109.5 Degrees
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 64
- Joined: Fri Sep 28, 2018 12:15 am
109.5 Degrees
I understand that a linear shape has 180 degrees, and a trigonal planar has 120. However, why does a tetrahedral have 109.5 degrees? Is there a specific mathematical or scientific reason behind this, or is this just a value we need to memorize?
Re: 109.5 Degrees
This is because it is a three dimensional shape so its wont just be 360 divided by the number of sides. This is a link to a mathematical proof laying out how to come to 109.5 degrees. http://www.ctralie.com/Teaching/Tetrahedron/
-
- Posts: 63
- Joined: Fri Sep 28, 2018 12:27 am
Re: 109.5 Degrees
If all the electron densities were on one plane, then the angle would be 90 degrees.
However, since it is a three dimensional object, it can take advantage and spread out even more into more than one plane. Since the electron densities want to be far away from each other as possible, it will do so, and make 109.5 degree bond angle, which is greater than a 90 degree one.
However, since it is a three dimensional object, it can take advantage and spread out even more into more than one plane. Since the electron densities want to be far away from each other as possible, it will do so, and make 109.5 degree bond angle, which is greater than a 90 degree one.
-
- Posts: 32
- Joined: Fri Sep 28, 2018 12:17 am
Re: 109.5 Degrees
Are tetrahedrals and molecules that are bent with four domains the only shapes that have bond angles of 109.5?
-
- Posts: 70
- Joined: Fri Sep 28, 2018 12:24 am
Re: 109.5 Degrees
Bent molecules actually do not have a bond angle of 109.5 degrees because the lone pairs are more electronegative and push the other atoms more closely together. This would cause the bond angle of a bent molecule to be less than 109.5, the exact value i believe was 104.5 degrees. I think the only molecule with 109.5 angle would be the tetrahedral or maybe some other larger molecules with the same sort of structure as a tetrahedral, but those would most likely have other angles as well.
Return to “Determining Molecular Shape (VSEPR)”
Who is online
Users browsing this forum: No registered users and 3 guests