# HOF Lewis structure

HOF (hypofluorous acid) has one hydrogen atom, one oxygen atom, and one fluorine atom.

In the HOF Lewis structure, there are two single bonds around the oxygen atom, with hydrogen and fluorine atoms attached to it. The oxygen atom has two lone pairs, and the fluorine atom has three lone pairs.

Contents

## Steps

Here’s how you can easily draw the HOF 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, hydrogen lies in group 1, oxygen lies in group 16, and fluorine lies in group 17.

Hence, hydrogen has one valence electron, oxygen has six valence electrons, and fluorine has seven valence electrons.

Since HOF has one hydrogen atom, one oxygen atom, and one fluorine atom, so…

Valence electrons of one hydrogen atom = 1 × 1 = 1
Valence electrons of one oxygen atom = 6 × 1 = 6
Valence electrons of one fluorine atom = 7 × 1 = 7

And the total valence electrons = 1 + 6 + 7 = 14

• Second, find the total electron pairs

We have a total of 14 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 = 14 ÷ 2 = 7

• Third, determine the central atom

Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.

Now we have to choose the central atom from oxygen and fluorine. Place the least electronegative atom at the center.

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

Therefore, place oxygen in the center and hydrogen and fluorine on either side.

• And finally, draw the rough sketch

### #2 Mention lone pairs on the atoms

Here, we have a total of 7 electron pairs. And two bonds are already marked. So we have to only mark the remaining five electron pairs as lone pairs on the sketch.

Also remember that hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell. Oxygen is a period 2 element, so it can not keep more than 8 electrons in its last shell. And fluorine is a period 3 element, so it can keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are hydrogen and fluorine. But no need to mark on hydrogen, because hydrogen already has two electrons.

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

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 hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

For oxygen atom, formal charge = 6 – 4 – ½ (4) = 0

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

Here, the atoms do not have charges, so no need to mark the charges.

In the above structure, you can see that the central atom (oxygen) forms an octet. The outside atom (fluorine) also forms an octet, and hydrogen forms a duet. Hence, the octet rule and duet rule are satisfied.

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