Animated Solution for Chemistry - s and p-Block Elements: The shape/structure of $[{\text{XeF}}_5{]}^{-}$ and ${\text{XeO}}_3{\text{F}}_2$, respectively, are

Welcome to the fascinating world of molecular geometry! Today, we are going to unravel the exact shapes of two intriguing xenon compounds: the $[{\text{XeF}}_5{]}^{-}$ ion and the ${\text{XeO}}_3{\text{F}}_2$ molecule. To do this, we will rely on the powerful VSEPR (Valence Shell Electron Pair Repulsion) theory and the concept of hybridization.

The first step in determining the shape of any molecule is to find its steric number ($H$), which tells us the number of hybridized orbitals. The formula we use is:

Here, $V$ is the number of valence electrons on the central atom, $M$ is the number of monovalent atoms attached, $C$ is the cationic charge, and $A$ is the anionic charge.

Let's start with our first candidate, $[{\text{XeF}}_5{]}^{-}$. Xenon, being a noble gas, has $8$ valence electrons ($V=8$). There are $5$ monovalent fluorine atoms attached ($M=5$). Since it's an anion with a $-1$ charge, we add $1$ ($A=1$). Plugging these into our formula:

A steric number of $7$ corresponds to an $s{p}^3{d}^3$ hybridization. The base geometry for this hybridization is pentagonal bipyramidal.

Now, we need to figure out how many lone pairs are present. The number of bond pairs is simply the number of attached atoms, which is $5$. Therefore, the number of lone pairs is $7-5=2$.

According to VSEPR theory, lone pairs require more space and will position themselves to minimize repulsion. In a pentagonal bipyramidal geometry, the axial positions offer the most space (at $90^{\circ }$ to the equatorial plane, compared to the tight $72^{\circ }$ angles within the plane). Thus, the two lone pairs occupy the axial positions, leaving the five fluorine atoms to form a flat pentagon in the equatorial plane. The resulting shape is pentagonal planar.

Next, let's look at ${\text{XeO}}_3{\text{F}}_2$. Again, Xenon has $8$ valence electrons ($V=8$). When counting monovalent atoms, we only consider the $2$ fluorine atoms ($M=2$). Oxygen is divalent (forms double bonds) and is therefore ignored in this specific formula. The molecule is neutral, so $C=0$ and $A=0$.

A steric number of $5$ indicates an $s{p}^{3}d$ hybridization, which corresponds to a trigonal bipyramidal base geometry.

Let's check for lone pairs. We have $3$ oxygen atoms and $2$ fluorine atoms, making a total of $5$ bond pairs. Since our steric number is also $5$, there are exactly $0$ lone pairs on the central xenon atom ($5-5=0$).

Because there are no lone pairs, the shape is identical to the base geometry. However, we must apply Bent's Rule, which states that more electronegative atoms prefer axial positions. Therefore, the highly electronegative fluorine atoms will occupy the axial positions, while the oxygen atoms will form double bonds in the equatorial plane. The final shape is trigonal bipyramidal.

By carefully applying VSEPR theory and hybridization rules, we have determined that $[{\text{XeF}}_5{]}^{-}$ is pentagonal planar and ${\text{XeO}}_3{\text{F}}_2$ is trigonal bipyramidal. This perfectly matches our first option. Always remember to account for lone pairs and electronegativity differences when predicting molecular shapes!