Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Which of the following has maximum number of lone pairs associated with Xe?

Finding Lone Pairs in Xenon Compounds

When we think of noble gases, we usually picture them as aloof, unreactive elements perfectly content with their full octets. However, Xenon ($extXe$) is a bit of a rebel. Because of its large atomic size and relatively lower ionization energy, it can be coaxed into forming compounds, especially with highly electronegative bullies like Fluorine ($extF$) and Oxygen ($extO$).

In this problem, we are tasked with playing detective. We need to find out which of the given Xenon compounds hides the maximum number of lone pairs on the central Xenon atom.

Before we dive into the molecules, let's arm ourselves with a simple but powerful tool. The number of lone pairs ($LP$) on a central atom can be calculated using the formula:

Here, $V$ represents the number of valence electrons of the central atom. Since Xenon is a noble gas, it proudly sports $8$ valence electrons. $B$ represents the number of electrons shared by the surrounding atoms to form bonds.

Let's put our formula to the test, one molecule at a time.

1. Analyzing ${\text{XeO}}_3$ Oxygen is a divalent atom, meaning it demands two electrons to form a double bond. Since there are three oxygen atoms, they will collectively share $3\times 2=6$ electrons with Xenon.

Subtracting these from Xenon's total: $8-6=2$ electrons remaining. Dividing by two gives us exactly $1$ lone pair.

2. Analyzing ${\text{XeF}}_4$ Fluorine is much simpler; it's monovalent and forms single bonds. Four fluorine atoms will share exactly $4$ electrons.

Subtracting these: $8-4=4$ electrons remaining. This translates to $2$ lone pairs.

3. Analyzing ${\text{XeF}}_6$ Following the same logic, six fluorine atoms will share $6$ electrons.

Subtracting these: $8-6=2$ electrons remaining. This leaves Xenon with just $1$ lone pair.

4. Analyzing ${\text{XeF}}_2$ Finally, we look at ${\text{XeF}}_2$. Two fluorine atoms will share only $2$ electrons.

Subtracting these: $8-2=6$ electrons remaining. Dividing by two reveals a whopping $3$ lone pairs!

After our systematic investigation, the winner is clear. ${\text{XeF}}_2$ holds the maximum number of lone pairs, boasting three of them around the central Xenon atom.

The Power of VSEPR

Why does this matter? In the world of chemistry, lone pairs are not just passive bystanders; they are the invisible sculptors of molecular geometry. According to VSEPR theory, lone pairs repel each other strongly.

In ${\text{XeF}}_2$, the central atom has $2$ bond pairs and $3$ lone pairs, giving it a steric number of $5$ (Trigonal Bipyramidal geometry). To minimize the intense repulsion between themselves, these three lone pairs strategically occupy the equatorial positions (at $120^{\circ }$ to each other). This forces the two fluorine atoms into the axial positions, resulting in a perfectly linear shape for the molecule.

So, the next time you count lone pairs, remember: you are actually uncovering the hidden forces that shape the microscopic world!