lone pairs on H2O

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Mrudula Akkinepally
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lone pairs on H2O

Postby Mrudula Akkinepally » Fri Nov 27, 2020 6:39 pm

Hi everyone! In lecture 21, I noticed that the lone pairs for an H2O molecule are right next to each other on the O atom, while two lone pairs are on opposite ends of a xenon atom in a XeF4 molecule. I was wondering why that is. Does it have to do with the fact that Xe is a noble gas?

Joanne Yuh 3I
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Re: lone pairs on H2O

Postby Joanne Yuh 3I » Fri Nov 27, 2020 7:15 pm

I'm not entirely sure if my response is correct but I think it has to do with the shape of the H2O molecule. The water molecule is a bent shape so it makes more sense for the two lone pairs to be next to each other. And I think since XeF4 has a nonpolar structure, it has its 2 lone pairs opposite from each other.

Nathan Lao 2I
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Re: lone pairs on H2O

Postby Nathan Lao 2I » Fri Nov 27, 2020 7:40 pm

I think it has to do with the fact that the regions of electron density create a tetrahedral structure (the molecule is bent but the electron geometry is tetrahedral). If you switch any of the positions of the atoms/lone pairs, the structure will essentially be the same thing again. Dr. Lavelle describes this concept in lecture #21 at 15:00.

In terms of XeF4, it has six regions of electron density where two of them are lone pairs. The lone pairs have to be on opposite sides of the structure to minimize electron repulsion. This creates the most stable structure.

Mary Shih 3J
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Re: lone pairs on H2O

Postby Mary Shih 3J » Fri Nov 27, 2020 8:02 pm

I would suggest looking at the geometry instead of the lewis structure. the lewis structures are not a good indicator oh the location of the lone pairs. Although Xef2 and h2o have a diff number of lone pairs, they are both linear because the lone pairs want to repel each other

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Re: lone pairs on H2O

Postby Ian_Lee_1E » Fri Nov 27, 2020 8:25 pm

If you have watched the VSEPR Notation that Professor Lavelle showed in Lecture 21, you would get the glimpse of how many outer atoms would lead to what shape.

For oxygen, since it is AX2E2, the shape would be tetrahedral - 2, which would give you a bent shape. You can take out any of the bond pairs and it would not matter since all the bond pairs are far apart equally.

However, for the XeF4, you would take out the top two because taking out the axial bond pairs would give you the least amount of repulsion since they are farthest apart from each other.

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