H2SO4 Lewis structure

H2SO4 Lewis Structure
H2SO4 Lewis structure | Image: Learnool

H2SO4 (sulfuric acid) has two hydrogen atoms, one sulfur atom, and four oxygen atoms.

In the H2SO4 Lewis structure, there are two double bonds and two single bonds around the sulfur atom, with four oxygen atoms attached to it, and on each oxygen atom, there are two lone pairs. And the oxygen atoms having a single bond are attached with one hydrogen atom.

Rough sketch

  • First, determine the total number of valence electrons
Periodic table | Image: Learnool

In the periodic table, hydrogen lies in group 1, and both sulfur and oxygen lie in group 16.

Hence, hydrogen has one valence electron, and both sulfur and oxygen have six valence electrons.

Since H2SO4 has two hydrogen atoms, one sulfur atom, and four oxygen atoms, so…

Valence electrons of two hydrogen atoms = 1 × 2 = 2
Valence electrons of one sulfur atom = 6 × 1 = 6
Valence electrons of four oxygen atoms = 6 × 4 = 24

And the total valence electrons = 2 + 6 + 24 = 32

  • Second, find the total electron pairs

We have a total of 32 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 32 ÷ 2 = 16

  • Third, determine the central atom

Here hydrogen can not be 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.

Now we have to choose the central atom from sulfur and oxygen. Place the least electronegative atom at the center.

Since sulfur is less electronegative than oxygen, assume that the central atom is sulfur.

Therefore, place sulfur in the center and hydrogen and oxygen on either side.

  • And finally, draw the rough sketch
H2SO4 Lewis Structure (Step 1)
Rough sketch of H2SO4 Lewis structure | Image: Learnool

Lone pair

Here, we have a total of 16 electron pairs. And six bonds are already marked. So we have to only mark the remaining ten electron pairs as lone pairs on the sketch.

Also remember 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.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are hydrogens and oxygens. But no need to mark on hydrogen, because each hydrogen has already two electrons.

So for top oxygen and bottom oxygen, there are three lone pairs, for left oxygen and right oxygen, there are two lone pairs, and for sulfur, there is zero lone pair because all ten electron pairs are over.

Mark the lone pairs on the sketch as follows:

H2SO4 Lewis Structure (Step 2)
Lone pairs marked on H2SO4 Lewis structure | Image: Learnool

Formal charge

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For each hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

For sulfur atom, formal charge = 6 – 0 – ½ (8) = +2

For top oxygen and bottom oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

For left oxygen and right oxygen atom, formal charge = 6 – 4 – ½ (4) = 0

Here, both sulfur and oxygen atoms have charges, so mark them on the sketch as follows:

H2SO4 Lewis Structure (Step 3)
Formal charges marked on H2SO4 Lewis structure | Image: Learnool

The above structure is not a stable Lewis structure because both sulfur and oxygen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

Convert a lone pair of the top oxygen atom to make a new S — O bond with the sulfur atom as follows:

H2SO4 Lewis Structure (Step 4)
Lone pair of top oxygen is converted, but still there are charges | Image: Learnool

Since there are charges on sulfur and oxygen atoms, again convert a lone pair of the bottom oxygen atom to make a new S — O bond with the sulfur atom as follows:

H2SO4 Lewis Structure (Step 5)
Lone pair of bottom oxygen is converted, and got the stable Lewis structure of H2SO4 | Image: Learnool

Final structure

The final structure of H2SO4 comprises a central sulfur atom connected to four oxygen atoms, with two of those oxygen atoms each bonded to a hydrogen atom. In this configuration, the sulfur atom forms double bonds with the two terminal oxygen atoms and single bonds with the two hydroxyl (-OH) groups, utilizing an expanded octet. Each oxygen atom fulfills its octet—the double-bonded oxygens retain two lone pairs, while the single-bonded oxygens also maintain two lone pairs alongside their bonds to hydrogen. This arrangement is the most stable because it minimizes formal charges across the molecule, representing the most energetically favorable state. As a result, this specific electronic distribution serves as the definitive and most accurate Lewis representation of sulfuric acid.

Next: Ethanol 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.

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