ClO2 (chlorine dioxide) has one chlorine atom and two oxygen atoms.
In the Lewis structure of ClO2, there are two double bonds around the chlorine atom, with two oxygen atoms attached to it. Each oxygen atom has two lone pairs, and the chlorine atom has one lone pair and one unpaired electron.
Steps
Here’s how you can easily draw the Lewis structure of ClO2 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
#4 Minimize formal charges by converting lone pairs of the atoms, and try to get a stable Lewis structure
#5 Repeat step 4 again if needed, until all charges are minimized
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, chlorine lies in group 17, and oxygen lies in group 16.
Hence, chlorine has seven valence electrons and oxygen has six valence electrons.
Since ClO2 has one chlorine atom and two oxygen atoms, so…
Valence electrons of one chlorine atom = 7 × 1 = 7
Valence electrons of two oxygen atoms = 6 × 2 = 12
And the total valence electrons = 7 + 12 = 19
Learn how to find: Chlorine valence electrons and Oxygen valence electrons
- Second, find the total electron pairs
We have a total of 19 valence electrons. And when we divide this value by two, we get the value of total electron pairs.
But 19 can not be divided by two. Hence, there are a total of 9 electron pairs and one unpaired electron.
- Third, determine the central atom
We have to place the least electronegative atom at the center.
Since chlorine is less electronegative than oxygen, assume that the central atom is chlorine.
Therefore, place chlorine in the center and oxygens on either side.
- And finally, draw the rough sketch
#2 Mention lone pairs on the atoms
Here, we have 9 electron pairs and one unpaired electron. And two Cl — O bonds are already marked. So we have to only mark the remaining seven electron pairs and one unpaired electron as lone pairs on the sketch.
Also remember that chlorine is a period 3 element, so it can keep more than 8 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.
Always start to mark the lone pairs from outside atoms. Here, the outside atoms are oxygens.
So for each oxygen, there are three lone pairs, and for chlorine, there is one lone pair and one unpaired electron.
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 chlorine atom, formal charge = 7 – 3 – ½ (4) = +2
For each oxygen atom, formal charge = 6 – 6 – ½ (2) = -1
Here, both chlorine and oxygen atoms have charges, so mark them on the sketch as follows:
The above structure is not a stable Lewis structure because both chlorine and oxygen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.
#4 Minimize formal charges by converting lone pairs of the atoms
Convert a lone pair of the oxygen atom to make a new Cl — O bond with the chlorine atom as follows:
#5 Since there are charges on atoms, repeat step 4 again
Since there are charges on chlorine and oxygen atoms, again convert a lone pair of the oxygen atom to make a new Cl — O bond with the chlorine atom as follows:
In the above structure, you can see that the central atom (chlorine) forms an octet. And the outside atoms (oxygens) also form an octet. Hence, the octet rule is satisfied.
Also, the above structure is more stable than the previous structures. Therefore, this structure is the stable Lewis structure of ClO2.
Next: C4H10 Lewis structure
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Deep
Learnool.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.