VSEPR Practice

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

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VSEPR Practice

Postby Chem_Mod » Wed Oct 21, 2015 2:44 pm

Hey guys,

Here are a few practice exercises to try for fun. For each of the following please...
1. Draw Lewis structure
2. Identify electron pair geometry and molecular geometry
3. Identify hybridization of the central atom
4. Determine if the molecule is polar or non polar

BrF4+
SO2
SO32-
HClO4
XeOF4

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Re: VSEPR Practice

Postby Chem_Mod » Thu Oct 22, 2015 12:00 am

IMG_8923.JPG

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Re: VSEPR Practice

Postby Chem_Mod » Thu Oct 22, 2015 12:00 am

IMG_8922.JPG

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Re: VSEPR Practice

Postby Chem_Mod » Thu Oct 22, 2015 10:08 am

Hey guys, here is another set of practice problems you can do for fun.

image.jpeg

Priscila Coronado 1C
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Re: VSEPR Practice

Postby Priscila Coronado 1C » Thu Oct 22, 2015 7:56 pm

For BrFl4+, the answer says polar but I remember Dr. Lavelle saying that if you usually have four of the same element surrounding another element, it's usually nonpolar. Can someone explain why in this case it's polar?

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Re: VSEPR Practice

Postby Chem_Mod » Fri Oct 23, 2015 8:12 am

My apologies, Dr. Lavelle has informed me that net dipole moments do not apply to cations and anions. Please disregard that question.

Brenda Melano 3H
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Re: VSEPR Practice

Postby Brenda Melano 3H » Sat Oct 24, 2015 9:54 am

Lewis structure of SO2.JPG
lewis structure of SO2
I came up with another lewis structure for SO2 that has a formal charge of 0 for all atoms. However, when I checked my answer, I noticed it was not included. Is this a valid lewis structure? I thought it would be better, in general, to always answer with a lewis structure that has the lowest energy (as little formal charge as possible)?

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Re: VSEPR Practice

Postby Chem_Mod » Sat Oct 24, 2015 8:44 pm

That is most certainly a valid Lewis structure. Dr. Lavelle mentioned in lecture that due to the difference in electronegativity, the resonance would be better representation. However, there is no way you could really determine that. I know I have gotten that wrong before. The best thing to do is just memorize that SO2 is best represented by the resonance form and not the one you have drawn. Again, there is no way you can really know that (it was determined experimentally) so just memorize it.

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Re: VSEPR Practice

Postby Chem_Mod » Sat Oct 24, 2015 9:04 pm

image.jpeg

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Re: VSEPR Practice

Postby Chem_Mod » Sat Oct 24, 2015 9:05 pm

image.jpeg

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Re: VSEPR Practice

Postby Chem_Mod » Sat Oct 24, 2015 9:43 pm

Here is one more practice excercise so that you guys have something to do Saturday night.

image.jpeg

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Re: VSEPR Practice

Postby Chem_Mod » Sun Oct 25, 2015 5:18 pm

Here is the solution for you guys. Again, I want to emphasize these practice exercises are not meant to take place of your assigned problems. Rather, they are meant to supplement them and give you a chance to apply what you already know. But you should definitely be studying additional problems as well.

FullSizeRender-1.jpg

Anna Nordstrom 1A
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Re: VSEPR Practice

Postby Anna Nordstrom 1A » Mon Oct 26, 2015 2:46 pm

Why is XeO3F2 nonpolar? Isn't there an extra O that doesn't "cancel out" with the others?

Adriana Juarez 1
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Re: VSEPR Practice

Postby Adriana Juarez 1 » Mon Oct 26, 2015 3:30 pm

Hello!

I had a question about the first set of practice questions.

For XeOF4, why is it necessary to give Xe a lone pair and O a double bond? Why not use that lone pair to complete Oxygen's octet and eliminate the need for the double bond?

-Adriana

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Re: VSEPR Practice

Postby Chem_Mod » Mon Oct 26, 2015 4:11 pm

Anna Nordstrom 1A wrote:Why is XeO3F2 nonpolar? Isn't there an extra O that doesn't "cancel out" with the others?

Anna, the three oxygens will "cancel each other out" while the two fluorines will do the same. Remember all three of them are on the same plane in a circular shape around the central atom.

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Re: VSEPR Practice

Postby Chem_Mod » Mon Oct 26, 2015 4:16 pm

Adriana Juarez 1 wrote:Hello!

I had a question about the first set of practice questions.

For XeOF4, why is it necessary to give Xe a lone pair and O a double bond? Why not use that lone pair to complete Oxygen's octet and eliminate the need for the double bond?

-Adriana


Adriana, oxygen does have a complete octet. If there was a single bond, xenon would have a formal charge and so would oxygen. In the solution presented, oxygen has a formal charge of zero and a full octet, and xenon has a formal charge of zero and an expanded octet.

AlexNguyen15
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Re: VSEPR Practice

Postby AlexNguyen15 » Mon Oct 26, 2015 6:10 pm

Could the lewis structure for (S03)2- be drawn with 3 single bonds and a lone pair? Or for the quiz should we always draw the lewis structure that has the most formal charges of 0?

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Re: VSEPR Practice

Postby Chem_Mod » Mon Oct 26, 2015 6:21 pm

AlexNguyen15 wrote:Could the lewis structure for (S03)2- be drawn with 3 single bonds and a lone pair? Or for the quiz should we always draw the lewis structure that has the most formal charges of 0?


You would always want the lowest energy Lewis structure, so that would be the one with the minimized formal charges.

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Re: VSEPR Practice

Postby AlexNguyen15 » Mon Oct 26, 2015 6:35 pm

What does "greater eff. nuclear charge" mean in relation to ionization energy and electron affinity trends? Would that be a good enough explanation for the quiz?

