## 7th edition 2E.5

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

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

### 7th edition 2E.5

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

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.