Zinc ion precipitation and solubility tests

Steps:

• Identify Zn²⁺ from ZnCl₂ reacts with NH₃ to form white Zn(OH)₂ precipitate.
• Note excess NH₃ dissolves it as [Zn(NH₃)₄]²⁺ complex.
• Evaluate B: Cu turnings with ZnCl₂ yield no reaction, as Zn is more reactive but no displacement occurs without Cu²⁺.
• Evaluate C and D: Test for halide or anion precipitates, irrelevant to Zn²⁺.

Why A is correct:

• Zn²⁺ + 2NH₃ + 2H₂O → Zn(OH)₂ (white ppt) + 2NH₄⁺; excess NH₃ forms soluble [Zn(NH₃)₄]²⁺ per coordination chemistry.

Why the others are wrong:

• B: No zinc precipitate forms; Cu cannot reduce Zn²⁺, and ammonia is absent.
• C: AgNO₃ with Cl⁻ gives white AgCl, not yellow (yellow for Br⁻/I⁻).
• D: Ba(NO₃)₂ tests for SO₄²⁻; no white BaCl₂ precipitate forms.

Final answer: A