Among Fe$$^{3+}$$, Pb$$^{2+}$$, Cu$$^{2+}$$ and Mn$$^{2+}$$, identify the one that gets precipitated out while passing H$$_2$$S in presence of NH$$_4$$OH as group reagent. The highest possible oxidation state of the corresponding metal is

The classical qualitative analysis scheme for basic radicals (cations) divides them into groups according to the reagent that first precipitates them.

Recollection of the relevant groups
Group II cations (e.g., $$Pb^{2+}, Cu^{2+}$$) give sulphide precipitates with $$H_2S$$ in an acidic medium.
Group III cations (e.g., $$Fe^{3+}$$) are precipitated as hydroxides with $$NH_4OH$$ in the presence of $$NH_4Cl$$; they are no longer present when $$H_2S$$ is later passed through the basic medium.
Group IV cations (e.g., $$Co^{2+}, Ni^{2+}, Zn^{2+}, Mn^{2+}$$) are not precipitated in acid, but they do give sulphide precipitates when $$H_2S$$ is passed through an ammoniacal (basic) solution containing $$NH_4OH/NH_4Cl$$.

Now compare the given ions:

• $$Fe^{3+}$$ → Group III → precipitates earlier as $$Fe(OH)_3$$; it will not appear when $$H_2S$$ is passed in base.
• $$Pb^{2+}$$, $$Cu^{2+}$$ → Group II → already removed as $$PbS$$, $$CuS$$ in acidic medium.
• $$Mn^{2+}$$ → Group IV → remains in solution until the ammoniacal stage and then precipitates as $$MnS$$ with $$H_2S$$.

Hence, the only ion that precipitates with $$H_2S$$ in the presence of $$NH_4OH$$ is $$Mn^{2+}$$.

The question then asks for the highest oxidation state of this corresponding metal, i.e., of manganese. Manganese exhibits oxidation states from $$+2$$ up to $$+7$$; the highest, $$+7$$, is found in compounds like $$KMnO_4$$ and $$Mn_2O_7$$.

Therefore, the required highest oxidation state is $$+7$$.

Option D which is: $$+7$$