I.ZnSO4(aq)+2NH4OH(aq)⟶Zn(OH)2(s)+NH42SO4(aq) AII.FeCl3(aq)+3NH4OH(aq)⟶Fe2O3⋅xH2O(s)+NH4Cl(aq) B The colours of compounds A and B respectively are

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I. ${ZnSO}_{4} (aq) + 2 {NH}_{4} OH (aq) ⟶ Zn (OH)_{2} (s) + {({NH}_{4})}_{2} {SO}_{4} (aq) A$
II. ${FeCl}_{3} (aq) + 3 {NH}_{4} OH (aq) ⟶ {Fe}_{2} O_{3} \cdot {xH}_{2} O (s) + {NH}_{4} Cl (aq) B$
The colours of compounds A and B respectively are

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brown and white

brown and brown

white and brown

white and white

answer is C.

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Detailed Solution

(i) ${ZnSO}_{4} (aq) + 2 {NH}_{4} OH (aq) ⟶ \underset{(white ppt.)}{Zn (OH)_{2} (s) + {({NH}_{4})}_{2} {SO}_{4} (aq)}$
(ii) ${FeCl}_{3} (aq) + {NH}_{4} OH (aq) ⟶ \underset{(Brown ppt.)}{{Fe}_{2} O_{3} \cdot {xH}_{2} O (s) + {NH}_{4} Cl (aq)}$