## In Class Example, Sulfate

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

Fatemah Yacoub 1F
Posts: 114
Joined: Thu Jul 11, 2019 12:16 am

### In Class Example, Sulfate

When computing the formal charges for SO4^2- in class he said that by adding more bonds and making all the oxygens have a 0 formal charge it would make sulfur have a 2- charge and we would rather have the sulfur have a 0 charge than the oxygens? Why would the compound be more stable with oxygen having a negative formal charge rather than sulfur?

Daniel Honeychurch1C
Posts: 109
Joined: Thu Jul 11, 2019 12:15 am

### Re: In Class Example, Sulfate

Molecules are more stable when the atoms with a higher electronegativity have a negative formal charge. Since oxygen is more electronegative than sulfur, SO4-2 is more stable when oxygen has a -1 formal charge instead of sulfur.

Victoria Li 1L
Posts: 51
Joined: Sat Jul 20, 2019 12:15 am

### Re: In Class Example, Sulfate

It has to do with electron affinity. Atoms with lower affinities are less likely to have a stable negative formal charge because they're less likely to want to hold onto an added electron. Since electron affinity tends to decrease down a group, sulfur has a lower affinity than oxygen, and therefore should not have a lower formal charge than oxygen.

Sean Cheah 1E
Posts: 105
Joined: Wed Sep 18, 2019 12:20 am

### Re: In Class Example, Sulfate

After getting the formal charge on each atom as close to zero as possible, place the remaining negative formal charge(s), if present, on the more electronegative atom(s) and the positive formal charge(s), if present on the less electronegative atom(s) to get the most stable structure. The reason that this is done has to do with the fact that atoms with higher electronegativity will attract bonding electrons more strongly and are therefore more likely to have a negative formal charge.

Note, however, that electronegativity is not the same thing as electron affinity, which describes the energy released when an electron is added to a neutral atom of the element in the gas phase. I know that Lavelle made certain claims about trends in electron affinity during lecture but, honestly, I think that the trends that he pointed out are better used to describe electronegativity than electron affinity (at least for elements in periods 1 - 4). If you look up an image of electron affinities arranged in the likeness of a periodic table, you can pretty quickly find all kinds of weirdness like how chlorine actually has a higher electron affinity than fluorine and, more relevantly to your particular question, sulfur has a higher electron affinity than oxygen. And, for reasons I don't fully understand myself, this is all somehow consistent with the fact that fluorine still does have a higher electronegativity than chlorine, and oxygen has a higher electronegativity than sulfur!

Tl;dr: For maximum stability, residual negative formal charges are assigned to the more electronegative elements. Electronegativity =/= electron affinity