Lewis structure of HCOOH

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Lewis structure of HCOOH
Lewis structure of HCOOH

The Lewis structure of HCOOH contains one double bond and three single bonds, with carbon in the center, and two hydrogens and two oxygens on either side. There are two lone pairs on each oxygen atom, and carbon atom and hydrogen atom do not have any lone pair.

Steps

By using the following steps, you can easily draw the Lewis structure of HCOOH.

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Complete octet on central atom
#5 Calculate formal charge and check stability

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, carbon is a group 14 element, and oxygen is a group 16 element. Hence, hydrogen has one valence electron, carbon has four valence electrons, and oxygen has six valence electrons.

Now HCOOH has two hydrogen atoms, one carbon atom, and two oxygen atoms.

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

Therefore, the total number of valence electrons = 2 + 4 + 12 = 18

  • 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 carbon 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 carbon and oxygen as follows:

Electronegativity value of carbon = 2.55
Electronegativity value of oxygen = 3.44

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

So now, put carbon in the center and two oxygens and two hydrogens on either side. And draw the rough skeleton structure for the Lewis structure of HCOOH something like this:

Skeleton structure for Lewis structure of HCOOH

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since carbon is surrounded by two oxygens and two hydrogens, use eight electrons to show four chemical bonds as follows:

Four chemical bonds shown between atoms

#3 Mark lone pairs

As calculated earlier, we have a total of 18 valence electrons. And in the above structure, we have already used eight valence electrons. Hence, ten valence electrons are remaining.

Two valence electrons represent one lone pair. So ten valence electrons = five lone pairs.

Note that hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell. And both (carbon and oxygen) are the period 2 elements, so they can not keep more than 8 electrons in their 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 top oxygen will get three lone pairs and right oxygen will get two lone pairs. And the central atom (carbon) will not get any lone pair, because all five lone pairs are used.

Now draw the Lewis structure of HCOOH something like this:

Lone pairs marked on Lewis structure of HCOOH

In the above structure, you can see that the octet is completed on outside atoms. But, the central atom (carbon) doesn’t form an octet.

So in the next step, we have to complete the octet on the central atom.

#4 Complete octet on central atom

Remember that carbon is a period 2 element, so it can not keep more than 8 electrons in its last shell.

Now carbon already has six valence electrons. Hence, carbon needs two more valence electrons to complete its octet.

Which means that we have to convert one lone pair. But the question is, which lone pair should we convert to get the best Lewis structure?

There are two possible ways we can convert lone pair: 

1) Convert one lone pair from the top oxygen atom.
2) Convert one lone pair from the right oxygen atom.

In order to get the best Lewis structure, we have to make the formal charges on atoms closer to zero. And we can do so by choosing the 1).

So convert one lone pair from the top oxygen atom to make a new bond with the carbon atom. And then, the Lewis structure of HCOOH looks something like this:

Lone pair of top oxygen is converted, and octet is completed on atoms

In the above structure, you can see that the octet is completed on the central atom (carbon), and also on the outside atoms. Therefore, the octet rule is satisfied.

After completing the octet, one last thing we need to do is, calculate the formal charge and check the stability of the above structure.

#5 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 carbon atom

Valence electrons = 4
Nonbonding electrons = 0
Bonding electrons = 8

Formal charge = 4 – 0 – ½ (8) = 0

  • For each 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 HCOOH looks something like this:

Formal charges are calculated, and got the stable Lewis structure of HCOOH

In the above structure, you can see that the formal charges of all atoms are zero. Therefore, this is the stable Lewis structure of HCOOH.

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

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Deep

Rootmemory.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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