At 143 K, the reaction of `XeF_4` with `O_2F_2` produces a xenon compound Y. The total number of lone pair(s) of electrons present on the whole molecule of Y is………….

Explanation

To solve the problem, we need to determine the total number of lone pairs of electrons present in the compound Y, which is formed by the reaction of `XeF_4` with `O_2F_2`.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between `XeF_4` and `O_2F_2` produces a xenon compound Y. The expected product is `XeF_6` and `O_2`.

2. **Determine the Valence Electrons**:
- **Xenon (Xe)**: Xenon is a noble gas with the electronic configuration of `[Kr] 5s² 5p⁶`, which means it has 8 valence electrons.
- **Fluorine (F)**: Fluorine has the electronic configuration of `1s² 2s² 2p⁵`, giving it 7 valence electrons.

3. **Count the Electrons in `XeF_6`**:
- In `XeF_6`, xenon is bonded to 6 fluorine atoms. Each bond involves the sharing of one electron from xenon and one from fluorine.
- Therefore, xenon shares 6 of its 8 electrons with the 6 fluorine atoms.

4. **Calculate Lone Pairs on Xenon**:
- After forming 6 bonds with fluorine, xenon has 2 electrons left. These 2 electrons form 1 lone pair on xenon.

5. **Calculate Lone Pairs on Fluorine Atoms**:
- Each fluorine atom has 7 valence electrons. When it forms a bond with xenon, it shares 1 electron. Thus, each fluorine atom has 6 electrons remaining.
- These 6 electrons on each fluorine atom can be grouped into 3 lone pairs (since 6 electrons / 2 = 3 lone pairs).

6. **Total Lone Pairs Calculation**:
- There are 6 fluorine atoms, each contributing 3 lone pairs:
- Total lone pairs from fluorine = 6 fluorine × 3 lone pairs/fluorine = 18 lone pairs.
- Adding the lone pair from xenon:
- Total lone pairs = 18 (from fluorine) + 1 (from xenon) = 19 lone pairs.

### Final Answer:
The total number of lone pairs of electrons present on the whole molecule of Y is **19**.