HSO4- Lewis structure

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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.

Steps

To properly draw the HSO₄^– Lewis structure, follow these steps:

#1 Draw a rough sketch of the structure
#2 Next, indicate lone pairs on the atoms
#3 Indicate formal charges on the atoms, if necessary
#4 Minimize formal charges by converting lone pairs of the atoms
#5 Repeat step 4 if necessary, until all charges are minimized

Let’s break down each step in more detail.

#1 Draw a rough sketch of the structure

• 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

#2 Next, indicate lone pairs on the atoms

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:

#3 Indicate formal charges on the atoms, if necessary

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.

#4 Minimize formal charges by converting lone pairs of the atoms

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

#5 Repeat step 4 (minimize charges again)

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:

In the above structure, you can see that the central atom (sulfur) forms an octet. The outside atoms (oxygens) also form an octet, and hydrogen forms a duet. Hence, the octet rule and duet rule are satisfied.

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

This is okay, because the structure with a negative charge on the most electronegative atom is the best Lewis structure. And in this case, the most electronegative element is oxygen.

Also, the above structure is more stable than the previous structures. Therefore, this structure is the most stable Lewis structure of HSO₄^–.

And since the HSO₄^– has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: CCl₂F₂ Lewis structure

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External links

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