Final answer:
Compounds that will not form a yellow precipitate with excess iodine and NaOH are those not yielding iodide ions or reacting differently, such as sodium sulphate with copper(II) chloride or sodium cyanide with HBr.
Step-by-step explanation:
The student query asks for the identification of compounds that will not form a yellow precipitate when treated with excess iodine in the presence of NaOH. Typically, a yellow precipitate is associated with the presence of iodide ions reacting with lead or silver ions. However, certain compounds, like sodium sulphate reacting with copper(II) chloride, will not form a yellow precipitate. Instead, they may form solutions of different colors or no precipitate at all, such as the light blue solution in the case of sodium sulphate with copper(II) chloride.
Given the examples provided, substances like silver nitrate, when added to a solution containing chloride, bromide, or iodide ions, will result in precipitates of various colors. Silver chloride forms a white precipitate, silver bromide forms a cream-colored precipitate, and silver iodide forms a yellow precipitate.
In the context of a reaction with sodium hydroxide and iodine, compounds such as sodium cyanide (NaCN) when reacted with acids like HBr, will yield hydrogen cyanide gas, not a precipitate. Similarly, reacting (NH₄)₂SO₄ with barium nitrate (Ba(NO₃)₂) will result in a precipitate of barium sulfate, not a yellow precipitate. Therefore, the compounds that will not form a yellow precipitate in the presence of excess iodine and NaOH are those that do not yield iodide ions or those that react to form different compounds, like in the cases mentioned.