HO2- Lewis structure

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HO₂^– (hydroperoxyl) has one hydrogen atom and two oxygen atoms.

In the HO₂^– Lewis structure, there are two single bonds around the oxygen atom, with hydrogen and oxygen atoms attached to it, and on the right oxygen atom, there are three lone pairs.

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

Steps

Here’s how you can easily draw the HO₂^– 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, and oxygen lies in group 16.

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

Since HO₂^– has one hydrogen atom and two oxygen atoms, so…

Valence electrons of one hydrogen atom = 1 × 1 = 1
Valence electrons of two oxygen atoms = 6 × 2 = 12

Now the HO₂^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 1 + 12 + 1 = 14

Learn how to find: Hydrogen valence electrons and Oxygen valence electrons

• 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, there are only two atoms and both atoms are oxygen, so we can assume any one as the central atom.

Let’s assume that the central atom is center oxygen.

Therefore, place one oxygen in the center and hydrogen and other oxygen 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 oxygen is a period 2 element, so it can not keep more than 8 electrons in its last shell. And hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell.

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

So for right oxygen, there are three lone pairs, and for center 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 center oxygen atom, formal charge = 6 – 4 – ½ (4) = 0

For right oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

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

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

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

This is okay, 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 oxygen.

Therefore, this structure is the most stable Lewis structure of HO₂^–.

And since the HO₂^– has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: NH₂OH Lewis structure

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