HSO4- Lewis structure

HSO₄^– (hydrogen sulfate) has one hydrogen atom, one sulfur atom, and four oxygen atoms.

In the HSO₄^– Lewis structure, there are two single bonds and two double bonds around the sulfur atom, with four oxygen atoms attached to it. The oxygen atom with a double bond has two lone pairs, the left oxygen atom (with which the hydrogen atom is attached) also has two lone pairs, and the bottom oxygen atom with a single bond has three lone pairs.

Also, there is a negative (-1) charge on the bottom oxygen atom.

Alternative method: Lewis structure of HSO₄^–

Rough sketch

• First, determine the total number of valence electrons

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 HSO₄^– has one hydrogen atom, one sulfur atom, and four oxygen atoms, so…

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

Now the HSO₄^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 1 + 6 + 24 +1 = 32

Learn how to find: Hydrogen valence electrons, Sulfur valence electrons, and Oxygen valence electrons

• 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

Lone pair

Here, we have a total of 16 electron pairs. And five bonds are already marked. So we have to only mark the remaining eleven 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 hydrogen and oxygens. But no need to mark on hydrogen, because hydrogen already has two electrons.

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

Mark the lone pairs on the sketch as follows:

Formal charge

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

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

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

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

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

For left 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:

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 right oxygen atom to make a new S — O bond with the sulfur atom as follows:

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

Final structure

The final structure of HSO₄^– features a central sulfur atom connected to four oxygen atoms, one of which is bonded to a hydrogen atom. Within this layout, the sulfur atom utilizes an expanded valence shell to form double bonds with two oxygen atoms and single bonds with the remaining two. The oxygen atom bonded to the hydrogen and the two double-bonded oxygens result in a formal charge of zero, while the single-bonded oxygen atom that lacks a hydrogen carries a -1 formal charge. Each double-bonded oxygen retains two lone pairs, the hydroxyl oxygen maintains two lone pairs, and the single-bonded oxygen atom with the negative charge holds three lone pairs. Therefore, this specific electronic distribution serves as the definitive and most accurate Lewis representation of the hydrogen sulfate ion.

To complete the representation, draw square brackets around the entire Lewis structure and place a “-” or “-1” sign as a superscript outside the upper right bracket. This notation signifies that the negative charge is a property of the whole ion.

Next: CCl₂F₂ Lewis structure

External video

• HSO4- Lewis Structure: How to Draw the Lewis Structure for the Bisulfate Ion – YouTube • Wayne Breslyn

External links

• https://www.chemistryscl.com/general/HSO4–bisulfate-ion-lewis-structure/
• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/HSO4–lewis-structure.html