HClO4 Lewis Structure in 6 Steps (With Images)

So you have seen the above image by now, right?

Let me explain the above image in short.

HClO4 lewis structure has a Chlorine atom (Cl) at the center which is surrounded by three Oxygen atoms (O) and one O-H group. There are 3 double bonds between the Chlorine atom (Cl) & Oxygen atom (O) and the rest other atoms have a single bond.

If you haven’t understood anything from the above image of HClO4 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of HClO4.

So let’s move to the steps of drawing the lewis structure of HClO4.

Steps of drawing HClO4 lewis structure

Step 1: Find the total valence electrons in HClO4 molecule

In order to find the total valence electrons in a HClO4 molecule, first of all you should know the valence electrons present in hydrogen atom, chlorine atom as well as oxygen atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)

Here, I’ll tell you how you can easily find the valence electrons of hydrogen, chlorine as well as oxygen using a periodic table.

Total valence electrons in HClO4 molecule

→ Valence electrons given by hydrogen atom:

Hydrogen is group 1 element on the periodic table. ^[1] Hence the valence electron present in hydrogen is 1.

You can see that only 1 valence electron is present in the hydrogen atom as shown in the above image.

→ Valence electrons given by chlorine atom:

Chlorine is group 17 element on the periodic table. ^[2] Hence the valence electrons present in chlorine is 7.

You can see the 7 valence electrons present in the chlorine atom as shown in the above image.

→ Valence electrons given by oxygen atom:

Oxygen is group 16 element on the periodic table. ^[3] Hence the valence electrons present in oxygen is 6.

You can see the 6 valence electrons present in the oxygen atom as shown in the above image.

Hence, 

Total valence electrons in HClO4 molecule = valence electrons given by 1 hydrogen atom + valence electrons given by 1 chlorine atom + valence electrons given by 4 oxygen atoms = 1 + 7 + 6(4) = 32.

Step 2: Select the central atom

For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.

(Remember: If hydrogen is present in the given molecule, then always put hydrogen outside.)

Now here the given molecule is HClO4 and it contains hydrogen atom (H), chlorine atom (Cl) and oxygen atoms (O).

So as per the rule we have to keep hydrogen outside.

Now, you can see the electronegativity values of chlorine atom (Cl) and oxygen atom (O) in the above periodic table.

If we compare the electronegativity values of chlorine (Cl) and oxygen (O) then the chlorine atom is less electronegative.

So here the chlorine atom (Cl) is the center atom and the oxygen atoms (O) are the outside atom.

Step 3: Connect each atoms by putting an electron pair between them

Now in the HClO4 molecule, you have to put the electron pairs between the oxygen (O) & hydrogen (H) atom and between the oxygen (O) & chlorine (Cl) atoms.

This indicates that these atoms are chemically bonded with each other in a HClO4 molecule.

Step 4: Make the outer atoms stable

Now in this step, you have to check the stability of the outer atoms.

Here in the sketch of HClO4 molecule, you can see that the outer atoms are hydrogen and oxygen atoms.

These hydrogen and oxygen atoms are forming a duplet and octet respectively and hence they are stable.

Also, in step 1 we have calculated the total number of valence electrons present in the HClO4 molecule.

The HClO4 molecule has a total 32 valence electrons and all these valence electrons are used in the above sketch of HClO4.

Hence there are no remaining electron pairs to be kept on the central atom.

So now let’s proceed to the next step.

Step 5: Check the octet on the central atom

In this step, you have to check whether the central chlorine atom (Cl) is stable or not.

In order to check the stability of the central chlorine (Cl) atom, we have to check whether it is forming an octet or not.

You can see from the above picture that the chlorine atom is forming an octet. That means it has 8 electrons.

And hence the central chlorine atom is stable.

Now let’s proceed to the final step to check whether the lewis structure of HClO4 is stable or not.

Step 6: Check the stability of lewis structure

Now you have come to the final step in which you have to check the stability of lewis structure of HClO4.

The stability of lewis structure can be checked by using a concept of formal charge.

In short, now you have to find the formal charge on hydrogen (H) atom, chlorine (Cl) atom as well as oxygen (O) atoms present in the HClO4 molecule.

For calculating the formal charge, you have to use the following formula;

Formal charge = Valence electrons – (Bonding electrons)/2 – Nonbonding electrons

You can see the number of bonding electrons and nonbonding electrons for each atom of HClO4 molecule in the image given below.

For Hydrogen (H) atom:
Valence electron = 1 (because hydrogen is in group 1)
Bonding electrons = 2
Nonbonding electrons = 0

For Chlorine (Cl) atom:
Valence electrons = 7 (because chlorine is in group 17)
Bonding electrons = 8
Nonbonding electrons = 0

For Oxygen (O) atom:
Valence electrons = 6 (because oxygen is in group 16)
Bonding electrons = 2
Nonbonding electrons = 6

For Oxygen (O) atom (of O-H group):
Valence electrons = 6 (because oxygen is in group 16)
Bonding electrons = 4
Nonbonding electrons = 4

From the above calculations of formal charge, you can see that the chlorine (Cl) atom has +3 charge and the three oxygen (O) atoms have -1 charges.

Because of this reason, the above obtained lewis structure of HClO4 is not stable.

So we have to minimize these charges by shifting the electron pairs towards the chlorine atom.

After shifting the electron pairs from oxygen atoms to chlorine atom, the lewis structure of HClO4 becomes more stable.

In the above lewis dot structure of HClO4, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of HClO4.

I hope you have completely understood all the above steps.

For more practice and better understanding, you can try other lewis structures listed below.

Try (or at least See) these lewis structures for better understanding:

About author

Jay Rana

Jay is an educator and has helped more than 100,000 students in their studies by providing simple and easy explanations on different science-related topics. He is a founder of Pediabay and is passionate about helping students through his easily digestible explanations.

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