Lewis Structure + Molecular shapes

[whozum]

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 that's for a square planar.

Help? My test is next week.

 

[dextercioby]

It can't be square planar.It's a square pyramid.I hope u see why.


Daniel.

 

[whozum]

In all honesty I can't 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?

 

[whozum]

Also the answer sheet says its square planar.

 

[dextercioby]

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.

 

[whozum]

I have no idea of what what you just said means.

 

[dextercioby]

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.

 

[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).

 

[Gokul43201]

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.

 

[dextercioby]

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.

 

[whozum]

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?

 

[dextercioby]

Yes,it should be octaedrical,typical for $\mbox{sp}^{3}\mbox{d}^{2}$ hybridization.

Daniel.

 

[Gokul43201]

The electron geometry would be the x-y-z looking one, octahedral?

I'm not sure if this is what you mean by "x-y-z looking one" but essentially, you have one electron pair pointing along each of the 6 rectangular co-ordinate axes directions (+x, -x, +y, -y, +z, -z). The alternate way of picturing it is in terms of a pair of square pyramids stuck to each other; or otherwise, 4 pairs pointing towards the corners of a square (with the central atom in the middle of the square) and 2 pairs pointing normally up and down.

 

[whozum]

xyz looking one means the structure where it lokos like the xyz axes on a graph. I believe we are talking about the same one, octahedral. I just need to look over the names of the geometries.

Thanks to both of you.

 

[atomicblast]

It would be [c] because looking at I itself, it has 2 lone pair (4 valence electrons; don't forget the -ve charge on the molecule) and 4 arms sticking out to hold the Cl atoms to it.

Don't panic, try to visualise it this way...a piece of square cardboard with the I atom in the middle, 4 Cl atoms at the corners, and then you have 2 big lobes (2 lone paris) sticking out from the top and bottom of the cardboard.

Hope this helps!

 

[atomicblast]

Some additional notes to help you prepare (try to memorise them!)

Bonding domains (BD) - bonded to another atom
Non-bonding domains (NBD) - signify lone pairs

[2 pairs]
Linear - 2 BD, 0 NBD e.g. CO2

[3 pairs] incld lone pairs
Trigonal planar - 3 BD, 0 NBD
Bent - 2 BD, 1 NBD

[4 pairs] incld lone pairs
Tetrahedral - 4 BD, 0 NBD
Trigonal pyramidal - 3 BD, 1 NBD
Bent (tetrahedral) - 2 BD, 2 NBD e.g. H20

[5 pairs] incld lone pairs
Trigonal bipyrimidal - 5 BD, 0 NBD
Seesaw - 4 BD, 1 NBD
T-shaped - 3 BD, 2 NBD
Linear (trigonal bipyramidal) - 2 BD, 3 NBD *this one has 3 lobes (3 lone pairs) surrounding the central atom on a plane and 2 atom attached to it from the top and bottom.*

[6 pairs] incld lone pairs
Octahedral - 6 BD, 0 NBD
Square pyrimidal - 5 BD, 1 NBD
Square planar - 4 BD, 2 NBD

 

[dextercioby]

Hold on a second.Isn't $\mbox{CO}_{2}$

$$\left|\bar{O}=C=\bar{O}\right|$$

,with 4 pairs ...?


Daniel.

 

[atomicblast]

CO2 has 2 bonding domains, the lone pairs are on the oxygen atoms and not the central carbon atom. Each double bond counts as one electron domain.

No of electron domains = No of atoms bonded to central atom + No of non-bonding pairs on the central atom.

 

[dextercioby]

Okay,i see.I wasn't familiar with this terminology.

Daniel.

 

[jd527]

where does the extra electron come from that makes the entire molecule negative?