Pick out the correct statement (s):

Options

A. Golden yellow `PbI_(2)` dissolves in hot water to give is colourless solution
B. `Ba^(2+) and Ca^(2+)` ions can be sepurated by adding `SO_(4)^(2-)` ion in acetic acid medium
C. Salt of calcium copper and nickel give a green flame colour
D. The sulphide ion gives with alkline sodium nitroprtasside ,a violet colour

Answer

To solve the question of identifying the correct statements regarding the properties of certain compounds and ions, we will analyze each statement one by one.

### Step-by-Step Solution:

1. **Statement A: "Golden yellow lead iodide dissolves in hot water to give a colorless solution."**
- Lead iodide (PbI2) is known to be yellow in color. When heated, it can dissolve in hot water, and the resulting solution is indeed colorless. Therefore, this statement is **correct**.

2. **Statement B: "Ba²⁺ and Ca²⁺ ions can be separated by adding sulfate ion in acidic medium."**
- When sulfate ions (SO₄²⁻) are added to a solution containing Ba²⁺ and Ca²⁺ ions, barium sulfate (BaSO₄) precipitates out as it is insoluble in acidic conditions, while calcium sulfate (CaSO₄) remains soluble. This allows for the separation of barium from calcium ions. Thus, this statement is **correct**.

3. **Statement C: "Salts of calcium, copper, and nickel give a green flame."**
- Calcium salts produce an orange-red flame, copper salts produce a green flame, and nickel salts produce a brown flame. Therefore, the claim that all three give a green flame is **incorrect**. Thus, this statement is **wrong**.

4. **Statement D: "The sulfide ion gives alkaline sodium nitroprusside a violet compound."**
- When sulfide ions (S²⁻) react with sodium nitroprusside (Na₂[Fe(CN)₅NO]), a purple or violet compound is formed. This reaction is well-documented in qualitative analysis. Therefore, this statement is **correct**.

### Conclusion:
The correct statements are A, B, and D. Statement C is incorrect.

### Final Answer:
The correct statements are A, B, and D.

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