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Topic: VSEPR HELP  (Read 2831 times)

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Offline devilangelgod

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VSEPR HELP
« on: January 10, 2010, 10:59:30 PM »
I need help with this vsepr theory worksheet, finished it already but im not sure my answers are even right.

here is the worksheet.

http://henry.mpls.k12.mn.us/sites/40ffb804-cceb-41db-92d5-ed0b7a6eb987/uploads/VSEPR_Worksheet.pdf

Offline savy2020

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Re: VSEPR HELP
« Reply #1 on: January 11, 2010, 05:41:40 AM »
Post your answers. We'll tell if they are right. ;)
:-) SKS

Offline devilangelgod

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Re: VSEPR HELP
« Reply #2 on: January 11, 2010, 04:21:24 PM »
1. im not sure.
2. bent, tetrahedral, trigonal pyramidal.
3. linear, trigonal planar, tetrahedral.
4. decreases.
5. H2O
6.increases.
7. i think i know those.
8. tetrahedral, trigonal planar, tetrahedral.

Offline savy2020

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Re: VSEPR HELP
« Reply #3 on: January 11, 2010, 04:57:46 PM »
1. Not Sure!! OK. In a line- VSEPR says that electron pairs around the central atom arrange themselves in such a geometry where the repulsions are minimum.
    Refer http://en.wikipedia.org/wiki/VSEPR_theory
2.Correct
3.SnCl2 - Is it trigonal planar? How do you justify that?. . It is bent {For determining molecular geometry ,lone pairs aren't considered -only the bond pairs}. Remaining two are correct
4.Right.
5.How?? Apply your answer for the 4th question here. It's pretty straight forward.
6.The question is a bit ambigious. I don't think a general trend exists but generally i think the bond angles increases as no. of bonds increase.
7.OK
8. In order to find the molecular geometry consider the no. of lone pairs and bond pairs on the central atom.
For Eg: Say in TeCl4 there are 4 bond pairs(obvious from no. of Cl's) and 1 lone pair (how? - Te has 6 valence e-s 4 have been used in bonding). So total of 5 e- pairs surrounding Te. These e- pairs will be arranged in a trigonal bi-pyramidal geometry (This is the electron pair arrangement geometry not the molecular geometry) . But since one of those is a lone pair, in considering molecular geometry we do not consider that electron pair. Now the question arises - in the trigonal bi-pyramidal structure where will the lone pair be- in the axial position or the equatorial? The lone pair would be in the equatorial position only since in that case the repulsion are less that if the lone pair was in axial position (..try to reason why the repulsions are less in equatorial..)
Reason out in the similar way for ICl3 and XeF4 also.

:-) SKS

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