109.5 degrees

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

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Michelle Pham_3H
Posts: 26
Joined: Fri Sep 29, 2017 7:07 am

109.5 degrees

Postby Michelle Pham_3H » Tue Nov 14, 2017 10:15 am

I am a bit confused about where this value came from. Was it just calculated and given to us, or are we able to determine this value ourselves? Thanks in advance!

Vasiliki G Dis1C
Posts: 53
Joined: Fri Sep 29, 2017 7:04 am

Re: 109.5 degrees

Postby Vasiliki G Dis1C » Tue Nov 14, 2017 10:31 am

The bond angles have been experimentally determined, and you can use the VSEPR theory/model to coordinate the shape/geometry with the bond angle that has already been determined for that shape. Hope this helps!

Ryan Sydney Beyer 2B
Posts: 82
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time

Re: 109.5 degrees

Postby Ryan Sydney Beyer 2B » Tue Nov 14, 2017 12:49 pm

You can also calculate the exact value of this tetrahedral angle using a cube and it's vertices. The exact value comes out to be



I've attached the link below but I think for the purposes of this class, we should know that if there is a central molecule with 4 bonds then the angle is going to be 109.5. If there is a central molecule with 4 regions of electron density and 3 of them are bonds, then the angle will be slightly less than 109.5 and the same for a central molecule with 4 regions of electron density and 2 of them are bonds.

http://www.ctralie.com/Teaching/Tetrahedron/

Maddie Hong 1I
Posts: 21
Joined: Fri Sep 29, 2017 7:04 am

Re: 109.5 degrees

Postby Maddie Hong 1I » Tue Nov 14, 2017 7:13 pm

Just to add, 109.5 degrees is only for tetrahedral molecules with no lone pairs!


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