## Lewis Structure + Molecular shapes

3. Which of the following is the molecular shape of the ion, ICl4-?

A. Tetrahedral
B. See saw
C. Square planar
D. Square pyramidal
E. Trigonal pyramidal

I drew the lewis structure, 4 chlorines bound to an Iodine. Iodine has 7, Each clorine has 7 with 35 total, the bonds account for 4, so distributing the rest of the 31 electrons, I put 6 on each chloride, and 6 on the iodine, with the extra electron floating since its a negative ion.

This is kinda bugging me since Iodine has 8 electrons from the bonds alone, and now the extra electrons put it up to 14. I must have drawn it wrong, but anyway:

Molecular shape is decided by electron domains, and since we have 4 from bonds and 3 from electron pairs, 7 total. Thats way too many, we only went up to 6 electron domains in class, and thats for a square planar.

Help? My test is next week.
 PhysOrg.com chemistry news on PhysOrg.com >> New method for producing clean hydrogen>> Making ice-cream more nutritious with meat left-overs>> Non-wetting fabric drains sweat
 Blog Entries: 9 Recognitions: Homework Help Science Advisor It can't be square planar.It's a square pyramid.I hope u see why. Daniel.
 In all honesty I cant relate the names to the geometries. I know the seesaw one is a vertical (z) axis, and on the xy plane, 3 branches 120 degrees apart. I believe square planar is shaped like an x,y,z axis is set up. Do you know if I did my lewis structure right?

## Lewis Structure + Molecular shapes

Also the answer sheet says its square planar.
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Yes,one electron does provide the mesomerical (Pauling) structures...But it's okay.Just the geometry that u found was incorrect.Maybe u'll learn about Gillespie's models,too. Daniel.
 I have no idea of what what you jsut said means.
 Blog Entries: 9 Recognitions: Homework Help Science Advisor I'm sorry,you're right.The hybridization is not $\mbox{sp}^{3}$,so it's not a pyramid (somewhat similar to methane).The "d" orbitals of Iodine participate,too.I think it's $\text{sp}^{2}\text{d}$ who's got a planar structure. Daniel.
 Recognitions: Gold Member Science Advisor Staff Emeritus The hybridization is $sp^3d^2$. The structure is square planar. There will be two non-bonding pairs oriented normal to the plane (along +z and -z).

Recognitions:
Gold Member
Staff Emeritus
 Quote by whozum Molecular shape is decided by electron domains, and since we have 4 from bonds and 3 from electron pairs, 7 total.
Here's your mistake. There are only 6 pairs : 4 bonding pairs with the chlorines, (that leaves 3 electrons on I, plus the extra electron for the negative charge), and 2 non-bonding pairs.

Blog Entries: 9
Recognitions:
Homework Help
 Quote by Gokul43201 The hybridization is $sp^3d^2$. The structure is square planar. There will be two non-bonding pairs oriented normal to the plane (along +z and -z).
That means two things

1.I screwed it up real badly.
2.The structure should be octaedrical.But it's square planar,because of the 2 nonbonding pairs which don't count.Reminds me of water molecule with 2 non bonding pairs.

Daniel.

 Quote by Gokul43201 Here's your mistake. There are only 6 pairs : 4 bonding pairs with the chlorines, (that leaves 3 electrons on I, plus the extra electron for the negative charge), and 2 non-bonding pairs.
Ok so there are two electron pair domains on the iodine aside from the bonds. So the chlorines iwll all be one a plane. The extra electron pairs will go perpendicular to this plane, and since they arent molecular, have no structure.

The electron geometry would be the x-y-z looking one, octahedral?
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Yes,it should be octaedrical,typical for $\mbox{sp}^{3}\mbox{d}^{2}$ hybridization. Daniel.

Recognitions:
Gold Member
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Hold on a second.Isn't $\mbox{CO}_{2}$ $$\left|\bar{O}=C=\bar{O}\right|$$ ,with 4 pairs ...? Daniel.