Angles of Shapes  [ENDORSED]

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

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Sydni Stewart
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Angles of Shapes

Postby Sydni Stewart » Sun Nov 11, 2018 3:08 am

For a shape like a tetrahedral or something more complicated, how do we know the angle, is that something that needs to memorized?

Olivia L 4E
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Re: Angles of Shapes

Postby Olivia L 4E » Sun Nov 11, 2018 9:44 am

Dr. Lavelle hasn't gone over this with us yet so this could be a part of Wednesday's lecture or it may not be something we learn how to calculate in this class. For now, I would suggest just memorizing (it will probably be on the formula page though) until further notice from Lavelle.

Ibrahim Malik 1H
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Re: Angles of Shapes

Postby Ibrahim Malik 1H » Sun Nov 11, 2018 11:26 am

VSEPR structures require a certain amount of memorization in order to remember the bond angle. For example, a Trigonal Pyramid has bond angles of <109 degrees. Dr. Lavelle hasn't mentioned whether we are required to memorize the angles yet or not, but I would assume so.

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Re: Angles of Shapes  [ENDORSED]

Postby Chem_Mod » Sun Nov 11, 2018 11:29 am

Yes the ideal bond angles for each shape will be covered in class and need to be known.

We have already done:

Linear (bond angle 180)
Trigonal Planar (bond angle 120)
Tetrahedral (bond angle 109.5)

David S
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Re: Angles of Shapes

Postby David S » Sun Nov 11, 2018 12:03 pm

While it is true that we probably should memorize the bond angles pertaining to the different electron geometries, you dont have to (and probably shouldn't memorize) molecular geometries.

Overall it's best I think to memorize the electron geometries (linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral) and use their idealized bond angles as a reference point. Then we use what we know about electron repulsion and lone pairs to figure out the actual bond angles.

Ex: NH3

The central atom N has 4 electron pairs around it (bonding + non-bonding). This would mean NH3 has electron geometry of tetrahedral, and the optimal repulsion between the electron pairs will result in a bond angle of 109.5 degrees.

However since one of the four pairs is a lone pair, and we know that lone pairs cause more repulsion, we know that the actual bond angle will be less than the ideal one.

Thus, we just need to say that the molecular gemometry bond angles (actual bond angle) is slightly less than 109.5°. The precise value of bond angles will vary between molecules, and we will NOT be expected to know them as far as I know.

Aside from that, you should be aware that for a molecule with electron geometry of trigonal bipyramidal, lone pairs are placed in the midsection that has electron regions in a trigonal planar configuration. For most other molecular geometries, lone pair placement doesn't matter.

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