CHF3 (fluoroform) has one carbon atom, one hydrogen atom, and three fluorine atoms.
In the CHF3 Lewis structure, there are four single bonds around the carbon atom, with one hydrogen atom and three fluorine atoms attached to it, and on each fluorine atom, there are three lone pairs.
Steps
To properly draw the CHF3 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
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 fluorine lies in group 17.
Hence, carbon has four valence electrons, hydrogen has one valence electron, and fluorine has seven valence electrons.
Since CHF3 has one carbon atom, one hydrogen atom, and three fluorine atoms, so…
Valence electrons of one carbon atom = 4 × 1 = 4
Valence electrons of one hydrogen atom = 1 × 1 = 1
Valence electrons of three fluorine atoms = 7 × 3 = 21
And the total valence electrons = 4 + 1 + 21 = 26
Learn how to find: Carbon valence electrons, Hydrogen valence electrons, and Fluorine valence electrons
- Second, find the total electron pairs
We have a total of 26 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 = 26 ÷ 2 = 13
- 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 fluorine. Place the least electronegative atom at the center.
Since carbon is less electronegative than fluorine, assume that the central atom is carbon.
Therefore, place carbon in the center and hydrogen and fluorine on either side.
- And finally, draw the rough sketch
#2 Next, indicate lone pairs on the atoms
Here, we have a total of 13 electron pairs. And four bonds are already marked. So we have to only mark the remaining nine electron pairs as lone pairs on the sketch.
Also remember that both (carbon and fluorine) 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 hydrogen and fluorines. But no need to mark on hydrogen, because hydrogen already has two electrons.
So for each fluorine, there are three lone pairs, and for carbon, there is zero lone pair because all nine 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 carbon atom, formal charge = 4 – 0 – ½ (8) = 0
For hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0
For each fluorine 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 atoms (fluorines) also form an octet, and hydrogen forms a duet. Hence, the octet rule and duet rule are satisfied.
Therefore, this structure is the stable Lewis structure of CHF3.
Next: CF3Cl Lewis structure
External links
- https://techiescientist.com/chf3-lewis-structure/
- https://lambdageeks.com/chf3-lewis-structure/
- https://sciedutut.com/chf3-lewis-structure/
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.