Lewis structure of H2SO3

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The Lewis structure of H₂SO₃ contains one double bond and four single bonds, with sulfur in the center, and two hydrogens and three oxygens on either side. There are two lone pairs on each oxygen atom, one lone pair on the sulfur atom, and the hydrogen atom does not have any lone pair.

Contents

Steps

By using the following steps, you can easily draw the Lewis structure of H₂SO₃.

#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)
#5 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… hydrogen is a group 1 element, and both sulfur and oxygen are the group 16 elements. Hence, hydrogen has one valence electron, and both sulfur and oxygen have six valence electrons.

Now H₂SO₃ has two hydrogen atoms, one sulfur atom, and three oxygen atoms.

So the total number of valence electrons = (valence electrons of hydrogen atom × 2) + valence electrons of sulfur atom + (valence electrons of oxygen atom × 3)

Therefore, the total number of valence electrons = 2 + 6 + 18 = 26

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.

Therefore, choose the central atom from sulfur and oxygen.

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 sulfur is the central atom.

So now, put sulfur in the center and hydrogens and oxygens on either side. And draw the rough skeleton structure for the Lewis structure of H₂SO₃ something like this:

Skeleton structure for Lewis structure of H₂SO₃

Also read: How to draw Lewis structure of CCl₂F₂ (4 steps)

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since sulfur is surrounded by two hydrogens and three oxygens, use ten electrons to show five chemical bonds as follows:

#3 Mark lone pairs

As calculated earlier, we have a total of 26 valence electrons. And in the above structure, we have already used ten valence electrons. Hence, sixteen valence electrons are remaining.

Two valence electrons represent one lone pair. So sixteen valence electrons = eight lone pairs.

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

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

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

So top oxygen will get three lone pairs, and left oxygen and right oxygen will get two lone pairs. And the central atom (sulfur) will get one lone pair.

Now draw the Lewis structure of H₂SO₃ something like this:

Lone pairs marked on Lewis structure of H₂SO₃

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

Now calculate the formal charge and check the stability of the above structure.

Also read: How to draw Lewis structure of HSO₄^– (5 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 each hydrogen atom

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

Formal charge = 1 – 0 – ½ (2) = 0

For sulfur atom

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

Formal charge = 6 – 2 – ½ (6) = +1

For top oxygen atom

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

Formal charge = 6 – 6 – ½ (2) = -1

For left oxygen and right 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 H₂SO₃ looks something like this:

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.

Also read: How to draw Lewis structure of C₂H₂Cl₂ (5 steps)

#5 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 convert one lone pair from the top oxygen atom to make a new bond with the sulfur atom. And then, the Lewis structure of H₂SO₃ looks something like this:

Lone pair of top oxygen is converted, and octet is completed on atoms

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 each hydrogen atom

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

Formal charge = 1 – 0 – ½ (2) = 0

For sulfur atom

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

Formal charge = 6 – 2 – ½ (8) = 0

For each 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 H₂SO₃ looks something like this:

Formal charges are calculated, and got the stable Lewis structure of H₂SO₃

In the above structure, you can see that the formal charges of all atoms are zero. Therefore, this is the stable Lewis structure of H₂SO₃.

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

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External links

H2SO3 | Lewis Structure, Properties & Uses – Study.com
Sulfurous Acid (H2SO3) Lewis Structure – Chemistry School
Chemical Bonding: H2SO3 Lewis Structure – The Geoexchange
H2SO3 Lewis Structure in 6 Steps (With Images) – Pediabay
Sulfurous Acid| Formula & Lewis Structure – What’s Insight
if sulfur can form only two single bonds since it has 6 valence electrons, then why does the lewis structure of H2SO3 look like this? – Reddit
Sulphurous Acid – H2SO3, Structure, Properties and Uses – Turito
Why is H2SO3 an acid when the oxygens have lone pairs? – Laurence Lavelle

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.