Question 4.13 a

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

Moderators: Chem_Mod, Chem_Admin

Kendall Schemmer 1I
Posts: 20
Joined: Fri Sep 29, 2017 7:03 am

Question 4.13 a

Postby Kendall Schemmer 1I » Mon Nov 13, 2017 1:55 pm

Determine the VSEPR formula and molecular shape of I₃-

I understand why the VSEPR formula would be: AX₂E₃

But why would the molecular shape be linear if there is an odd number of lone pairs? I would imagine that the two lone pairs above the central I atom would be more "powerful" than the one lone pair below.

Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times

Re: Question 4.13 a

Postby Chem_Mod » Mon Nov 13, 2017 2:03 pm

It seems you assume that two lone pairs would be axial and one lone pair would be equatorial, so that one bond would be equatorial, one axial.

If we see three lone pairs as equatorial, and the two bonds as axial, it would make sense that the molecule is linear.

Hope this helps.

Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

Re: Question 4.13 a

Postby Cooper1C » Mon Nov 13, 2017 3:41 pm

If you look on page 118 in the textbook, there are really good diagrams that show the shapes. If you think about a molecule with 5 regions of electron density and no lone pairs, the shape is trigonal bipyramidal. Then if you replace the 3 atoms in the plane with 3 lone pairs, the shape is linear.

Return to “Determining Molecular Shape (VSEPR)”

Who is online

Users browsing this forum: No registered users and 3 guests