CH3NCO Lewis structure

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CH₃NCO (methyl isocyanate) has two carbon atoms, three hydrogen atoms, one nitrogen atom, and one oxygen atom.

In the CH₃NCO Lewis structure, there is one single bond and one double bond around the nitrogen atom, with two carbon atoms attached to it. The left carbon is attached with three hydrogen atoms, and the right carbon is attached with one oxygen atom. The nitrogen atom has one lone pair, and the oxygen atom has two lone pairs.

Steps

Use these steps to correctly draw the CH₃NCO Lewis structure:

#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required
#4 Convert lone pairs of the atoms, and minimize formal charges
#5 Repeat step 4 if needed, until all charges are minimized, to get a stable Lewis structure

Let’s discuss each step in more detail.

#1 First draw a rough sketch

• 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 CH₃NCO has two carbon atoms, three hydrogen atoms, one nitrogen atom, and one oxygen atom, so…

Valence electrons of two carbon atoms = 4 × 2 = 8
Valence electrons of three hydrogen atoms = 1 × 3 = 3
Valence electrons of one nitrogen atom = 5 × 1 = 5
Valence electrons of one oxygen atom = 6 × 1 = 6

And the total valence electrons = 8 + 3 + 5 + 6 = 22

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 22 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 = 22 ÷ 2 = 11

• 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.

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

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

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

• And finally, draw the rough sketch

#2 Mark lone pairs on the atoms

Here, we have a total of 11 electron pairs. And six 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, left carbon, nitrogen, 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 nitrogen, there are two lone pairs, and for carbons, there is zero lone pair because all five electron pairs are over.

Mark the lone pairs on the sketch as follows:

#3 Calculate and mark formal charges on the atoms, if required

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 (4) = +2

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

For nitrogen atom, formal charge = 5 – 4 – ½ (4) = -1

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

Here, all three (carbon, nitrogen, and oxygen) atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable Lewis structure because all three (carbon, nitrogen, and oxygen) atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

#4 Convert lone pairs of the atoms, and minimize formal charges

Convert a lone pair of the nitrogen atom to make a new C — N bond with the right carbon atom as follows:

#5 Repeating step 4 to get a stable Lewis structure

Since there are charges on carbon and oxygen atoms, again 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 (right carbon) forms an octet. The outside atoms (left carbon, nitrogen, 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 CH₃NCO.

Next: XeCl₄ Lewis structure

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External links

• https://oneclass.com/homework-help/chemistry/6934365-for-the-following-skeletal-stru.en.html
• https://studysoup.com/tsg/100775/organic-chemistry-6-edition-chapter-1-problem-1-59