Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Which among the following species has unequal bond lengths ?

When we dive into the microscopic world of molecules, we often expect perfect symmetry. However, the reality is that molecules are dynamic, and their shapes are dictated by the invisible forces of electron repulsion. In this problem, we are tasked with finding which of the given molecules—${\text{XeF}}_4$, ${\text{SiF}}_4$, ${\text{SF}}_4$, or ${\text{BF}}_4^{-}$—breaks the rules of perfect symmetry and exhibits unequal bond lengths. To solve this, we must rely on the Valence Shell Electron Pair Repulsion (VSEPR) theory.

Let's first examine the molecules that maintain their symmetry.

Take Xenon tetrafluoride (${\text{XeF}}_4$). Xenon, a noble gas, has $8$ valence electrons. When it bonds with $4$ fluorine atoms, the steric number becomes $\frac{1}{2}(8+4)=6$, which corresponds to $s{p}^3{d}^2$ hybridization. This geometry is octahedral. However, because there are only $4$ bond pairs, the remaining $2$ electron pairs are lone pairs. To minimize repulsion, these lone pairs position themselves exactly opposite to each other (axially), leaving the $4$ fluorine atoms in a perfectly symmetrical square planar arrangement. Because of this symmetry, all four $\text{Xe-F}$ bonds are perfectly equal.

Similarly, Silicon tetrafluoride (${\text{SiF}}_4$) and the Tetrafluoroborate ion (${\text{BF}}_4^{-}$) both have a steric number of $4$, leading to $s{p}^3$ hybridization. With zero lone pairs, they both adopt a perfect tetrahedral geometry. In a perfect tetrahedron, every bond angle is exactly $109.5^{\circ }$, and every bond length is identical.

Now, let's look at the rebel of the group: Sulfur tetrafluoride (${\text{SF}}_4$).

Sulfur has $6$ valence electrons. Bonding with $4$ fluorine atoms gives a steric number of $\frac{1}{2}(6+4)=5$, which means it has $s{p}^{3}d$ hybridization. The base geometry for $s{p}^{3}d$ is trigonal bipyramidal.

Here is where the catch lies: ${\text{SF}}_4$ has $4$ bond pairs and $1$ lone pair. In a trigonal bipyramidal structure, the equatorial positions offer more space ($120^{\circ }$ apart) compared to the axial positions ($90^{\circ }$ from the equatorial plane). To minimize the intense lone pair-bond pair repulsion, the lone pair smartly occupies an equatorial position.

This creates a see-saw shape. The lone pair pushes strongly against the two axial $\text{S-F}$ bonds, forcing them to bend slightly and, more importantly, elongating them. As a result, the axial bonds are significantly longer than the equatorial bonds.

By systematically applying VSEPR theory, we can confidently conclude that the presence of an equatorial lone pair in $s{p}^{3}d$ hybridized ${\text{SF}}_4$ breaks the symmetry, making it the only species among the choices with unequal bond lengths.