The compound that is white in color is

Step 1: Analyze option (a)

Ammonium sulphide ($\ce{(NH4)2S}$) is practically known in solution form. Typically, "yellow ammonium sulphide" (which contains polysulphides $\ce{(NH4)2S_x}$) is commonly used in qualitative analysis. It is distinctively yellow.

Step 2: Analyze option (b)

Lead sulphate ($\ce{PbSO4}$) is formed as an insoluble precipitate when sulfuric acid reacts with a lead salt (like lead nitrate). $\ce{Pb^{2+} + SO4^{2-} -> PbSO4 v}$ This precipitate is white.

Step 3: Analyze option (c)

Lead iodide ($\ce{PbI2}$), formed upon reacting lead salts with a soluble iodide, is a characteristic bright yellow precipitate (Golden spangles).

Step 4: Analyze option (d)

Ammonium arsinomolybdate ($\ce{(NH4)3[As(Mo3O10)4]}$ or similar complexes) is the precipitate formed when an arsenate reacts with ammonium molybdate. It produces a characteristic canary yellow precipitate.

Step 5: Conclusion

The only compound out of the choices that is white is lead sulphate.

Key Points to Remember:

• $\ce{PbSO4}$ and $\ce{PbCl2}$ are white precipitates of lead.
• $\ce{PbCrO4}$ and $\ce{PbI2}$ are yellow precipitates.
• $\ce{PbS}$ is a black precipitate.