Silver iodide remains as yellow precipitate after ammonia treatment

Steps:

• Dissolve mixture: Cl⁻ and I⁻ ions from NaCl and NaI in water.
• Add excess AgNO₃: Forms white AgCl and yellow AgI precipitates; X is their mixture.
• Add excess concentrated NH₃: AgCl dissolves to [Ag(NH₃)₂]⁺ (soluble); AgI remains insoluble.
• Shake: Y is undissolved AgI precipitate.

Why D is correct:

• AgI has a distinct yellow color due to its composition and crystal structure.

Why the others are wrong:

• A: X contains yellow AgI with white AgCl, so not pure white.
• B: Y is AgI, but "pure" cannot be confirmed without mixture proportions; test detects I⁻, not purity.
• C: No reduction occurs; Y is AgI, not metallic silver.

Final answer: D