Bond Angles

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

Moderators: Chem_Mod, Chem_Admin

Dustin Shin 2I
Posts: 64
Joined: Fri Sep 28, 2018 12:26 am

Bond Angles

Postby Dustin Shin 2I » Wed Nov 28, 2018 4:21 am

For the most part, are bond angles for the different structures supposed to be memorized or will we be given a chart of some sorts on the test? This might be a dumb question but I am just curious.

Posts: 64
Joined: Fri Sep 28, 2018 12:17 am

Re: Bond Angles

Postby taline_n » Wed Nov 28, 2018 8:47 am

Unfortunately there's no chart, you have to have all the bond angles memorized.

Posts: 31
Joined: Fri Sep 28, 2018 12:24 am

Re: Bond Angles

Postby katietietsworth_3c » Wed Nov 28, 2018 10:34 am

There is no chart, but it helped me to make my own chart with bond angles, shapes, electron domains, hybridization and polarity and memorize that for the test. Good luck!

Dana Wilks 3I
Posts: 30
Joined: Fri Sep 28, 2018 12:23 am

Re: Bond Angles

Postby Dana Wilks 3I » Wed Nov 28, 2018 11:37 am

Many of the bond angles are intuitive, while others are less so and should probably be memorized. For example, the bond angles for molecules with trigonal planar shape is 120 degrees (which is just the degrees of a circle- 360 degrees- divided by 3). If there is a lone pair, then it exerts more repulsive force, making the bond angle slightly less than 120. Molecules with tetrahedral shape have bond angles of 109.5, but if there's a lone pair, then you know that the bond angles will be slightly less than 109.5.

Kassidy Tran 1E
Posts: 77
Joined: Fri Sep 28, 2018 12:15 am

Re: Bond Angles

Postby Kassidy Tran 1E » Wed Nov 28, 2018 11:54 am

You can generally determine the bond angles if you know the shape too. For example if you know the shape is trigonal planar, then you can just divide 360 by 3 and get 120 degrees. This also applies to the trigonal bipyramidal, as that is essentially a trigonal planar shape intersected with a linear shape. I think the only one that you really need to know is the tetrahedral, which is 109.5 degrees. If there's a lone pair, you just need to know that the repulsive force is a bit stronger, so the bond angles would be in general less than it would be if the shape had no lone pairs.

Madison Gil 3D
Posts: 32
Joined: Fri Sep 28, 2018 12:23 am

Re: Bond Angles

Postby Madison Gil 3D » Wed Nov 28, 2018 1:17 pm

There is no chart, but it helps me to just remember the main numbers and their possible variations
Linear = 180
Trigonal Planar = 120
Tetrahedral = 109.5
Trigonal Bipyramidal = 90 and 120
Octahedral = 90
and if lone pairs are present on the molecule, the angles will be less.

Return to “Determining Molecular Shape (VSEPR)”

Who is online

Users browsing this forum: Aaina 1D and 2 guests