Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The number of lone pairs on Xe in ${\text{XeF}}_2$, ${\text{XeF}}_4$ and ${\text{XeF}}_6$ respectively, are

For a long time, chemists believed that noble gases were completely inert. They had a full octet, so why would they react?

But Xenon, being a large atom with loosely held outer electrons, proved them wrong. When exposed to highly electronegative elements like Fluorine, Xenon is forced to share its electrons.

This leads to the formation of fascinating compounds like ${\text{XeF}}_2$, ${\text{XeF}}_4$, and ${\text{XeF}}_6$.

In this problem, we are tasked with finding the number of lone pairs on the central Xenon atom in each of these molecules.

To find the lone pairs, we don't need to draw complex Lewis structures every time. We can use a powerful and simple formula.

First, we calculate the total number of Electron Pairs (EP) around the central atom.

The formula is:

Here, $V$ represents the number of valence electrons on the central atom. $M$ represents the number of monovalent atoms (like Hydrogen or Halogens) attached to it.

Once we have the total electron pairs, finding the lone pairs is straightforward. We simply subtract the number of Bond Pairs (BP), which is equal to the number of attached atoms.

Let's apply our master formula to our first molecule, ${\text{XeF}}_2$.

Xenon is a noble gas, so it naturally has $8$ valence electrons. Thus, $V=8$.

It is bonded to $2$ Fluorine atoms, which are monovalent. So, $M=2$.

Plugging these into our formula:

There are $5$ total electron pairs around Xenon. Since it is bonded to $2$ Fluorine atoms, there are $2$ bond pairs.

Subtracting the bond pairs from the total pairs gives us the lone pairs:

So, ${\text{XeF}}_2$ has $3$ lone pairs on the central Xenon atom.

Next, let's look at ${\text{XeF}}_4$.

Again, Xenon has $8$ valence electrons ($V=8$). But this time, it is bonded to $4$ Fluorine atoms ($M=4$).

Calculating the total electron pairs:

With $6$ total electron pairs and $4$ bond pairs, we can easily find the lone pairs.

Therefore, ${\text{XeF}}_4$ has $2$ lone pairs.

Finally, we examine ${\text{XeF}}_6$.

The valence electrons remain $8$ ($V=8$), and now there are $6$ Fluorine atoms attached ($M=6$).

The total electron pairs are:

Subtracting the $6$ bond pairs from the $7$ total electron pairs:

This leaves ${\text{XeF}}_6$ with exactly $1$ lone pair.

We have successfully calculated the lone pairs for all three molecules.

For ${\text{XeF}}_2$, we found $3$ lone pairs. For ${\text{XeF}}_4$, we found $2$ lone pairs. And for ${\text{XeF}}_6$, we found $1$ lone pair.

This gives us the sequence: $3,2,1$.

These lone pairs are not just numbers; they dictate the physical shape of the molecules through VSEPR theory.

The $3$ lone pairs in ${\text{XeF}}_2$ force it into a linear shape. The $2$ lone pairs in ${\text{XeF}}_4$ create a square planar geometry. And the single lone pair in ${\text{XeF}}_6$ results in a distorted octahedral structure.

Understanding these invisible electron clouds allows us to visualize the true architecture of the chemical world!