`E` represents an element belonging to carbon family.
`E+2X_(2) to EX_(4) (X=F, Cl, Br, l)`

Options

A. Stability of `EX_(4)` decreases down the `14^(th)` group
B. `PbL_(4)` does not exist
C. `Ge` & `Pb` forms `EX_(2)` as well.
D. All are correct

Answer

To solve the problem, we need to analyze the behavior of an element `E` from the carbon family (Group 14) when reacting with halogens (X = F, Cl, Br, I) to form compounds of the type `EX4` and `EX2`.

### Step-by-Step Solution:

1. **Identify the Elements in the Carbon Family**:
The carbon family consists of the following elements: Carbon (C), Silicon (Si), Germanium (Ge), Tin (Sn), and Lead (Pb).

2. **Understand the Oxidation States**:
- In Group 14, elements can exhibit oxidation states of +2 and +4.
- The stability of the +4 oxidation state decreases down the group due to the inert pair effect, while the stability of the +2 oxidation state increases.

3. **Analyze the Formation of EX4**:
- For the element `E`, when it reacts with halogens (X), it can form the compound `EX4`.
- However, as we move down the group, the stability of `EX4` decreases. For example, while carbon and silicon can form stable tetrahalides (like CF4, SiF4), germanium can form GeX4, but it is less stable than SiX4.
- Lead (Pb) does not form a stable PbX4 due to the inert pair effect, and it is more stable in the +2 oxidation state (PbX2).

4. **Check the Existence of EX2**:
- Germanium (Ge) and Lead (Pb) can form stable compounds of the type `EX2` (e.g., GeX2 and PbX2).
- This is significant because while PbX4 does not exist, PbX2 is stable.

5. **Conclusion on the Options**:
- Based on the above analysis:
- The stability of EX4 decreases down the group (correct).
- PbX4 does not exist (correct).
- Both Ge and Pb can form EX2 (correct).
- Therefore, all statements are correct.

### Final Answer:
All options are correct.

---