Which radical of group `IV^(th)` gives bluish white / white precipitate with excess `K_(4)[Fe(CN)_(6)]`?

Answer

To solve the question regarding which radical of group IV gives a bluish white or white precipitate with excess \( K_4[Fe(CN)_6] \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Group IV Radicals**:
- The radicals in group IV typically include \( Zn^{2+} \), \( Pb^{2+} \), and \( Cd^{2+} \).

2. **Understand the Reaction**:
- We need to consider the reaction of these radicals with \( K_4[Fe(CN)_6] \) (potassium ferrocyanide). This compound can react with certain metal ions to form precipitates.

3. **Focus on \( Zn^{2+} \)**:
- Among the group IV radicals, \( Zn^{2+} \) is known to react with \( K_4[Fe(CN)_6] \). The reaction can be represented as:
\[ Zn^{2+} + K_4[Fe(CN)_6] \rightarrow K_2[Zn_3[Fe(CN)_6]_2] \downarrow \]
- This reaction produces a precipitate.

4. **Determine the Color of the Precipitate**:
- The precipitate formed from the reaction of \( Zn^{2+} \) with \( K_4[Fe(CN)_6} \) is described as bluish white or white in color.

5. **Conclusion**:
- Therefore, the radical from group IV that gives a bluish white or white precipitate with excess \( K_4[Fe(CN)_6] \) is \( Zn^{2+} \).

### Final Answer:
The radical of group IV that gives a bluish white or white precipitate with excess \( K_4[Fe(CN)_6] \) is \( Zn^{2+} \).