Lewis structure of HSO4-

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The Lewis structure of HSO₄^– contains two double bonds and three single bonds, with sulfur in the center, and hydrogen and four oxygens on either side. The top oxygen atom, left oxygen atom, and right oxygen atom have two lone pairs. The bottom oxygen atom has three lone pairs, and sulfur atom and hydrogen atom do not have any lone pair.

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

Steps

By using the following steps, you can easily draw the Lewis structure of HSO₄^–.

#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 HSO₄^– has one hydrogen atom, one sulfur atom, and four oxygen atoms.

So the total number of valence electrons = valence electrons of hydrogen atom + valence electrons of sulfur atom + (valence electrons of oxygen atom × 4)

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

Therefore, the total number of valence electrons = 1 + 6 + 24 + 1 = 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 hydrogen and oxygens on either side. And draw the rough skeleton structure for the Lewis structure of HSO₄^– something like this:

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

#2 Show chemical bond

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

#3 Mark lone pairs

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

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

Now draw the Lewis structure of HSO₄^– something like this:

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 C₂^2- (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 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, top oxygen, and bottom oxygen atom

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

Formal charge = 6 – 6 – ½ (2) = -1

• For 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 HSO₄^– 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 SeO₃^2- (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 top oxygen atom, and one lone pair from left oxygen atom to make a new bond with the sulfur atom. And then, the Lewis structure of HSO₄^– looks something like this:

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 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 top oxygen, left oxygen, and right oxygen atom

Valence electrons = 6
Nonbonding electrons = 4
Bonding electrons = 4

Formal charge = 6 – 4 – ½ (4) = 0

• For bottom oxygen atom

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

Formal charge = 6 – 6 – ½ (2) = -1

Mention the formal charges of atoms on the structure. So the Lewis structure of HSO₄^– looks something like this:

In the above structure, you can see that the formal charges of atoms are closer to zero. Therefore, this is the most stable Lewis structure of HSO₄^–.

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

Now HSO₄^– is an ion having a negative (-1) charge, so draw brackets around the above Lewis structure and mention that charge on the top right corner. And then, the Lewis structure of HSO₄^– looks something like this:

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Related

• Lewis structure of C₂H₂Cl₂
• Lewis structure of C₂^2-
• Lewis structure of SeO₃^2-
• Lewis structure of P₄
• Lewis structure of ClF

External links

• What is the Lewis dot structure of (HSO4) ^-? What is the formal charge on each atom? – Quora
• Bisulfate ion (HSO4-) Ion Lewis Structure – Chemistry School
• HSO4- Lewis Structure in 6 Steps (With Images) – Pediabay
• Chemical Bonding: HSO4- Lewis Structure – The Geoexchange
• HSO4– Structure, Properties and more – Geometry of Molecules
• Draw a Lewis structure for HSO₄^– – Homework.Study.com
• Draw Lewis structure of hydrogen sulfate ion (HSO4∗) by adding nonbonding electrons, shared electrons (or bonds) and formal charges – Chegg
• Draw the Lewis structure for HSO₄^– – Quizlet
• What is the Lewis structure for HSO₄^–? – OneClass
• HSO4- (Hydrogen Sulfate Ion) Oxidation Number – ChemicalAid