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### Formal Charge

Posted: Tue Oct 18, 2016 7:21 pm
1) I don't get how turning the electron dots into dashes (one dash represents an electron pair) makes a bond "more stable".

2) Also what does "formal charge" mean? For example, one sulfur atom in the middle (it's in the middle because ionization energy is lower than oxygen's ionization energy) surrounded by two oxygen atoms shares one electron pair with each oxygen atom. However, it turns into +1 when another dash is created in place of two shared electrons, why does it change into +/- values??

3) One more question, in the "most stable" state, sulfur has two dots and 4 dashes.. how is this possible since outter shell only hold 8 electrons yet these dots and dashes total 10 electrons????

If you only have the answer to one or two questions, it's okay, any clarification is appreciated (:

### Re: Formal Charge  [ENDORSED]

Posted: Tue Oct 18, 2016 7:38 pm
For the first question, the reason why its more stable is because for whatever element you're using, when you calculate the formal charge of it, it's closer to 0 then it would've been if you didn't add the dashes. The closer it is to 0, the more stable it is.

### Re: Formal Charge

Posted: Tue Oct 18, 2016 9:46 pm
For the second question, the formal charge of an atom indicates either a gain or loss of electrons when forming a covalent bond. The equation for formal charge is FC = V - (L + S/2) where V is the number of valence electrons, L is the number of lone pair electrons, and S is the number of shared electrons. For your example when there is a double bond between the sulfur and an oxygen atom, the sulfur's formal charge is FC = 6 - (2 + 6/2) = -1.

### Re: Formal Charge

Posted: Wed Oct 19, 2016 11:28 am
Third question: Sulfur can hold more electrons in it's valence shell because it borrows space from a d-orbital. When the electrons don't fit in the p-orbital anymore, it just adds the electrons to the next available orbital (3d). However, you can only do this if the atom is in the 3rd period or lower since the 1st and 2nd period don't have d-orbitals.