At 143 K, the reaction of XeF4 with O2F2 produces a xenon compound Y. The total number of lone pair(

Q: Why does Xenon react at all if it is a noble gas?

While noble gases are generally inert, larger ones like Xenon have lower ionization energies. Highly electronegative atoms like Fluorine and Oxygen can force Xenon to share its electrons and form stable compounds.

Q: What is the exact shape of the XeF6 molecule?

It has a distorted octahedral geometry. The six fluorine atoms form an octahedron, but the single stereochemically active lone pair on Xenon pushes them, causing a distortion.

Q: Why is O2F2 used in this reaction?

O2F2 is an extremely powerful fluorinating and oxidizing agent. It is capable of fluorinating XeF4 to XeF6 even at very low temperatures like 143 K.

Q: Do we always have to count the lone pairs on the surrounding atoms?

It depends entirely on the wording of the question! If it asks for lone pairs on the 'central atom', you only count Xenon's. Because this question specified the 'whole molecule', you must include the fluorines.

Q: Is there a quick trick to find valence electrons?

Yes! For p-block elements, simply take their group number and subtract 10. Xenon is in Group 18 (8 valence electrons), and Fluorine is in Group 17 (7 valence electrons).

Question

At $143	ext{ K}$, the reaction of $	ext{XeF}_4$ with $	ext{O}_2	ext{F}_2$ produces a xenon compound $	ext{Y}$. The total number of lone pair(s) of electrons present on the whole molecule of $	ext{Y}$ is _______.

Sigma Percentile · Accepted Answer

### The Mystery Compound Y

Imagine you are in a chemistry lab, working at a chilling $143	ext{ K}$. You mix xenon tetrafluoride ($	ext{XeF}_4$) with dioxygen difluoride ($	ext{O}_2	ext{F}_2$). What happens?

$	ext{O}_2	ext{F}_2$ is an incredibly aggressive fluorinating agent. It doesn't just sit there; it actively forces its fluorine atoms onto the xenon compound. The reaction proceeds as follows:

$$ 	ext{XeF}_4 + 	ext{O}_2	ext{F}_2 \xrightarrow{143	ext{ K}} 	ext{XeF}_6 + 	ext{O}_2 $$

Through this fluorination, the $	ext{XeF}_4$ is upgraded to xenon hexafluoride ($	ext{XeF}_6$), while oxygen gas is released. Thus, our mysterious compound $	ext{Y}$ is identified as $	ext{XeF}_6$.

### Analyzing the Central Atom

Now that we know we are dealing with $	ext{XeF}_6$, the question asks for the **total number of lone pairs** on the *entire* molecule. To do this systematically, we must first look at the central atom: Xenon.

Xenon is a noble gas located in Group 18 of the periodic table. This means it has a full octet, giving it $8$ valence electrons in its outermost shell.

In the $	ext{XeF}_6$ molecule, Xenon forms single covalent bonds with six fluorine atoms. Each single bond requires one electron from Xenon.

$$ 	ext{Electrons used in bonding} = 6 $$

Subtracting these from the total valence electrons gives us the non-bonding electrons:

$$ 	ext{Remaining electrons} = 8 - 6 = 2 $$

These $2$ remaining electrons pair up to form exactly **$1$ lone pair** on the central Xenon atom. This lone pair is stereochemically active, pushing the six fluorine atoms and distorting the octahedral geometry into a *distorted octahedral* shape.

### Don't Forget the Fluorines!

A common trap in such questions is to stop after finding the lone pairs on the central atom. However, the question explicitly asks for the lone pairs on the **whole molecule**. We must account for the fluorine atoms!

Fluorine belongs to Group 17 (the halogens) and has $7$ valence electrons.

For each fluorine atom, $1$ electron is shared with Xenon to form the covalent bond. This leaves $6$ non-bonding electrons per fluorine atom.

$$ 	ext{Remaining electrons per F} = 7 - 1 = 6 $$

These $6$ electrons arrange themselves into **$3$ lone pairs** on each fluorine atom.

### The Final Calculation

Now, we simply tally up all the lone pairs across the entire molecule.

We have $1$ lone pair on the central Xenon atom.
We have $6$ fluorine atoms, and each carries $3$ lone pairs.

$$ 	ext{Total Lone Pairs} = 1 + (6 	imes 3) $$

$$ 	ext{Total Lone Pairs} = 1 + 18 = 19 $$

The grand total is **$19$ lone pairs**. By breaking the molecule down into its central and surrounding atoms, a seemingly complex counting problem becomes a straightforward exercise in valence electron bookkeeping!

Animated Solution for Chemistry - s and p-Block Elements: At $143 K$ , the reaction of $XeF_{4}$ with $O_{2} F_{2}$ produces a xenon compound $Y$ . The total number of lone pair(s) of electrons present on the whole molecule of $Y$ is _______.

Enter Numerical Value:

Visualized Solution

$Reaction of XeF_{4} with O_{2} F_{2}$

$Valence Electrons of Xenon$

$Lone Pairs on Xenon$

$Lone Pairs on Fluorine$

$Total Lone Pairs$

The Sigma Insight: Group 18 Elements

The Mystery Compound Y

Analyzing the Central Atom

Don't Forget the Fluorines!

The Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reaction of XeF4​ with O2​F2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Valence Electrons of Xenon

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lone Pairs on Xenon

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lone Pairs on Fluorine

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Lone Pairs