CH3CH2I Lewis structure

The information on this page is ✔ fact-checked.

CH₃CH₂I (ethyl iodide) has two carbon atoms, five hydrogen atoms, and one iodine atom.

In CH₃CH₂I 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 iodine atom and two hydrogen atoms. And on the iodine atom, there are three lone pairs.

Steps

Here’s how you can easily draw the CH₃CH₂I 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

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, and iodine lies in group 17.

Hence, carbon has four valence electrons, hydrogen has one valence electron, and iodine has seven valence electrons.

Since CH₃CH₂I has two carbon atoms, five hydrogen atoms, and one iodine atom, so…

Valence electrons of two carbon atoms = 4 × 2 = 8
Valence electrons of five hydrogen atoms = 1 × 5 = 5
Valence electrons of one iodine atom = 7 × 1 = 7

And the total valence electrons = 8 + 5 + 7 = 20

Learn how to find: Carbon valence electrons and Hydrogen valence electrons

• Second, find the total electron pairs

We have a total of 20 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 = 20 ÷ 2 = 10

• 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 iodine. Place the least electronegative atom at the center.

Since carbon is less electronegative than iodine, 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 iodine on either side.

• And finally, draw the rough sketch

#2 Mention lone pairs on the atoms

Here, we have a total of 10 electron pairs. And seven bonds are already marked. So we have to only mark the remaining three electron pairs as lone pairs on the sketch.

Also remember that carbon is a period 2 element, so it can not keep more than 8 electrons in its last shell. Hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell. And iodine is a period 5 element, so it can keep more than 8 electrons in its last shell.

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

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

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

For iodine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, the atoms do not have charges, so no need to mark the charges.

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

Therefore, this structure is the stable Lewis structure of CH₃CH₂I.

Next: C₂H₄Br₂ Lewis structure

📝 Your feedback matters. Visit our contact page.