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Q.
The reagents, $ N{{H}_{4}}Cl $ and aqueous $ N{{H}_{3}} $ will precipitate (1) $ F{{e}^{3+}} $ (2) $ A{{l}^{3+}} $ (3) $ M{{g}^{2+}} $ (4) $ Z{{n}^{2+}} $
BHUBHU 2008
Solution:
$ N{{H}_{4}}Cl $ and aq $ N{{H}_{3}} $ are the group reagents for IIIrd group of qualitative inorganic analysis. Hence, they precipitate the radicals belongs to IIIrd group ie,
$ F{{e}^{3+}},A{{l}^{3+}},C{{r}^{3+}} $
$ F{{e}^{3+}}+N{{H}_{4}}OH\xrightarrow{{}}\underset{brown\text{ }ppt.}{\mathop{Fe{{(OH)}_{3}}\downarrow }}\,+NH_{4}^{+} $
$ A{{l}^{3+}}+N{{H}_{4}}OH\xrightarrow{{}}\underset{whit\text{ }ppt.}{\mathop{Al{{(OH)}_{3}}\downarrow }}\,+NH_{4}^{+} $