Lewis structure of H2SO4

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Lewis structure of H2SO4
Lewis structure of H2SO4 | Image: Root Memory

The Lewis structure of H2SO4 contains two double bonds and four single bonds, with sulfur in the center, and two hydrogens and four oxygens on either side. There are two lone pairs on each oxygen atom, and sulfur atom and hydrogen atom do not have any lone pair.

Steps

By using the following steps, you can easily draw the Lewis structure of H2SO4:

#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 H2SO4 has two hydrogen atoms, one sulfur atom, and four 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 × 4)

Therefore, the total number of valence electrons = 2 + 6 + 24 = 32

  • 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 H2SO4 something like this:

Skeleton structure for Lewis structure of H2SO4 | Image: Root Memory

#2 Show chemical bond

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

Six chemical bonds shown between atoms | Image: Root Memory

#3 Mark lone pairs

As calculated earlier, we have a total of 32 valence electrons. And in the above structure, we have already used twelve valence electrons. Hence, twenty valence electrons are remaining.

Two valence electrons represent one lone pair. So twenty valence electrons = ten 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 left oxygen and right oxygen will get three lone pairs, and top oxygen and bottom oxygen will get two lone pairs. And the central atom (sulfur) will not get any lone pair, because all ten lone pairs are used.

Now draw the Lewis structure of H2SO4 something like this:

Lone pairs marked on Lewis structure of H2SO4 | Image: Root Memory

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.

#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 = 0
Bonding electrons = 8

Formal charge = 6 – 0 – ½ (8) = +2

  • For left oxygen and right oxygen atom

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

Formal charge = 6 – 6 – ½ (2) = -1

  • For top oxygen and bottom 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 H2SO4 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.

#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 left oxygen atom, and one lone pair from right oxygen atom to make a new bond with the sulfur atom. And then, the Lewis structure of H2SO4 looks something like this:

Lone pair of left and right 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 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 = 0
Bonding electrons = 12

Formal charge = 6 – 0 – ½ (12) = 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 H2SO4 looks something like this:

Formal charges are calculated, and got the stable Lewis structure of H2SO4 | Image: Root Memory

In the above structure, you can see that the formal charges of all atoms are zero. Therefore, this is the stable Lewis structure of H2SO4.

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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