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

Moderators: Chem_Mod, Chem_Admin

Posts: 108
Joined: Thu Jul 25, 2019 12:16 am


Postby JamieVu_2C » Sun Nov 17, 2019 5:14 pm

For I3-, the shape is linear with 3 lone pairs around the central I atom. Why does the bond angle equal 180 degrees instead of being slightly below 180 degrees when there are 3 lone pairs of electrons around the central I atom? Wouldn't those lone pairs slightly push down on the bond angles?

Brian J Cheng 1I
Posts: 115
Joined: Thu Jul 11, 2019 12:15 am

Re: 2E.13

Postby Brian J Cheng 1I » Sun Nov 17, 2019 5:23 pm

I would try to visualize this in 3 dimensions. If there were no lone pairs, the molecule would have a trigonal bipyramidal shape with 180 degree bond angle through the axis and 120 degrees equatorial. But because there are 3 lone pairs, you remove the three equatorial atoms, causing enough repulsions to preserve the 180 degree bond angle. It's tricky but sometimes you just need to hard code the tendencies into your head XD

Return to “Determining Molecular Shape (VSEPR)”

Who is online

Users browsing this forum: No registered users and 2 guests