# SbF5 Lewis structure

SbF5 (antimony pentafluoride) has one antimony atom and five fluorine atoms.

In the SbF5 Lewis structure, there are five single bonds around the antimony atom, with five fluorine atoms attached to it, and on each fluorine atom, there are three lone pairs.

Contents

## Steps

Use these steps to correctly draw the SbF5 Lewis structure:

#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required

Let’s discuss each step in more detail.

### #1 First draw a rough sketch

• First, determine the total number of valence electrons

In the periodic table, antimony lies in group 15, and fluorine lies in group 17.

Hence, antimony has five valence electrons and fluorine has seven valence electrons.

Since SbF5 has one antimony atom and five fluorine atoms, so…

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

And the total valence electrons = 5 + 35 = 40

• Second, find the total electron pairs

We have a total of 40 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 = 40 ÷ 2 = 20

• Third, determine the central atom

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

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

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

• And finally, draw the rough sketch

### #2 Mark lone pairs on the atoms

Here, we have a total of 20 electron pairs. And five Sb — F bonds are already marked. So we have to only mark the remaining fifteen electron pairs as lone pairs on the sketch.

Also remember that antimony is a period 5 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 antimony, there is zero lone pair because all fifteen electron pairs are over.

Mark the lone pairs on the sketch as follows:

### #3 Calculate and mark formal charges on the atoms, if required

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

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

For antimony atom, formal charge = 5 – 0 – ½ (10) = 0

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

Here, both antimony and fluorine atoms do not have charges, so no need to mark the charges.

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

Therefore, this structure is the stable Lewis structure of SbF5.