Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: In which of the following molecules/ions, all the bonds are not equal?

Imagine you are an architect tasked with designing microscopic structures. Your building blocks are atoms, and your mortar is the chemical bond. Most of the time, you want your structures to be perfectly symmetrical, with every pillar (bond) bearing the exact same length and strength. But nature is full of quirks, and sometimes, the invisible forces of electron repulsion force these structures to warp, resulting in unequal bond lengths.

In this problem, we are presented with four molecular candidates: ${\text{SF}}_4$, ${\text{SiF}}_4$, ${\text{XeF}}_4$, and ${\text{BF}}_4^{-}$. Our mission is to identify the rebel—the molecule that refuses to maintain equal bond lengths. To do this, we must dive deep into the elegant world of VSEPR (Valence Shell Electron Pair Repulsion) Theory.

VSEPR Theory and Steric Numbers

The core principle of VSEPR theory is beautifully simple: electron pairs, whether they are shared in a bond or sitting alone as a lone pair, negatively charge their local space. Because like charges repel, these electron pairs will arrange themselves in three-dimensional space to be as far apart from each other as physically possible.

To predict a molecule's geometry, we first calculate its Steric Number (SN), which is the sum of the number of atoms bonded to the central atom (Bond Pairs, BP) and the number of lone pairs (LP) on the central atom:

This steric number dictates the fundamental hybridization and the base geometry of the molecule.

Tetrahedral and Square Planar Geometries

Let's evaluate our first two suspects: ${\text{SiF}}_4$ and ${\text{BF}}_4^{-}$.

Silicon is in Group 14, possessing 4 valence electrons. It forms 4 single bonds with 4 fluorine atoms, leaving zero lone pairs. Similarly, Boron in ${\text{BF}}_4^{-}$ has 3 valence electrons, plus 1 extra from the negative charge, giving it 4 electrons to form 4 bonds with zero lone pairs.

For both molecules:

A steric number of 4 corresponds to $s{p}^3$ hybridization. The electron pairs arrange themselves at the corners of a perfect tetrahedron, with bond angles of exactly $109.5^{\circ }$. Because all four positions are identical and occupied by the same type of atom (Fluorine), the repulsion is perfectly balanced. Consequently, all four bonds are absolutely equal in length.

Now, let's look at ${\text{XeF}}_4$. Xenon is a noble gas with 8 valence electrons. It uses 4 of these to bond with 4 fluorine atoms, leaving 4 electrons, which pair up to form 2 lone pairs.

A steric number of 6 means $s{p}^3{d}^2$ hybridization, which has an octahedral base geometry. However, lone pairs are bulky and highly repulsive. To minimize their repulsion with each other, the two lone pairs take up positions exactly opposite to each other (at $180^{\circ }$). This leaves the four fluorine atoms to occupy the equatorial plane, forming a square planar shape. Because the repulsive forces from the lone pairs cancel each other out perfectly from above and below the plane, the four Xe-F bonds remain perfectly symmetrical and equal in length.

SF4 and the See-Saw Shape

Finally, we arrive at ${\text{SF}}_4$. Sulfur is in Group 16, with 6 valence electrons. It forms 4 bonds with fluorine, leaving 2 electrons, or 1 lone pair.

A steric number of 5 corresponds to $s{p}^{3}d$ hybridization. The base geometry for this is trigonal bipyramidal. This geometry is unique and notoriously tricky because, unlike a tetrahedron or an octahedron, not all positions are equivalent. It consists of two distinct types of positions: 1. Equatorial positions: Three positions lying in a central plane, separated by $120^{\circ }$. 2. Axial positions: Two positions pointing straight up and down, perpendicular ($90^{\circ }$) to the equatorial plane.

According to Bent's Rule and VSEPR theory, lone pairs demand more space than bond pairs. If a lone pair were to occupy an axial position, it would suffer severe $90^{\circ }$ repulsions from three equatorial bonds. However, if it occupies an equatorial position, it only suffers $90^{\circ }$ repulsions from the two axial bonds. Therefore, the lone pair strictly prefers the equatorial position.

With the lone pair sitting in the equatorial plane, the molecule adopts a see-saw shape.

Here is where the magic happens. The bulky lone pair in the equatorial plane exerts a massive repulsive force on the two axial fluorine atoms. To escape this intense repulsion, the axial bonds bend slightly away from the lone pair and, more importantly, they elongate.

The repulsion experienced by the axial bonds is significantly greater than the repulsion experienced by the remaining two equatorial bonds. As a direct physical consequence, the axial S-F bonds are pushed further out, making them measurably longer than the equatorial S-F bonds.

Thus, ${\text{SF}}_4$ is the rebel. It is the molecule where the delicate balance of symmetry is broken by a lone pair, resulting in unequal bond lengths.

Key Takeaway: Whenever you encounter $s{p}^{3}d$ hybridization (steric number 5), be on high alert. The inherent asymmetry between axial and equatorial positions almost always guarantees that the bond lengths will not be equal!