## Lewis structures

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

Rambod Meshgi 3J
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### Lewis structures

I understand how to do Lewis structure and how to calculate formal charge, but in some instances the Lewis structure I draw doesn't have the most favorable formal charge for each atom. Are there any go to ways to get formal charge closer to 0 for the atoms of a molecule?

Rachel Hunsucker 3L
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### Re: Lewis structures

The best way to go about that would be to add another bond such as a double bond or a triple bond and take away a lone pair or vice-versa.

Vanessa A 3F
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### Re: Lewis structures

If the formal charges you're getting aren't 0, try to go back and either add bonds, or lone pairs (if applicable), to make the formal charge more favorable. Most of the time, using a double or triple bond instead of a single bond solves the problem.

604735966
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### Re: Lewis structures

So should we be calculating the formal charge for every problem from now on? I took AP Chem in high school and have never even seen this before!

dayannaramirez_1J
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### Re: Lewis structures

So the bond is the most stable when the FC is 0?

David Julfayan 1F
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### Re: Lewis structures

dayannaramirez_1E wrote:So the bond is the most stable when the FC is 0?

Thats how i understood it to be. I think of the zero being a neutral number between positives and negatives so there is no imbalance and therefore is "stable."

Jocelyn Sandoval 3B
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### Re: Lewis structures

At one of the review sessions, the UA said that you're better off when the FC of the center atom is 0.

Aashi_Patel_3B
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### Re: Lewis structures

To my knowledge, the goal is to get at least the central atom to a FC of 0. The rest should be "balanced" to the best you can, meaning one side of the atom shouldn't have an excessive charge compared to the other atoms.

This is one issue that my friends and I were discussing when we got together today to study: If there is no molecule where the FC is 0, then how do you compare the Lewis structures to find the "lowest" FC?

Chem_Mod
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### Re: Lewis structures

604735966 wrote:So should we be calculating the formal charge for every problem from now on? I took AP Chem in high school and have never even seen this before!

Good that you are learning new concepts in my class!
If uncertain about your Lewis structure, then yes calculate formal charge as I have done in class many times.

Chem_Mod
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### Re: Lewis structures

Aashi_Patel_4E wrote:To my knowledge, the goal is to get at least the central atom to a FC of 0. The rest should be "balanced" to the best you can, meaning one side of the atom shouldn't have an excessive charge compared to the other atoms.

This is one issue that my friends and I were discussing when we got together today to study: If there is no molecule where the FC is 0, then how do you compare the Lewis structures to find the "lowest" FC?

I think you are confusing total charge (also called net charge) with formal charge for each atom.

The sum of all the individual atom formal charges = total charge

Therefore SO42- must have at least two atoms with formal charge equal to -1.
As discussed in class two of the oxygen atoms must have FC = -1.
This gives total charge = -2

Oxygen has higher electronegativity than sulfur which is why it has the FC = -1.

404757006
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### Re: Lewis structures

I missed the lecture where he talked about calculating the formal charges and I didn't do it in high school... Can someone explain its purpose?

Chem_Mod
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### Re: Lewis structures

The goal of formal charge is to determine how much more positive or negative an atom is in comparison to its neutral state (ex. Oxygen normally has 6 valence electrons, which represents its neutral state, but it can gain negative or positive charge based on the number of electrons it holds in lone pairs or shares in bonds, respectively). The further it is from the neutral state, the more charge it has to hold, making it more unstable.

The formula for formal charge is FC = # valence electrons in atom - [# lone pair electrons + 1/2(# bond electrons)]. Each bond and pair hold 2 electrons (double bonds hold 4 and so on).

Ex. Oxygen in a bond has 3 lone pairs and a single bond. To calculate formal charge, we take oxygen's normal number of valence electrons (6) and subtract the number of lone pair electrons (3 pairs = 6 electrons) and half of the bond electrons (1 single bond = 2 electrons, but we want half so 2/2 =1). So we get 6 - (6+1) = -1.

Hope this helps!