Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The molecular geometry of $S{F}_6$ is octahedral. What is the geometry of $S{F}_4$ (including lone pair(s) of electrons, if any)?

Unmasking the Geometry of $S{F}_4$

Molecules are like microscopic architectural marvels, and understanding their shapes requires us to look closely at the invisible forces holding them together. In this problem, we are tasked with finding the geometry of the $S{F}_4$ molecule. But there is a subtle trap hidden in the wording of the question. Let's break it down step by step.

To determine the geometry of any molecule, our first stop is always the central atom. In $S{F}_4$, the central atom is Sulfur ($S$).

Sulfur is located in Group 16 of the periodic table, which means it has 6 valence electrons in its outermost shell. In the $S{F}_4$ molecule, Sulfur forms four single bonds with four Fluorine ($F$) atoms.

This accounts for 4 of its valence electrons. But what about the remaining two? Those two electrons pair up to form exactly one lone pair.

Now that we know the number of bonds and lone pairs, we can use VSEPR (Valence Shell Electron Pair Repulsion) theory to find the steric number. The steric number tells us how many electron domains are surrounding the central atom.

Substituting our values:

A steric number of 5 means the central Sulfur atom needs 5 hybrid orbitals to accommodate all its electron domains. This corresponds to $s{p}^{3}d$ hybridization.

For an $s{p}^{3}d$ hybridized atom, the base electron geometry is Trigonal Bipyramidal. Imagine a triangle lying flat, with one axis pointing straight up and another pointing straight down.

But where does the lone pair go? In a trigonal bipyramid, there are two types of positions: axial (top and bottom) and equatorial (the flat triangle). Lone pairs are bulky and highly repulsive. If we place a lone pair in an axial position, it suffers three severe $90^{\circ }$ repulsions from the equatorial bonds.

To minimize this repulsion, the lone pair always occupies an equatorial position, where it only experiences two $90^{\circ }$ repulsions. This specific arrangement pushes the remaining bonds slightly, resulting in a molecular shape known as a See-saw.

Here is where many students make a silly mistake. They calculate the shape as See-saw and immediately look for it in the options. However, the question explicitly asks for the geometry "including lone pair(s) of electrons".

When we include lone pairs in our spatial description, we are talking about the Electron Domain Geometry, not the molecular shape. Because the steric number is 5, the arrangement of all five domains (4 bonds + 1 lone pair) remains Trigonal Bipyramidal.

Always read the question carefully! If it asks for the "shape", ignore the lone pairs visually. If it asks for the "geometry including lone pairs", look at the base hybridization geometry. Therefore, the correct answer is Trigonal Bipyramidal.