The fluoride which does not exist is

Options

A. `CF_(4)`
B. `SF_(6)`
C. `HeF_(4)`
D. `XeF_(4)`

Explanation

To determine which fluoride does not exist among the given options (CF4, SF6, HeF4, and XeF4), we can analyze each compound step by step.

### Step 1: Analyze CF4
- **Structure**: CF4 (Carbon Tetrafluoride) has a tetrahedral geometry.
- **Existence**: CF4 is a well-known stable compound and exists in nature.

### Step 2: Analyze SF6
- **Structure**: SF6 (Sulfur Hexafluoride) has an octahedral geometry.
- **Existence**: SF6 is also a stable compound and is widely used in various applications, confirming that it exists.

### Step 3: Analyze HeF4
- **Structure**: HeF4 (Helium Tetrafluoride) would imply the formation of a compound between helium and fluorine.
- **Existence**: Helium has a very high ionization energy and is a noble gas, which means it does not readily form compounds. Therefore, HeF4 does not exist.

### Step 4: Analyze XeF4
- **Structure**: XeF4 (Xenon Tetrafluoride) has a square planar geometry.
- **Existence**: XeF4 is a stable compound and exists, as xenon can form compounds due to its ability to expand its octet.

### Conclusion
After analyzing all four fluorides, we conclude that the fluoride that does not exist is **HeF4**.

### Final Answer
The fluoride which does not exist is **HeF4**.

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