Lewis structure of S2O

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Lewis structure of S2O
Lewis structure of S2O | Image: Root Memory

The Lewis structure of S2O contains two double bonds, with one sulfur in the center, and one other sulfur and oxygen on either side. The left sulfur atom has two lone pairs, the center sulfur atom has one lone pair, and the oxygen atom also has two lone pairs.

Steps

By using the following steps, you can easily draw the Lewis structure of S2O.

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Complete octet on central atom
#5 Calculate formal charge and check stability
#6 Convert lone pair and calculate formal charge again (if formal charges are not closer to zero)

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… both sulfur and oxygen are the group 16 elements. Hence, both sulfur and oxygen have six valence electrons.

Now S2O has two sulfur atoms and one oxygen atom.

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

Therefore, the total number of valence electrons = 12 + 6 = 18

  • Now decide the central atom

The atom with the least electronegative value is placed at the center. By looking at the periodic table, we get the electronegativity values for sulfur and oxygen as follows:

Electronegativity value of sulfur = 2.58
Electronegativity value of oxygen = 3.44

Obviously, sulfur is less electronegative than oxygen. Hence, assume that center sulfur is the central atom (as there are two sulfurs).

So now, put one sulfur in the center and one other sulfur and oxygen on either side. And draw the rough skeleton structure for the Lewis structure of S2O something like this:

Skeleton structure for Lewis structure of S2O | Image: Root Memory

#2 Show chemical bond

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

Two chemical bonds shown between atoms | Image: Root Memory

#3 Mark lone pairs

As calculated earlier, we have a total of 18 valence electrons. And in the above structure, we have already used four valence electrons. Hence, fourteen valence electrons are remaining.

Two valence electrons represent one lone pair. So fourteen valence electrons = seven lone pairs.

Note that sulfur is 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.

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

The outside atoms are left sulfur and oxygen, so left sulfur and oxygen will get three lone pairs. And the central atom (center sulfur) will get one lone pair.

So the Lewis structure of S2O looks something like this:

Lone pairs marked on Lewis structure of S2O | Image: Root Memory

In the above structure, you can see that the octet is completed on outside atoms. But, the central atom (center sulfur) doesn’t form an octet.

So in the next step, we have to complete the octet on the central atom.

#4 Complete octet on central atom

Remember that sulfur is a period 3 element, so it can keep more than 8 electrons in its last shell.

Now center sulfur already has six valence electrons. Hence, center sulfur needs two more valence electrons to complete its octet.

So convert one lone pair from the left sulfur atom to make a new bond with the center sulfur atom. And then, the Lewis structure of S2O looks something like this:

Lone pair of left sulfur is converted, and octet is completed on atoms | Image: Root Memory

Here, the lone pair of left sulfur is converted (instead of oxygen). This is because sulfur is less electronegative than oxygen, and so it can give more electrons to share them.

In the above structure, you can see that the octet is completed on the central atom (center sulfur), 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.

#5 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 left sulfur atom

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

Formal charge = 6 – 4 – ½ (4) = 0

  • For center sulfur atom

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

Formal charge = 6 – 2 – ½ (6) = +1

  • For 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 S2O looks something like this:

Formal charges are not closer to zero | Image: Root Memory

In the above structure, you can see that the formal charges of atoms are not closer to zero. Therefore, convert lone pair and calculate formal charge again.

#6 Convert lone pair and calculate formal charge again

As mentioned earlier, sulfur is a period 3 element, so it can keep more than 8 electrons in its last shell.

So again convert one lone pair from the oxygen atom to make a new bond with the center sulfur atom. And then, the Lewis structure of S2O looks something like this:

Lone pair of oxygen is converted, and octet is completed on atoms | Image: Root Memory

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

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 left sulfur atom

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

Formal charge = 6 – 4 – ½ (4) = 0

  • For center sulfur atom

Valence electrons = 6
Nonbonding electrons = 2
Bonding electrons = 8

Formal charge = 6 – 2 – ½ (8) = 0

  • For oxygen atom

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

Formal charge = 6 – 4 – ½ (4) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of S2O looks something like this:

Formal charges are calculated, and got the stable Lewis structure of S2O | Image: Root Memory

In the above structure, you can see that the formal charges of both (sulfur and oxygen) are zero. Therefore, this is the stable Lewis structure of S2O.

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

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Deep

Rootmemory.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.

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