13.a)

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

Moderators: Chem_Mod, Chem_Admin

Magdalena Palavecino 1A
Posts: 54
Joined: Fri Sep 29, 2017 6:04 am

13.a)

Postby Magdalena Palavecino 1A » Mon Nov 13, 2017 7:48 pm

I don't understand why the Iodine chain would be linear if there are two lone pairs bonded to the central atom. Shouldn't it be bent/angular instead of linear?

Magdalena Palavecino 1A
Posts: 54
Joined: Fri Sep 29, 2017 6:04 am

Re: 13.a)

Postby Magdalena Palavecino 1A » Mon Nov 13, 2017 7:51 pm

I read the previous answer to this question, but in the answers it places both electron lone pairs on the same side of the Iodine central atom, not on opposite sides. Is this because the answer booklet placed the electrons arbitrarily? Or because they actually do go on one side of the Iodine.

Isaac Eyler 1E
Posts: 12
Joined: Fri Sep 29, 2017 6:04 am

Re: 13.a)

Postby Isaac Eyler 1E » Tue Nov 14, 2017 6:28 pm

I was wondering about the same question (4.13a). The solutions manual lists two lone pairs on the right side of the central Iodine atom and one lone pair on the left side. This drawing makes it look like it should be angular. Can someone explain why it's linear?

Priyanka Bhakta 1L
Posts: 50
Joined: Fri Sep 29, 2017 6:04 am

Re: 13.a)

Postby Priyanka Bhakta 1L » Tue Nov 14, 2017 9:31 pm

Hi! So this is kind of confusing, given how we haven't really gone over these more difficult situations.

I have attached a graphic I found on the internet for this example. I think it ends up being linear because the 3 lone pairs on the Iodine repel each other and push against each other from both directions that the angles between the central Iodine and the ones on either side are pressured into staying 180 degrees because of how strong the repulsions are from each side of the central atom.

Hope this helps!
Attachments
220i3.jpg


Return to “Determining Molecular Shape (VSEPR)”

Who is online

Users browsing this forum: No registered users and 1 guest