Urea Lewis structure

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CO(NH₂)₂ or CH₄N₂O (urea) has one carbon atom, four hydrogen atoms, two nitrogen atoms, and one oxygen atom.

In urea Lewis structure, there is one double bond and two single bonds around the carbon atom, with one oxygen atom and two nitrogen atoms attached to it, and each nitrogen is attached with two hydrogen atoms. There are two lone pairs on the oxygen atom, and each nitrogen atom has one lone pair.

Steps

Here’s how you can easily draw the urea Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms
#4 Minimize formal charges by converting lone pairs of the atoms, and try to get a stable Lewis structure
#5 Repeat step 4 again if needed, until all charges are minimized

Now, let’s take a closer look at each step mentioned above.

#1 Draw a rough skeleton structure

• First, determine the total number of valence electrons

In the periodic table, carbon lies in group 14, hydrogen lies in group 1, nitrogen lies in group 15, and oxygen lies in group 16.

Hence, carbon has four valence electrons, hydrogen has one valence electron, nitrogen has five valence electrons, and oxygen has six electrons.

Since urea (CH₄N₂O) has one carbon atom, four hydrogen atoms, two nitrogen atoms, and one oxygen atom, so…

Valence electrons of one carbon atom = 4 × 1 = 4
Valence electrons of four hydrogen atoms = 1 × 4 = 4
Valence electrons of two nitrogen atoms = 5 × 2 = 10
Valence electrons of one oxygen atom = 6 × 1 = 6

And the total valence electrons = 4 + 4 + 10 + 6 = 24

Learn how to find: Carbon valence electrons, Hydrogen valence electrons, Nitrogen valence electrons, and Oxygen valence electrons

• Second, find the total electron pairs

We have a total of 24 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 = 24 ÷ 2 = 12

• 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 carbon, nitrogen, and oxygen. Place the least electronegative atom at the center.

Since carbon is less electronegative than nitrogen and oxygen, assume that the central atom is carbon.

Therefore, place carbon in the center and hydrogen, nitrogen, and oxygen on either side.

• And finally, draw the rough sketch

#2 Mention lone pairs on the atoms

Here, we have a total of 12 electron pairs. And seven bonds are already marked. So we have to only mark the remaining five electron pairs as lone pairs on the sketch.

Also remember that all three (carbon, nitrogen, and oxygen) are the period 2 elements, so they can not keep more than 8 electrons in their last shell. And hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are hydrogens, nitrogens, and oxygen. But no need to mark on hydrogen, because each hydrogen has already two electrons.

So for oxygen, there are three lone pairs, for each nitrogen, there is one lone pair, and for carbon, there is zero lone pair because all five electron pairs are over.

Mark the lone pairs on the sketch as follows:

#3 If needed, mention formal charges on the atoms

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

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

For carbon atom, formal charge = 4 – 0 – ½ (6) = +1

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

For each nitrogen atom, formal charge = 5 – 2 – ½ (6) = 0

For oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

Here, both carbon and oxygen atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable Lewis structure because both carbon 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 oxygen atom to make a new C — O bond with the carbon atom as follows:

In the above structure, you can see that the central atom (carbon) forms an octet. The outside atoms (nitrogens and oxygen) also form an octet, and all hydrogens form a duet. Hence, the octet rule and duet rule are satisfied.

Also, the above structure is more stable than the previous structures. Therefore, this structure is the stable Lewis structure of urea.

Next: N₂O₂ Lewis structure

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External links

• https://www.chemistryscl.com/general/urea-lewis-structure/
• https://study.com/academy/answer/what-is-the-urea-lewis-structure.html
• https://www.bartleby.com/questions-and-answers/draw-lewis-structures-for-urea-nh2conh2-and-methanol-ch3oh/a21d5771-8a7d-4120-bb0c-99bf056c0163
• https://www.chegg.com/homework-help/questions-and-answers/urea-lewis-structure-drawn-right-hybridization-nitrogen-atom-labeled-lewis-diagram-urea-b–q89826621