## Lecture O Formal Charge Ex

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

Alison Perkins 2B
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### Lecture O Formal Charge Ex

In lecture when Prof. Lavell was calculating the individual formal charges of S and O in SO4 I was very confused by the FC of the Oxygen in the molecule.Can someone explain how he did that ?

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### Re: Lecture O Formal Charge Ex

The formula for formal charge is the number of valence electrons minus the number of bonding pairs (or the shared electrons/2) minus the number of lone electrons. In the scenario where there was a single bond between the oxygen and the sulfur, the formal charge would be

FC= valence electrons - bonding pairs (or shared electrons/2)- lone electrons
FC= 6-1-6=-1

This is because oxygen has 6 valence electrons, 1 bonding pair due to the single bond, and 6 lone electrons to complete the octet.

On the other hand, with a double bond, the formal charge would be

FC= valence electrons - bonding pairs - lone electrons
FC= 6-2-4=0

Similarly, the formal charge of sulfur with only single bonds with oxygen is

FC= valence electrons - bonding pairs - lone electrons
FC= 6-4-0=+2

However, when it has two double bonds and two single bonds with the oxygen atoms, the formal charge is

FC= valence electrons - bonding pairs - lone electrons
FC= 6-6-0=0

By having two double bonds, there are two oxygen atoms left with a formal charge of -1 and the central atom, sulfur has a formal charge of 0, instead of four oxygen atoms with a formal charge of -1 and sulfur with a formal charge of +2. A formal charge of 0 is more stable, so the double bonds reduce the formal charges (not the net amount but all the separate formal charges), making the overall molecule more stable. Since sulfur is in the 3p orbital, it can expand its octet and therefore have 6 bonds instead of the usual 4, which allows this arrangement to be possible.

Hope this helps!
Last edited by Stuti Pradhan 2J on Wed Nov 04, 2020 3:02 pm, edited 1 time in total.

Ashlen Bullock 1H
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### Re: Lecture O Formal Charge Ex

So the equation for formal charge is FC= V - (S/2+ L). Oxygen has 6 valence electrons because it has 8 electrons in total minus the two inner electrons which equals six. In the drawing, that particular oxygen shared one bond with S. That means it shared two electrons because two electrons are shared per bond. So S would equal 2. That oxygen has 6 Lone electrons because those are the electrons depicted outside of the O element that are not bonded to the S element. So L would equal 6. The filled in equation would look like this, FC = 6 - (2/2 + 6) = -1 Hope this helps!

Marylyn Makar 1B
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### Re: Lecture O Formal Charge Ex

Hi! So formal charge is calculated by calculating the number of valence electrons minus the sum of the number of lone pair electrons plus the number of shared electrons divided by 2, or FC= V- (L + S/2)
So oxygen has 6 valence electrons since it is in the 6th group excluding the d-block. In SO4, oxygen has 4 lone pair electrons and 4 shared electrons ( 2 bonds). So, the formal charge would be calculated by
6- (4+ 4/2)= 0
Hope this helps!

Eliana Carney 3E
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### Re: Lecture O Formal Charge Ex

Hey Alison!

The first thing to know when calculating formal charge is the formal charge equation. The equation is as follows:
formal charge = V (number of valence electrons for the bound atom) - ( (shared electrons/2) + L (number of lone pair electrons))

In this lecture, Dr. Lavelle calculated the formal charge of oxygen in two different Lewis structures. First, I'll explain how he calculated the formal charge in the Lewis structure for the sulfate ion that did not have any double bonds. First of all, we know that the oxygen has six valence electrons by looking at its electron configuration. Then, looking at the Lewis structure we see that oxygen has one bond. This means that there are two shared electrons between the oxygen atom and the sulfur atom. Lastly, we see that there are six dots, or electrons, around the oxygen atom so we know that L is equal to six. Using this information, we just plug the numbers into the formal charge equation. It should look something like this: formal charge = (6) - ( (2/2) + 6) = -1. Therefor, we know that the formal charge of oxygen in the Lewis stricture for sulfate that contains no double bonds is -1.

The second Lewis structure we had for the sulfate ion had two double bonds and two single bonds between sulfur and oxygen and. The formal charge of the oxygen atoms with the single bond to sulfur are the same as in the example above, so the formal charge is equal to negative one. However, the formal charge of the oxygen atoms with the double bond to sulfur are going to be different. First of all, we know that oxygen still has six valence electrons. That doesn't change with different Lewis structures. We also know that there are two bonds between the oxygen atom and the sulfur atom, therefor there is four shared electrons. Lastly, we can see that there are four dots, or electrons, around this oxygen atom so we know L is equal to four. Using this information, we once again plug these numbers into the formal charge equation. It should look something like this: formal charge = (6) - ( (4/2) + 4) = 0. Therefor, we know that the formal charge of the oxygen in this Lewis structure that has a double bond to sulfur is 0.

Hope this helps!