CH3CHO Lewis structure

CH3CHO (acetaldehyde) has two carbon atoms, four hydrogen atoms, and one oxygen atom.

In CH3CHO Lewis structure, there is a single bond between the two carbon atoms. The left carbon is attached with three hydrogen atoms, and the right carbon is attached with one oxygen atom and one hydrogen atom. And on the oxygen atom, there are two lone pairs.

Contents

Steps

To properly draw the CH3CHO 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, carbon lies in group 14, hydrogen lies in group 1, and oxygen lies in group 16.

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

Since CH3CHO has two carbon atoms, four hydrogen atoms, and one oxygen atom, so…

Valence electrons of two carbon atoms = 4 × 2 = 8
Valence electrons of four hydrogen atoms = 1 × 4 = 4
Valence electrons of one oxygen atom = 6 × 1 = 6

And the total valence electrons = 8 + 4 + 6 = 18

• Second, find the total electron pairs

We have a total of 18 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 = 18 ÷ 2 = 9

• 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 and oxygen. Place the least electronegative atom at the center.

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

Here, there are two carbon atoms, so we can assume any one as the central atom.

Let’s assume that the central atom is left carbon.

Therefore, place carbons 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 9 electron pairs. And six bonds are already marked. So we have to only mark the remaining three electron pairs as lone pairs on the sketch.

Also remember that both (carbon 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, oxygen, and right carbon. But no need to mark on hydrogen, because each hydrogen has already two electrons.

So for oxygen, there are three lone pairs, and for right carbon, there is zero lone pair because all three 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 left carbon atom, formal charge = 4 – 0 – ½ (8) = 0

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

For each hydrogen atom, formal charge = 1 – 0 – ½ (2) = 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 right carbon atom as follows:

In the above structure, you can see that the central atom (left carbon) forms an octet. The outside atoms (right carbon 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 CH3CHO.

Next: BBr3 Lewis structure