Lewis structure of HO2-

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The Lewis structure of HO₂^– contains two single bonds, with one oxygen in the center, and hydrogen and one other oxygen on either side. The center oxygen atom has two lone pairs, the right oxygen atom has three lone pairs, and the hydrogen atom does not have any lone pair.

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

Steps

By using the following steps, you can easily draw the Lewis structure of HO₂^–.

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Calculate formal charge and check stability (if octet is already completed on central atom)

Let’s one by one discuss each step in detail.

#1 Draw skeleton

In this step, first calculate the total number of valence electrons. And then, decide the central atom.

• Let’s calculate the total number of valence electrons

We know that… hydrogen is a group 1 element and oxygen is a group 16 element. Hence, hydrogen has one valence electron and oxygen has six valence electrons.

Now HO₂^– has one hydrogen atom and two oxygen atoms.

So the total number of valence electrons = valence electrons of hydrogen atom + (valence electrons of oxygen atom × 2)

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

Therefore, the total number of valence electrons = 1 + 12 + 1 = 14

• Now decide the central atom

We can not assume hydrogen as 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.

Hence, assume that center oxygen is the central atom (as there are two oxygens).

So now, put oxygen in the center and hydrogens on either side. And draw the rough skeleton structure for the Lewis structure of H₃O⁺ something like this:

So now, put one oxygen in the center and one other oxygen and hydrogen on either side. And draw the rough skeleton structure for the Lewis structure of HO₂^– something like this:

Also read: How to draw Lewis structure of TeF₄ (4 steps)

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since center oxygen is surrounded by one other oxygen and hydrogen, use four electrons to show two chemical bonds as follows:

Also read: How to draw Lewis structure of SO (5 steps)

#3 Mark lone pairs

As calculated earlier, we have a total of 14 valence electrons. And in the above structure, we have already used four valence electrons. Hence, ten valence electrons are remaining.

Two valence electrons represent one lone pair. So ten valence electrons = five lone pairs.

Note that hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell. And oxygen is a period 2 element, so it can not keep more than 8 electrons in its last shell.

Also, make sure that you start marking these lone pairs on outside atoms first. And then, on the central atom.

The outside atoms are hydrogen and right oxygen. But hydrogen can not keep more than 2 electrons in its last shell. Hence, don’t mark the lone pairs on hydrogen.

So right oxygen will get three lone pairs. And the central atom (center oxygen) will get two lone pairs.

Now draw the Lewis structure of HO₂^– something like this:

In the above structure, you can see that the octet is completed on the central atom (center oxygen), and also on the outside atoms. Therefore, the octet rule is satisfied.

After completing the octet, one last thing we need to do is, calculate the formal charge and check the stability of the above structure.

Also read: How to draw Lewis structure of CH₂ (4 steps)

#4 Calculate formal charge and check stability

The following formula is used to calculate the formal charges on atoms:

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

Collect the data from the above structure and then, write it down below as follows:

• For hydrogen atom

Valence electrons = 1
Nonbonding electrons = 0
Bonding electrons = 2

Formal charge = 1 – 0 – ½ (2) = 0

• For center oxygen atom

Valence electrons = 6
Nonbonding electrons = 4
Bonding electrons = 4

Formal charge = 6 – 4 – ½ (4) = 0

• For right oxygen atom

Valence electrons = 6
Nonbonding electrons = 6
Bonding electrons = 2

Formal charge = 6 – 6 – ½ (2) = -1

Mention the formal charges of atoms on the structure. So the Lewis structure of HO₂^– looks something like this:

In the above structure, you can see that the formal charges of atoms are closer to zero. Therefore, this is the most stable Lewis structure of HO₂^–.

And each horizontal line drawn in the above structure represents a pair of bonding valence electrons.

Now HO₂^– is an ion having a negative (-1) charge, so draw brackets around the above Lewis structure and mention that charge on the top right corner. And then, the Lewis structure of HO₂^– looks something like this:

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Related

• Lewis structure of TeF₄
• Lewis structure of SO
• Lewis structure of CH₂
• Lewis structure of AlF₃
• Lewis structure of Cl

External links

• HO2- Lewis Structure in 6 Steps (With Images) – Pediabay
• Draw the Lewis structure for the polyatomic hydroperoxyl (HO2) anion – Chegg
• Draw the lewis structure for the polyatomic hydroperoxyl ho2- anion – Brainly
• The Lewis structure of HO2- is – Homework.Study.com
• Draw the lewis structure for the polyatomic hydroperoxyl ho2- anion – Numerade
• Draw the Lewis structure for the polyatomic hydroperoxyl (HO2-) anion – Bartleby
• ho2- lewis structure resonance – OneClass
• Hydroperoxy radical | HO2 | CID 520535 – National Institutes of Health (.gov)
• HO2- (Hydrogenperoxyl) Oxidation Number – ChemicalAid