Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The molecular shapes of $S{F}_4,C{F}_4$ and $Xe{F}_4$ are

Have you ever wondered why molecules, despite having the exact same number of surrounding atoms, take on completely different shapes in 3D space? It is like giving three architects the exact same number of bricks and getting a pyramid, a cube, and a flat patio. This architectural marvel at the atomic level is governed by a beautiful concept called the Valence Shell Electron Pair Repulsion (VSEPR) theory.

In this problem, we are given three molecules: $S{F}_4$, $C{F}_4$, and $Xe{F}_4$. All three have a central atom bonded to exactly four fluorine atoms. Yet, their geometries are wildly different. Why? The secret lies in the invisible ghosts of the atomic world: Lone Pairs.

VSEPR Theory

VSEPR theory states that electron pairs around a central atom will arrange themselves as far apart as possible to minimize repulsion. To find the shape, we first need to calculate the number of lone pairs (LP) on the central atom. The formula is elegantly simple:

Let's apply this master tool to our three candidates and watch their true shapes reveal themselves.

The See-Saw of $S{F}_4$

Let's start with Sulfur Tetrafluoride ($S{F}_4$). Sulfur is a Group 16 element, meaning it brings 6 valence electrons to the table. It forms four single bonds with four fluorine atoms, utilizing 4 of its electrons.

What remains? We have $6-4=2$ electrons left over. These two electrons pair up to form 1 lone pair.

Now, the central sulfur atom has 4 bond pairs and 1 lone pair, making a total of 5 electron domains. This corresponds to an $s{p}^{3}d$ hybridization, which fundamentally arranges itself in a trigonal bipyramidal geometry. However, lone pairs demand more space than bond pairs. To minimize repulsion, the lone pair occupies an equatorial position. The resulting molecular shape, when we only look at the atoms, looks exactly like a playground See-saw.

The Perfect Symmetry of $C{F}_4$

Next up is Carbon Tetrafluoride ($C{F}_4$). Carbon is the quintessential Group 14 element, possessing exactly 4 valence electrons. It forms four single bonds with the four fluorine atoms, using up all 4 of its valence electrons.

How many electrons are left? Exactly zero. This means carbon has 0 lone pairs.

With 4 bond pairs and 0 lone pairs, we have 4 electron domains. This gives us an $s{p}^3$ hybridization. Because there are no lone pairs to distort the symmetry, the four fluorine atoms push each other equally, resulting in a perfectly symmetrical Tetrahedral shape with bond angles of $109.5^{\circ }$.

The Noble Exception of $Xe{F}_4$

Finally, we arrive at Xenon Tetrafluoride ($Xe{F}_4$). Xenon is a noble gas from Group 18. Traditionally, noble gases are inert, but highly electronegative fluorine can force xenon to react. Xenon starts with a full octet of 8 valence electrons.

It uses 4 electrons to form single bonds with the four fluorine atoms. This leaves a whopping $8-4=4$ electrons behind. These four electrons pair up to form 2 lone pairs.

With 4 bond pairs and 2 lone pairs, xenon has 6 electron domains, leading to an $s{p}^3{d}^2$ hybridization. The base geometry for 6 domains is octahedral. To minimize the intense repulsion between the two bulky lone pairs, they position themselves exactly opposite to each other (axially). The four fluorine atoms are forced into the equatorial plane, creating a flat, Square Planar shape.

By simply counting the valence electrons, we unlocked the 3D structures of these molecules. $S{F}_4$ has 1 lone pair (See-saw), $C{F}_4$ has 0 lone pairs (Tetrahedral), and $Xe{F}_4$ has 2 lone pairs (Square Planar).

Their shapes are entirely different, driven by the varying number of lone pairs: 1, 0, and 2 respectively. This perfectly matches option (a). The atomic world is never random; it is a masterclass in spatial optimization!