Homework 4.15 B

(Polar molecules, Non-polar molecules, etc.)

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Homework 4.15 B

Postby SammiOrsini_1B » Thu May 31, 2018 9:24 am

In problem number 4.15 b, we need to find the lewis structure and shape of TeCl4. When you calculate the number of valence electrons you get 34, so when you draw the lewis structure Te has four bonds and also has a lone pair. I thought the shape would be a tetrahedral however the answer is see saw. Is this see saw shape due to the fact that Te has a lone pair? Also how do we know that the bond angles for a see saw are 120 and 90? Is there a way to calculate this or are the bond angles experimentally determined and we just need to memorize the angles? Thank you for any help!

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Re: Homework 4.15 B

Postby AnnaYan_1l » Thu May 31, 2018 11:04 am

Hi Sammi! TeCl4 has a see saw shape. Professor Lavelle used a similar example in class using SF4 (I believe on Wednesday of Week 8 if you want to refer to your notes). The reason why TeCl4 and SF4 have a see saw shape is because they have the VSEPR notation of AX4E (4 bonding pairs and 1 lone pair).

The shape with the least amount of repulsion (and thus more stable) with the notation AX4E has these parts:
- 5 pairs of electrons located in a trigonal bypyramidal geometry
- Lone pair in equatorial plane (more repulsive lone pair interacts with only 2 bonds at 90 degrees)
- Since only 4 of 5 positions = occupied by atoms, these 4 atoms form the shape of a seesaw

I attached a visual of what the see saw and trigonal bipyramidal shapes look like. You can see that the see saw looks very similar to a trigonal bipyramidal shape, with a 5th atom removed (note that it is more stable to remove the atom from the "middle" area than it is from the "top" or "bottom."

Furthermore, I would just memorize the angles. There is a post called "Helpful Chart," I believe, that shows an easy way to memorize the shapes and angles.
220px-Trigonal-bipyramidal-3D-balls.png (20.02 KiB) Viewed 160 times
VSEPR.jpg (6.67 KiB) Viewed 160 times

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