Given Problem: Draw the Lewis dot structures, identify the molecular geometry, and give the bond angle for the following compounds:
1. BeF2: The first step is to draw the Lewis dot structure. This can be done by identifying how many bonding groups are involved. A bonding group, counts as 1 lone electron, an electron pair, a single bond, a double bond, or a triple bond. For this compound, we have three elements, and two single bonds. It should be noted that the bonding motif for fluorine, or any other halogen is -X with its 3 electron pairs on the outside as shown.
This yields a linear molecular geometry for BeF2. If we were looking for the electron structure, we would also regard the lone pairs on each fluorine molecule. Because it’s a linear molecule, the bond angle between the groups is 180°.
2. Now we’ll go onto BCl3. Because there’s one boron to every chlorine, it’s safe to say that it’s probably the middle molecule.
3. Boron has 3 valence electrons, so those can bond with the valence electrons from the three chlorines in a chemically stable way.
4. The molecular geometry for this molecule is trigonal planar. I tried to draw a structure of a dummy molecule next to our BCl3 to show how the molecules are on the same plane. That, and the triangular shape obtained by reciprocal forces causes the 120° bond angle.
5. Now we can move on to the Sulfur hexaiodide. There are 7 bonding groups, but 6 bonds total. Just like the other halogens in this example, Iodine bonds naturally as a single bond with it’s remaining electron pairs on the “outside”.
6. Since sulfur has 6 valence electrons, each iodine can bond naturally. The six iodines can bond to the same sulfur.
7. I messed up my drawing (don’t do chemistry in Sharpie, kids!). This is a standard way to draw molecules. The hatched bonds indicate the Iodines are placed behind the sulfur molecule. The wedged bonds show that they are in front of the sulfur, almost as if they’re sticking out from the paper. These four are. however, on the same plane.
8. To show the different planes, I labeled them with red and blue.
9. Every bond angle in an octahedral molecule is 90°, intraplanar and interplanar.
Shoutout to chem.purdue.edu for the example problem since I forgot my workbook for break! Head on over there to check out more VSEPR theory based problems.
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