7th edition 2E.5

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

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Saachi_Kotia_4E
Posts: 68
Joined: Fri Sep 28, 2018 12:23 am

7th edition 2E.5

Postby Saachi_Kotia_4E » Sun Nov 11, 2018 11:25 pm

Can someone explain how to do this problem? what does the lewis structure look like and how can we determine the bond angle based off of that?

armintaheri
Posts: 68
Joined: Fri Sep 28, 2018 12:26 am

Re: 7th edition 2E.5

Postby armintaheri » Sun Nov 11, 2018 11:38 pm

First step of drawing a Lewis structure: count your valence electrons. 7 from the chlorine and 6 from each oxygen gives you 19. Since there is a charge of +1, you have 18 valence electrons. Normally, chlorine would only be able to form one bond, since it only has one unpaired electron. But since we have a charge of +1, one electron is missing. So chlorine has two unpaired electrons and can form two bonds, one with each oxygen. That gets rid of 4 valence electrons, two per bond. Each oxygen has 3 lone pairs, accounting for another 12 valence electrons. But now chlorine has an incomplete octet. So it's going to use its d-orbitals to form an additional bond with one of the oxygens (it doesn't matter which. It's a resonance structure). So you're left with one lone pair on the chlorine, getting rid of the last two valence electrons. Since we have two bonds and a lone pair, it's trigonal planar geometry. The bond angles in trigonal planar are 120. But since the lone pair repels more than a bond would, the angle between the bonds is actually slightly less than 120.


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