Before adding $NH X_{4} Cl$ and $NH X_{4} OH$ for group III, the filtrate from group II is boiled with a few drops of concentrated $HNO X_{3}$ . The main purpose is to

Aremove $NH X_{4} X^{+}$ ions from the solution
Bprecipitate $Al X^{3 +}$ as the nitrate
Coxidise $Fe X^{2 +}$ to $Fe X^{3 +}$ so that iron is precipitated completely as $Fe (OH) X_{3}$
Dreduce $Fe X^{3 +}$ to $Fe X^{2 +}$ which is precipitated more easily

Answer: C

Passing

H X_{2} S

in group II reduces

Fe X^{3 +}

Fe X^{2 +}

Fe (OH) X_{2}

has a much larger

K_{s p}

than

Fe (OH) X_{3}

and would not precipitate completely at the low

[OH X^{-}]

provided by the buffer, so concentrated

HNO X_{3}

re-oxidises it to

Fe X^{3 +}

(and boiling also expels

H X_{2} S