Step-by-step derivation:

1. Identify the reaction and the product Y. The reaction is between xenon tetrafluoride ($XeF_4$) and dioxygen difluoride ($O_2F_2$) at a low temperature of 143 K. $O_2F_2$ is a very strong fluorinating agent, capable of oxidizing xenon from its +4 oxidation state in $XeF_4$ to a higher oxidation state. The reaction is as follows: $XeF_4 + O_2F_2 \xrightarrow{143 K} XeF_6 + O_2$ In this reaction, Xenon is oxidized from +4 to +6, and Oxygen is reduced from +1 (in $O_2F_2$) to 0 (in $O_2$). Therefore, the xenon compound Y produced in the reaction is Xenon Hexafluoride ($XeF_6$).

2. Determine the number of lone pairs on the central Xenon (Xe) atom in Y ($XeF_6$).

   • The central atom is Xenon (Xe), which belongs to Group 18 of the periodic table. It has 8 valence electrons.
   • In the $XeF_6$ molecule, the Xe atom forms six single covalent bonds with six fluorine atoms.
   • The number of electrons from Xe used in bonding is 6.
   • The number of non-bonding electrons (lone pair electrons) on the Xe atom is the total valence electrons minus the bonding electrons: $8 - 6 = 2$ electrons.
   • The number of lone pairs on the Xe atom is the number of non-bonding electrons divided by 2: $2 / 2 = 1$ lone pair.

3. Determine the number of lone pairs on the terminal Fluorine (F) atoms in Y ($XeF_6$).

   • There are six terminal fluorine atoms.
   • Fluorine (F) belongs to Group 17 and has 7 valence electrons.
   • Each F atom forms one single bond with the central Xe atom, using 1 of its valence electrons.
   • The number of non-bonding electrons on each F atom is $7 - 1 = 6$ electrons.
   • The number of lone pairs on each F atom is $6 / 2 = 3$ lone pairs.
   • Since there are six F atoms, the total number of lone pairs on all the fluorine atoms is $6 \times 3 = 18$ lone pairs.

4. Calculate the total number of lone pairs in the entire molecule of Y ($XeF_6$). The total number of lone pairs in the molecule is the sum of the lone pairs on the central atom and the lone pairs on all terminal atoms. Total lone pairs = (Lone pairs on Xe) + (Total lone pairs on F atoms) Total lone pairs = $1 + 18 = 19$.

Verification using total valence electrons:

• Total number of valence electrons in $XeF_6$ = (Valence electrons of Xe) + 6 $\times$ (Valence electrons of F) Total valence electrons = $8 + 6 \times 7 = 8 + 42 = 50$ electrons.
• Number of bonding electrons in $XeF_6$ = 6 Xe-F single bonds $\times$ 2 electrons/bond = 12 electrons.
• Total number of non-bonding (lone pair) electrons = Total valence electrons - Bonding electrons Total non-bonding electrons = $50 - 12 = 38$ electrons.
• Total number of lone pairs = Total non-bonding electrons / 2 Total lone pairs = $38 / 2 = 19$.

Both methods yield the same result.

The total number of lone pair(s) of electrons present on the whole molecule of Y ($XeF_6$) is 19.