Lewis structure of H2PO4-

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The Lewis structure of H₂PO₄^– contains one double bond and five single bonds, with phosphorus in the center, and two hydrogens and four oxygens on either side. The left oxygen atom, top oxygen atom, and bottom oxygen atom have two lone pairs. The right oxygen atom has three lone pairs, and phosphorus atom and hydrogen atom do not have any lone pair.

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

Steps

By using the following steps, you can easily draw the Lewis structure of H₂PO₄^–.

#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, phosphorus is a group 15 element, and oxygen is a group 16 element. Hence, hydrogen has one valence electron, phosphorus has five valence electrons, and oxygen has six valence electrons.

Now H₂PO₄^– has two hydrogen atoms, one phosphorus atom, and four oxygen atoms.

So the total number of valence electrons = (valence electrons of hydrogen atom × 2) + valence electrons of phosphorus atom + (valence electrons of oxygen atom × 4)

And H₂PO₄^– has a negative (-1) charge, so we have to add one more electron.

Therefore, the total number of valence electrons = 2 + 5 + 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 phosphorus 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 phosphorus and oxygen as follows:

Electronegativity value of phosphorus = 2.19
Electronegativity value of oxygen = 3.44

Obviously, phosphorus is less electronegative than oxygen. Hence, assume that phosphorus is the central atom.

So now, put phosphorus in the center and hydrogens and oxygens on either side. And draw the rough skeleton structure for the Lewis structure of H₂PO₄^– something like this:

Also read: How to draw Lewis structure of C₂H₂O (5 steps)

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since phosphorus is surrounded by two hydrogens and four oxygens, use twelve electrons to show six 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 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. Phosphorus 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 (phosphorus) will not get any lone pair, because all ten lone pairs are used.

Now draw the Lewis structure of H₂PO₄^– something like this:

In the above structure, you can see that the octet is completed on the central atom (phosphorus), 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 HPO₄^2- (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 each hydrogen atom

Valence electrons = 1
Nonbonding electrons = 0
Bonding electrons = 2

Formal charge = 1 – 0 – ½ (2) = 0

• For phosphorus atom

Valence electrons = 5
Nonbonding electrons = 0
Bonding electrons = 8

Formal charge = 5 – 0 – ½ (8) = +1

• 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 H₂PO₄^– 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 C₆H₁₂ (3 steps)

#5 Convert lone pair and calculate formal charge again

As mentioned earlier, phosphorus is a period 3 element, so it can keep more than 8 electrons in its last shell.

So convert one lone pair from the left oxygen atom to make a new bond with the phosphorus atom. And then, the Lewis structure of H₂PO₄^– 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 each hydrogen atom

Valence electrons = 1
Nonbonding electrons = 0
Bonding electrons = 2

Formal charge = 1 – 0 – ½ (2) = 0

• For phosphorus atom

Valence electrons = 5
Nonbonding electrons = 0
Bonding electrons = 10

Formal charge = 5 – 0 – ½ (10) = 0

• For left oxygen, top oxygen, and bottom oxygen atom

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

Formal charge = 6 – 4 – ½ (4) = 0

• For right 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 H₂PO₄^– 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 H₂PO₄^–.

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

Now H₂PO₄^– 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 H₂PO₄^– looks something like this:

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Related

• Lewis structure of C₂H₂O
• Lewis structure of HPO₄^2-
• Lewis structure of C₆H₁₂
• Lewis structure of POF₃
• Lewis structure of H₃PO₃

External links

• Using the VSEPR theory, predict the molecular structure of dihydrogen phosphate ion, H2PO4- – Homework.Study.com
• Draw a Lewis structure for the anion H2PO4-, derived from phosphoric acid – Chegg
• Lewis structure for H2PO4- – Reddit
• lewis structure for H2PO4- – Chemical Forums
• What is the model of Lewis of H2PO4- – Course Hero
• Dihydrogenphosphate | H2O4P- | CID 1003 – National Institutes of Health (.gov)
• H2PO4- (Dihydrogenphosphate Ion) Oxidation Number – ChemicalAid