## 6th edition: 3.23

$FC=V-(L+\frac{S}{2})$

Katherine Grillo 1B
Posts: 47
Joined: Thu Sep 27, 2018 11:28 pm

### 6th edition: 3.23

How can chlorine exist in both positive and negative oxidation states? And how do you find the maximum and negative oxidation number that chlorine can have?

julia_lok_2K
Posts: 41
Joined: Thu Sep 27, 2018 11:25 pm

### Re: 6th edition: 3.23

I think it complies with the octet rule, as in taking away electrons or adding more electrons until you get the previous or next noble gas configuration. So to find the minimum oxidation number, you take away electrons from the atom's outer shell until you reach the preceding noble gas. In this case, you take away electrons from Cl until you reach Ne--by this point you would have taken away 7 electrons, so the oxidation number is +7. Same concept applies to maximum oxidation number.

Hope this helps!