Zinc (II) ion on reaction with NaOH first give a white precipitate which dissolves in excess of NaOH — d and f Block Elements Chemistry Question

💡 Solution & Explanation

Step 1: Adding a limited amount of sodium hydroxide (NaOH) to a zinc(II) solution precipitates white zinc hydroxide: Zn2+ + 2OH- -> Zn(OH)2(s). Step 2: Zinc hydroxide is amphoteric and reacts with excess hydroxide ions to form a soluble, colorless tetrahydroxozincate(II) complex. Step 3: The reaction is: Zn(OH)2 + 2OH- -> [Zn(OH)4]2-. Thus, the precipitate dissolves due to the formation of the sodium zincate complex ([Zn(OH)4]2-), matching option (c).