## Formal charge and most electronegative element

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

ZoeHahn1J
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### Formal charge and most electronegative element

I have in my notes that when calculating formal charge, the most electronegative element will be more stable with a negative charge. Why is this so/why not with 0 charge?
Thank you so much!

ami patel
Posts: 20
Joined: Thu Jul 27, 2017 3:00 am

### Re: Formal charge and most electronegative element

When you draw Lewis structures, the most stable form would have 0 formal charges. Sometimes, you might run into structures that have unavoidable formal charges. In these cases, there is a preference for the more electronegative atom to carry the negative formal charge, and it is more stable for the less electronegative atom to carry the positive formal charge.

ZoeHahn1J
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### Re: Formal charge and most electronegative element

Oooh okay! Thank you so much!

Gobinder Pandher 3J
Posts: 21
Joined: Fri Sep 29, 2017 7:07 am

### Re: Formal charge and most electronegative element

If a certain Lewis Structure has an uneven distribution of charges (uneven distribution defined as not all atoms having a formal charge of 0) then the most electronegative atom will typically have a negative formal charge. A TA explained that this would depict a more stable resonance as the more electronegative atom has the greatest pull on the electrons. This pull would most likely lead to a negative formal charge on the atom. However, this does not necessarily mean that the most electronegative atom ALWAYS needs to have a negative charge. If the Lewis Dot Structure can be constructed so each atom has a formal charge of zero, this is still said to be more favorable.

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