When `KNO_(2)` and `CH_(2)COOH` is added as `CoCl_(2)` solution, yellow ppt of `K_(4)[Cu(NO_(2))_(6)]` is formed.

Explanation

To solve the question regarding the formation of a yellow precipitate when `KNO2` and `CH3COOH` are added to a `CoCl2` solution, we will follow a systematic approach.

### Step-by-Step Solution:

1. **Identify the Reactants:**
- The reactants in this case are potassium nitrite (`KNO2`), acetic acid (`CH3COOH`), and cobalt(II) chloride (`CoCl2`).

2. **Understand the Reaction:**
- When `KNO2` (which provides `NO2-` ions) is added to `CoCl2`, the cobalt(II) ions (`Co^2+`) can react with the nitrite ions.
- The presence of acetic acid (`CH3COOH`) provides an acidic medium which can influence the reaction.

3. **Write the Expected Reaction:**
- The expected reaction can be represented as:
\[ Co^{2+} + 3 NO_2^{-} + 6 H^+ \rightarrow K_3[Co(NO_2)_6] + H_2O + NO \]
- Here, `K3[Co(NO2)6]` is the complex formed, and it is a precipitate.

4. **Analyze the Precipitate:**
- The statement claims that `K4[Cu(NO2)6]` is formed, which is incorrect.
- The correct complex formed is `K3[Co(NO2)6]`, which is a yellow precipitate.

5. **Conclusion:**
- The statement that a yellow precipitate of `K4[Cu(NO2)6]` is formed is false. Instead, the correct complex is `K3[Co(NO2)6]`.

### Final Answer:
The statement is **False**. The yellow precipitate formed is `K3[Co(NO2)6]`, not `K4[Cu(NO2)6]`.