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Re: VSEPR Practice

Postby Chem_Mod » Mon Oct 26, 2015 7:56 pm

AlexNguyen15 wrote:What does "greater eff. nuclear charge" mean in relation to ionization energy and electron affinity trends? Would that be a good enough explanation for the quiz?


For IE, the greater effective nuclear charge means that it is harder to pull off an electron because it is being held tighter. For EA, it means that it will be easier to add an electron bc it will be attracted more due to increased eff. nuc. charge. Both answers would be perfectly valid on a test because it is the correct reason for those answers.

Adriana Juarez 1
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Re: VSEPR Practice

Postby Adriana Juarez 1 » Mon Oct 26, 2015 8:06 pm

Will we need to know paramagnetism and diamagnetism for this quiz? It's in our practice quizzes but I'm not sure if we've gone over it yet.

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Re: VSEPR Practice

Postby Chem_Mod » Mon Oct 26, 2015 8:42 pm

Adriana Juarez 1 wrote:Will we need to know paramagnetism and diamagnetism for this quiz? It's in our practice quizzes but I'm not sure if we've gone over it yet.


Adriana, nope. Only material up to hybridization is fair game for the quiz.

Adriana Juarez 1
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Re: VSEPR Practice

Postby Adriana Juarez 1 » Mon Oct 26, 2015 8:52 pm

Perfect, thanks.

I also have another question... Whenever it asks us to write the electron configuration in the quiz workbook the answer looks a bit strange and I'm not sure we went over how to do this in class... (examples include #10 on Quiz 2 Fall 2013, and #6 and #7 on Fall 2014).

Do we need to know how to do this or can we write the traditional electron configuration? (for Cl's valence e- config I wrote 3s2-3p6)

AlexNguyen15
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Re: VSEPR Practice

Postby AlexNguyen15 » Mon Oct 26, 2015 9:04 pm

Will we need to know all of the VSEPR shapes or just the ones we covered in class? For example pentagonal bipyramidal, square pyramidal, and T-shaped?

Christine Pahel 1C
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Re: VSEPR Practice

Postby Christine Pahel 1C » Mon Oct 26, 2015 9:26 pm

How do you determine when to look at unpaired electrons in subshells or at the nuclear charge in order to figure out which atom has a greater ionization energy and/or electron affinity?

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Re: VSEPR Practice

Postby Chem_Mod » Mon Oct 26, 2015 9:30 pm

AlexNguyen15 wrote:Will we need to know all of the VSEPR shapes or just the ones we covered in class? For example pentagonal bipyramidal, square pyramidal, and T-shaped?

You should be able to answer any of them.

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Re: VSEPR Practice

Postby Chem_Mod » Mon Oct 26, 2015 9:33 pm

Christine Pahel 1C wrote:How do you determine when to look at unpaired electrons in subshells or at the nuclear charge in order to figure out which atom has a greater ionization energy and/or electron affinity?

Only in cases where there is a half/fully filled subshell (the exceptions) would the electrons matter (and you're comparing atoms that are in the same row and adjacent).

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Re: VSEPR Practice

Postby Chem_Mod » Mon Oct 26, 2015 9:34 pm

Hey guys, please try and post any questions in the appropriate forum category from now on. It is getting a little off topic in this thread.

AlexNguyen15
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Re: VSEPR Practice

Postby AlexNguyen15 » Mon Oct 26, 2015 9:50 pm

Do we need to know anything about Covalent versus Ionic bonds?

Rohan Singhal 2D
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Re: VSEPR Practice

Postby Rohan Singhal 2D » Wed Oct 28, 2015 4:38 pm

Brenda Melano 1A wrote:
Lewis structure of SO2.JPG
I came up with another lewis structure for SO2 that has a formal charge of 0 for all atoms. However, when I checked my answer, I noticed it was not included. Is this a valid lewis structure? I thought it would be better, in general, to always answer with a lewis structure that has the lowest energy (as little formal charge as possible)?


I believe one acceptable explanation that I was told is that in the lewis structure you have drawn, only the Oxygens have an octet, but in the structure with resonance you have a low formal charge, and all the molecules have formed octets.

Rohan Singhal 2D
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Re: VSEPR Practice

Postby Rohan Singhal 2D » Wed Oct 28, 2015 5:05 pm

I was wondering why the HClO4 molecule and the XeOF4 molecule are both polar. There are 4 Oxygen atoms attached directly to the Cl in the HClO4 molecule, so I thought they would have the same pull on the electrons. Or is it because of the extra H that theres a polarity? And is this the same explanation for XeOF4, except with an O attached to a F which is directly attached to the Xe?

LinaLi2E
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Re: VSEPR Practice

Postby LinaLi2E » Sun Nov 01, 2015 4:52 pm

Rohan Singhal 3H wrote:I was wondering why the HClO4 molecule and the XeOF4 molecule are both polar. There are 4 Oxygen atoms attached directly to the Cl in the HClO4 molecule, so I thought they would have the same pull on the electrons. Or is it because of the extra H that theres a polarity? And is this the same explanation for XeOF4, except with an O attached to a F which is directly attached to the Xe?

The reason why HClO4 molecule is polar is because of what you said, which is that there is a H atom attached to an O atoms, making the pull stronger. The reason why XeOF4 is polar is because it is a molecule with a square pyramidal shape. The O atom is not attached to a F atom, but to the central Xe atom. Then there is a lone pair attached to the Xeon the opposite side of the O atom. Since there is not an odd number of atoms attached to the central Xe, the dipoles do not cancel out and the molecule is polar.


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