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 xenon compound Y, which is formed by the reaction of `XeF4` with `O2F2`.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction of `XeF4` with `O2F2` produces a xenon compound, which is `XeF6` and oxygen gas (`O2`).

\[ \text{XeF}_4 + \text{O}_2\text{F}_2 \rightarrow \text{XeF}_6 + \text{O}_2 \]

2. **Determine the Valence Electrons**:
- Xenon (Xe) is a noble gas with the electronic configuration of \( [Kr] 5s^2 5p^6 \), giving it a total of 8 valence electrons.
- Each fluorine (F) atom has 7 valence electrons.

3. **Count the Electrons in `XeF6`**:
- In `XeF6`, xenon shares 6 of its electrons with 6 fluorine atoms.
- This means that xenon has used 6 of its 8 valence electrons for bonding, leaving it with 2 electrons as lone pairs.

4. **Determine Lone Pairs on Fluorine Atoms**:
- Each fluorine atom, after forming a bond with xenon, has 6 remaining electrons (7 original - 1 shared).
- These 6 electrons can be grouped into 3 lone pairs for each fluorine atom.

5. **Calculate Total Lone Pairs**:
- Since there are 6 fluorine atoms, the total number of lone pairs from fluorine is:
\[ 6 \text{ (fluorine atoms)} \times 3 \text{ (lone pairs per fluorine)} = 18 \text{ lone pairs} \]
- Adding the lone pair from xenon:
\[ 18 \text{ (from fluorine)} + 1 \text{ (from xenon)} = 19 \text{ total lone pairs} \]

### Final Answer:
The total number of lone pairs of electrons present on the whole molecule of Y (XeF6) is **19**.