Total Lone Pairs in Xenon Compound Y

To determine the total number of lone pairs of electrons in the xenon compound $\text{Y}$, we first identify the chemical reaction taking place.

Step 1: Chemical Reaction

When xenon tetrafluoride ($\text{XeF}_4$) reacts with dioxygen difluoride ($\text{O}_2\text{F}_2$) at $143\text{ K}$, $\text{O}_2\text{F}_2$ acts as a fluorinating agent: $\text{XeF}_4 + \text{O}_2\text{F}_2 \xrightarrow{143\text{ K}} \text{XeF}_6 + \text{O}_2$

Thus, the xenon compound $\text{Y}$ is xenon hexafluoride ($\text{XeF}_6$).

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Step 2: Calculate Lone Pairs on the Central Xenon Atom ($\text{Xe}$)

• Valence electrons of Xenon ($\text{Xe}$) = $8$
• Number of fluorine atoms bonded to $\text{Xe}$ = $6$
• Electrons used in forming $6$ single $\text{Xe}-\text{F}$ bonds = $6$
• Remaining valence electrons on $\text{Xe} = 8 - 6 = 2\text{ electrons}$
• Number of lone pairs on $\text{Xe} = \frac{2}{2} = 1\text{ lone pair}$

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Step 3: Calculate Lone Pairs on the Fluorine Atoms ($\text{F}$)

• Valence electrons of each Fluorine atom ($\text{F}$) = $7$
• Electrons used in forming $1$ single bond = $1$
• Non-bonding electrons on each $\text{F}$ atom = $7 - 1 = 6\text{ electrons}$ ($3\text{ lone pairs}$)
• Number of lone pairs on $6$ fluorine atoms = $6 \times 3 = 18\text{ lone pairs}$

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Step 4: Total Lone Pairs Across the Entire Molecule

$\text{Total Lone Pairs} = (\text{Lone pairs on Xe}) + (\text{Lone pairs on } 6\text{ F atoms})$ $\text{Total Lone Pairs} = 1 + 18 = 19$

The total number of lone pairs of electrons present across the whole molecule of $\text{Y}$ is 19.