Lewis Structure + Molecular shapes

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Discussion Overview

The discussion centers around the molecular shape of the ion ICl4-, specifically examining its Lewis structure and the implications for its geometry. Participants explore various molecular shapes, hybridization, and electron domains, with a focus on understanding the relationship between these concepts and the correct classification of ICl4-.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant describes drawing the Lewis structure for ICl4- and expresses confusion over the total number of electrons and electron domains, suggesting a possible error in their drawing.
  • Another participant asserts that the molecular shape cannot be square planar and claims it is a square pyramid, but does not provide a clear rationale.
  • Several participants discuss the hybridization of iodine in ICl4-, with some suggesting it is sp^3d^2 and others proposing sp^3d, indicating a lack of consensus on the hybridization model.
  • There is a debate about the number of electron pairs and domains, with one participant correcting another's count of electron pairs and asserting that there are only 6 pairs total, leading to confusion about the geometry.
  • One participant visualizes the structure as a square cardboard with chlorine atoms at the corners and lone pairs above and below, attempting to clarify the spatial arrangement.
  • Another participant provides a detailed breakdown of bonding and non-bonding domains for various molecular geometries, contributing to the understanding of the topic.
  • There is a question raised about the source of the extra electron that contributes to the negative charge of the ion, indicating a gap in understanding among participants.

Areas of Agreement / Disagreement

Participants express differing views on the molecular shape of ICl4-, with some supporting square planar and others suggesting square pyramidal. The discussion remains unresolved regarding the correct hybridization and the implications for molecular geometry.

Contextual Notes

Participants exhibit uncertainty regarding the terminology and concepts related to molecular geometry, hybridization, and electron domains. There are unresolved mathematical steps and definitions that contribute to the confusion.

Who May Find This Useful

This discussion may be useful for students preparing for exams in chemistry, particularly those studying molecular geometry, Lewis structures, and hybridization concepts.

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


Daniel.
 
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?
 
Also the answer sheet says its square planar.
 
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 just said means.
 
I'm sorry,you're right.The hybridization is not [itex]\mbox{sp}^{3}[/itex],so it's not a pyramid (somewhat similar to methane).The "d" orbitals of Iodine participate,too.I think it's [itex]\text{sp}^{2}\text{d}[/itex] who's got a planar structure.


Daniel.
 
The hybridization is [itex]sp^3d^2[/itex]. The structure is square planar. There will be two non-bonding pairs oriented normal to the plane (along +z and -z).
 
whozum said:
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.
 
  • #10
Gokul43201 said:
The hybridization is [itex]sp^3d^2[/itex]. 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. :frown:
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.
 
  • #11
Gokul43201 said:
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?
 
  • #12
Yes,it should be octaedrical,typical for [itex]\mbox{sp}^{3}\mbox{d}^{2}[/itex] hybridization.

Daniel.
 
  • #13
whozum said:
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.
 
  • #14
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.
 
  • #15
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! :smile:
 
  • #16
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
 
  • #17
Hold on a second.Isn't [itex]\mbox{CO}_{2}[/itex]

[tex]\left|\bar{O}=C=\bar{O}\right|[/tex]

,with 4 pairs ...?


Daniel.
 
  • #18
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.
 
  • #19
Okay,i see.I wasn't familiar with this terminology.

Daniel.
 
  • #20
where does the extra electron come from that makes the entire molecule negative?
 

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