Precipitate of Zn(OH)2 is soluble in:

Zinc hydroxide (Zn(OH)2) is a sparingly soluble compound, which means it partially dissolves in water to form a precipitate. However, its solubility can be influenced by the presence of certain substances. Let's analyze the solubility of Zn(OH)2 in different scenarios:

a) Excess of sodium hydroxide:
When sodium hydroxide (NaOH) is added in excess to a solution containing zinc ions (Zn2+), it forms a complex ion called sodium zincate, [Zn(OH)4]2-. This complex ion is soluble in water, and therefore, the precipitate of Zn(OH)2 dissolves. The reaction can be represented as follows:
Zn(OH)2 + 2NaOH → [Zn(OH)4]2- + 2Na+

b) Excess of ammonia solution:
Similar to sodium hydroxide, ammonia (NH3) also forms a complex ion with zinc ions. In the presence of excess ammonia, the complex ion formed is called tetraamminezinc(II) ion, [Zn(NH3)4]2+. This complex ion is also soluble in water, leading to the dissolution of the Zn(OH)2 precipitate. The reaction can be represented as follows:
Zn(OH)2 + 4NH3 → [Zn(NH3)4]2+ + 2OH-

c) Solutions of ammonium salts:
When Zn(OH)2 is added to solutions of ammonium salts, the ammonia produced from the salt reacts with the Zn(OH)2 to form the soluble complex ion [Zn(NH3)4]2+. This complex ion is again soluble in water, causing the Zn(OH)2 precipitate to dissolve. The reaction can be represented as follows:
Zn(OH)2 + 4NH4+ → [Zn(NH3)4]2+ + 4H2O

d) All of these:
From the explanations above, it is clear that the precipitate of Zn(OH)2 is soluble in excess sodium hydroxide, excess ammonia solution, and solutions of ammonium salts. Therefore, the correct answer is option 'D' - all of these.