion ($N{{i}^{2+}}$ ), chromium ion ($C{{r}^{3+}}$ ), zinc ion ($Z{{n}^{2+}}$ ), and manganese ion ($M{{n}^{2+}}$ ). Their sulphide and hydroxide are insoluble in an alkaline medium.
 (iv)- Group IV cations- These are calcium ion ($C{{a}^{2+}}$ ), strontium ion ($S{{r}^{2+}}$ ), and barium ion ($B{{a}^{2+}}$ ). There bicarbonates precipitate out.
(v)- Group V cations- These are magnesium ion ($M{{g}^{2+}}$ ), sodium ion ($N{{a}^{+}}$ ), potassium ion (${{K}^{+}}$), and ammonium ion ($N{{H}_{4}}^{+}$ ).
Since the ferric ion is precipitated at group III. Hence concentrated nitric acid is added to oxidize the${{H}_{2}}S$ in the solution, to convert $F{{e}^{2+}}\text{ }to\text{ }F{{e}^{3+}}$because it precipitate $Fe{{(OH)}_{3}}$ as in the solution.
So, the correct answer is “Option A”.

Note: The reagents used in the group three are ammonium hydroxide in the presence of ammonium chloride. The nitric acid is also acid to increase the ionization of ammonium hydroxide so that Fe is precipitated as $Fe{{(OH)}_{3}}$ not as $Fe{{(OH)}_{2}}$ because ferrous hydroxide is more soluble in water than ferric hydroxide. All the steps should be followed one by one.