CH3I Lewis structure

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CH3I Lewis Structure
CH3I Lewis structure

CH3I (iodomethane) has one carbon atom, three hydrogen atoms, and one iodine atom.

In CH3I Lewis structure, there are four single bonds around the carbon atom, with three hydrogen atoms and one iodine atom attached to it, and on the iodine atom, there are three lone pairs.

Steps

Here’s how you can easily draw the CH3I 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
Periodic table

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 CH3I has one carbon atom, three hydrogen atoms, and one iodine atom, so…

Valence electrons of one carbon atom = 4 × 1 = 4
Valence electrons of three hydrogen atoms = 1 × 3 = 3
Valence electrons of one iodine atom = 7 × 1 = 7

And the total valence electrons = 4 + 3 + 7 = 14

  • Second, find the total electron pairs

We have a total of 14 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 = 14 ÷ 2 = 7

  • 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.

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

  • And finally, draw the rough sketch
CH3I Lewis Structure (Step 1)
Rough sketch of CH3I Lewis structure

#2 Mention lone pairs on the atoms

Here, we have a total of 7 electron pairs. And four 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 and iodine. But no need to mark on hydrogen, because each hydrogen has already two electrons.

So for iodine, there are three lone pairs, and for carbon, there is zero lone pair because all three electron pairs are over.

Mark the lone pairs on the sketch as follows:

CH3I Lewis Structure (Step 2)
Lone pairs marked, and got the stable Lewis structure of CH3I

#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 (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 (carbon) forms an octet. The outside atom (iodine) also forms 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 CH3I.

Next: C2H4Cl2 Lewis structure

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Deep

Learnool.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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