Animated Solution for Chemistry - s and p-Block Elements: The type of hybridisation and number of lone pair(s) of electrons of Xe in ${\text{XeOF}}_4$, respectively, are

Chemical bonding can sometimes feel like a puzzle, especially when noble gases decide to break the rules and form compounds. Today, we are going to decode the structure of Xenon oxytetrafluoride, or ${\text{XeOF}}_4$. Our mission is to find its hybridization and the number of lone pairs sitting on the central atom.

Steric Number

To figure out the hybridization of any central atom, our best friend is the Steric Number ($H$) formula. It acts as a mathematical bridge to the molecule's geometry.

Here is what each letter stands for: - $V$: Number of valence electrons on the central atom. - $M$: Number of monovalent surrounding atoms (like Hydrogen or Halogens). - $C$: Charge of the cation (if it's a positive ion). - $A$: Charge of the anion (if it's a negative ion).

Let's break down our molecule, ${\text{XeOF}}_4$. The central atom is Xenon ($\text{Xe}$). Since Xenon is a noble gas, it has a full octet in its outermost shell. Therefore, its valence electrons, $V=8$.

Now, let's look at the surrounding atoms. We have 4 Fluorine atoms and 1 Oxygen atom. Fluorine needs only one electron to complete its octet, making it monovalent. So, $M=4$.

But what about Oxygen? Oxygen needs two electrons to complete its octet, meaning it forms a double bond. Because it is divalent, we strictly do not count it in $M$. This is a classic trap where many students make a silly mistake!

Since ${\text{XeOF}}_4$ is a neutral molecule, both $C$ and $A$ are zero. Let's substitute our values into the master equation:

A steric number of 6 tells us that the central Xenon atom needs 6 hybrid orbitals to accommodate its electron domains. Mixing one $s$, three $p$, and two $d$ orbitals gives us exactly six orbitals. Thus, the hybridization is ${\text{sp}}^3{\text{d}}^2$.

We know Xenon has 6 electron domains in total. But how many of these are actual bonds, and how many are invisible lone pairs?

To find out, we simply count the total number of surrounding atoms. Xenon is bonded to 4 Fluorines and 1 Oxygen, giving us a total of 5 bond pairs ($bp=5$).

The number of lone pairs ($lp$) is simply the steric number minus the bond pairs:

We have successfully cracked the code! The hybridization is ${\text{sp}}^3{\text{d}}^2$ and there is 1 lone pair on the Xenon atom.

If you visualize this, the 6 domains arrange themselves in an Octahedral geometry to minimize repulsion. However, because one of those domains is a lone pair, the physical shape we actually see is a Square Pyramidal. The four Fluorine atoms form the square base, the Oxygen atom sits at the peak, and the lone pair occupies the space directly opposite the Oxygen, silently pushing the bonds slightly closer together.