SF3- Lewis structure

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SF₃^– (sulfur trifluoride) has one sulfur atom and three fluorine atoms.

In the SF₃^– Lewis structure, there are three single bonds around the sulfur atom, with three fluorine atoms attached to it. Each fluorine atom has three lone pairs, and the sulfur atom has two lone pairs.

Also, there is a negative (-1) charge on the sulfur atom.

Steps

To properly draw the SF₃^– Lewis structure, follow these steps:

#1 Draw a rough sketch of the structure
#2 Next, indicate lone pairs on the atoms
#3 Indicate formal charges on the atoms, if necessary

Let’s break down each step in more detail.

#1 Draw a rough sketch of the structure

• First, determine the total number of valence electrons

In the periodic table, sulfur lies in group 16, and fluorine lies in group 17.

Hence, sulfur has six valence electrons and fluorine has seven valence electrons.

Since SF₃^– has one sulfur atom and three fluorine atoms, so…

Valence electrons of one sulfur atom = 6 × 1 = 6
Valence electrons of three fluorine atoms = 7 × 3 = 21

Now the SF₃^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 6 + 21 + 1 = 28

Learn how to find: Sulfur valence electrons and Fluorine valence electrons

• Second, find the total electron pairs

We have a total of 28 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 28 ÷ 2 = 14

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since sulfur is less electronegative than fluorine, assume that the central atom is sulfur.

Therefore, place sulfur in the center and fluorines on either side.

• And finally, draw the rough sketch

#2 Next, indicate lone pairs on the atoms

Here, we have a total of 14 electron pairs. And three S — F bonds are already marked. So we have to only mark the remaining eleven electron pairs as lone pairs on the sketch.

Also remember that sulfur is a period 3 element, so it can keep more than 8 electrons in its last shell. And fluorine is a period 2 element, so it can not keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are fluorines.

So for each fluorine, there are three lone pairs, and for sulfur, there are two lone pairs.

Mark the lone pairs on the sketch as follows:

#3 Indicate formal charges on the atoms, if necessary

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For sulfur atom, formal charge = 6 – 4 – ½ (6) = -1

For each fluorine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, the sulfur atom has a charge, so mark it on the sketch as follows:

In the above structure, you can see that the central atom (sulfur) forms an octet. And the outside atoms (fluorines) also form an octet. Hence, the octet rule is satisfied.

Now there is still a negative (-1) charge on the sulfur atom.

This is not okay, right? Because the structure with a negative charge on the most electronegative atom is the best Lewis structure. And in this case, the most electronegative element is fluorine.

But if we convert a lone pair of the sulfur atom to make a new S — F bond with the fluorine atom, and calculate the formal charge, then we do not get the formal charges on atoms closer to zero.

And the structure with the formal charges on atoms closer to zero is the best Lewis structure.

Therefore, this structure is the most stable Lewis structure of SF₃^–.

And since the SF₃^– has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: XeO₃ Lewis structure

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