Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The number of species below that have two lone pairs of electrons in their central atom is ……… . (Round off to the nearest integer) ${\text{SF}}_4,{\text{BF}}_4^{-},{\text{ClF}}_3,{\text{AsF}}_3,{\text{PCl}}_5,{\text{BrF}}_5,{\text{XeF}}_4,{\text{SF}}_6$

Mastering Lone Pair Calculations

When diving into the world of Chemical Bonding and Molecular Structure, one of the most crucial skills you can develop is the ability to quickly and accurately determine the number of lone pairs on a central atom. Lone pairs are the invisible architects of a molecule; they don't form bonds, but their repulsive forces dictate the entire 3D geometry of the species according to VSEPR theory.

In this problem, we are tasked with acting as molecular detectives. We have a lineup of eight different chemical species, and we need to interrogate each one to find out which of them are hiding exactly two lone pairs on their central atom.

Before we start our investigation, we need a reliable tool. The number of lone pairs ($LP$) can be calculated using a very straightforward formula:

Here, $V$ represents the number of valence electrons on the central atom. $S$ represents the number of electrons shared with the surrounding atoms. Since each surrounding halogen atom (like Fluorine or Chlorine) forms a single bond, it shares exactly one electron from the central atom. Finally, we divide by 2 because electrons come in pairs!

Let's systematically apply our formula to the lineup:

1. ${\text{SF}}_4$ (Sulfur tetrafluoride) Sulfur is in Group 16, so it has $V=6$ valence electrons. It forms 4 single bonds with Fluorine, so $S=4$. ${\text{SF}}_4$ has 1 lone pair (giving it a see-saw shape). Not our target.

2. ${\text{BF}}_4^{-}$ (Tetrafluoroborate ion) Boron is in Group 13, normally having 3 valence electrons. However, the negative charge means it has gained an extra electron, making $V=4$. It forms 4 bonds, so $S=4$. Zero lone pairs here.

3. ${\text{ClF}}_3$ (Chlorine trifluoride) Chlorine is a halogen (Group 17) with $V=7$ valence electrons. It forms 3 bonds with Fluorine, so $S=3$. Bingo! We found our first match. ${\text{ClF}}_3$ has exactly 2 lone pairs, which force the molecule into a distinctive T-shape.

4. ${\text{AsF}}_3$ (Arsenic trifluoride) Arsenic is in Group 15, so $V=5$. It forms 3 bonds, so $S=3$. Only 1 lone pair.

5. ${\text{PCl}}_5$ (Phosphorus pentachloride) Phosphorus (Group 15) has $V=5$. It forms 5 bonds, so $S=5$. Zero lone pairs.

6. ${\text{BrF}}_5$ (Bromine pentafluoride) Bromine (Group 17) has $V=7$. It forms 5 bonds, so $S=5$. Only 1 lone pair.

7. ${\text{XeF}}_4$ (Xenon tetrafluoride) Xenon is a noble gas (Group 18), boasting a full octet with $V=8$ valence electrons. It forms 4 bonds with Fluorine, so $S=4$. We have our second match! ${\text{XeF}}_4$ has 2 lone pairs. These lone pairs position themselves opposite to each other (axially) to minimize repulsion, resulting in a beautiful square planar geometry.

8. ${\text{SF}}_6$ (Sulfur hexafluoride) Sulfur has $V=6$. It forms 6 bonds, so $S=6$. Zero lone pairs.

After thoroughly examining all eight species, only ${\text{ClF}}_3$ and ${\text{XeF}}_4$ possess exactly two lone pairs on their central atoms. Therefore, the total number of species meeting the criteria is 2.