SF5- Lewis structure

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SF₅^– has one sulfur atom and five fluorine atoms.

In the SF₅^– Lewis structure, there are five single bonds around the sulfur atom, with five fluorine atoms attached to it. Each fluorine atom has three lone pairs, and the sulfur atom has one lone pair.

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

Steps

Here’s how you can easily draw the SF₅^– Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms

Now, let’s take a closer look at each step mentioned above.

#1 Draw a rough skeleton 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 SF₅^– has one sulfur atom and five fluorine atoms, so…

Valence electrons of one sulfur atom = 6 × 1 = 6
Valence electrons of five fluorine atoms = 7 × 5 = 35

Now the SF₅^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 6 + 35 + 1 = 42

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

• Second, find the total electron pairs

We have a total of 42 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 = 42 ÷ 2 = 21

• 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 Mention lone pairs on the atoms

Here, we have a total of 21 electron pairs. And five S — F bonds are already marked. So we have to only mark the remaining sixteen 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 is one lone pair.

Mark the lone pairs on the sketch as follows:

#3 If needed, mention formal charges on the atoms

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 – 2 – ½ (10) = -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 SF₅^–.

And since the SF₅^– has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: Lewis structure of SF₃⁺

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