The ratio of total number of lone pairs in `XeF_2` and `XeF_4` are:

Options

A. `3:2`
B. `9:14`
C. `14:19`
D. `9:19`

Explanation

To find the ratio of total number of lone pairs in `XeF_2` (xenon difluoride) and `XeF_4` (xenon tetrafluoride), we will follow these steps:

### Step 1: Determine the valence electrons of xenon and fluorine.
- **Xenon (Xe)** has 8 valence electrons (as it is a noble gas).
- **Fluorine (F)** has 7 valence electrons.

### Step 2: Draw the Lewis structure of `XeF_2`.
- In `XeF_2`, xenon bonds with 2 fluorine atoms.
- Each bond between xenon and fluorine uses 1 electron from xenon and 1 electron from fluorine.
- Therefore, 2 electrons from xenon are used for bonding (1 for each F).
- Remaining electrons in xenon = 8 (total) - 2 (bonding) = 6 electrons.
- These 6 electrons form 3 lone pairs.

### Step 3: Count the total number of lone pairs in `XeF_2`.
- Lone pairs on xenon = 3 (from the remaining 6 electrons).
- Each fluorine atom does not contribute any lone pairs to the xenon atom.
- Total lone pairs in `XeF_2` = 3.

### Step 4: Draw the Lewis structure of `XeF_4`.
- In `XeF_4`, xenon bonds with 4 fluorine atoms.
- Each bond uses 1 electron from xenon and 1 from fluorine.
- Therefore, 4 electrons from xenon are used for bonding.
- Remaining electrons in xenon = 8 (total) - 4 (bonding) = 4 electrons.
- These 4 electrons form 2 lone pairs.

### Step 5: Count the total number of lone pairs in `XeF_4`.
- Lone pairs on xenon = 2 (from the remaining 4 electrons).
- Each fluorine atom does not contribute any lone pairs to the xenon atom.
- Total lone pairs in `XeF_4` = 2.

### Step 6: Calculate the ratio of lone pairs in `XeF_2` to `XeF_4`.
- Lone pairs in `XeF_2` = 3.
- Lone pairs in `XeF_4` = 2.
- Ratio = 3:2.

### Final Answer:
The ratio of total number of lone pairs in `XeF_2` to `XeF_4` is **3:2**.

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