SF5Cl (sulfur chloride pentafluoride) has one sulfur atom, five fluorine atoms, and one chlorine atom.
In SF5Cl Lewis structure, there are six single bonds around the sulfur atom, with one chlorine atom and five fluorine atoms attached to it. And on chlorine and each fluorine atom, there are three lone pairs.
Steps
Use these steps to correctly draw the SF5Cl 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
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, sulfur lies in group 16, and both fluorine and chlorine lie in group 17.
Hence, sulfur has six valence electrons, and both fluorine and chlorine have seven valence electrons.
Since SF5Cl has one sulfur atom, five chlorine atoms, and one chlorine atom, so…
Valence electrons of one sulfur atom = 6 × 1 = 6
Valence electrons of five chlorine atoms = 7 × 5 = 35
Valence electrons of one chlorine atom = 7 × 1 = 7
And the total valence electrons = 6 + 35 + 7 = 48
Learn how to find: Sulfur valence electrons, Fluorine valence electrons, and Chlorine valence electrons
- Second, find the total electron pairs
We have a total of 48 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 = 48 ÷ 2 = 24
- Third, determine the central atom
We have to place the least electronegative atom at the center.
Since sulfur is less electronegative than fluorine and chlorine, assume that the central atom is sulfur.
Therefore, place sulfur in the center and fluorine and chlorine on either side.
- And finally, draw the rough sketch
#2 Mark lone pairs on the atoms
Here, we have a total of 24 electron pairs. And six bonds are already marked. So we have to only mark the remaining eighteen electron pairs as lone pairs on the sketch.
Also remember that both (sulfur and chlorine) are the period 3 elements, so they can keep more than 8 electrons in their last shell. And fluorine is a period 2 element, so it can not keep more than 8 electrons in its last shell.
Always start to mark the lone pairs from outside atoms. Here, the outside atoms are chlorine and fluorines.
So for chlorine and each fluorine, there are three lone pairs, and for sulfur, there is zero lone pair because all eighteen 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 sulfur atom, formal charge = 6 – 0 – ½ (12) = 0
For each fluorine and chlorine 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 (sulfur) forms an octet. And the outside atoms (chlorine and fluorines) also form an octet. Hence, the octet rule is satisfied.
Therefore, this structure is the stable Lewis structure of SF5Cl.
Next: PF2Cl3 Lewis structure
External links
- https://www.chegg.com/homework-help/questions-and-answers/draw-lewis-structure-sf5cl-i3-window-answer-questions-follow-q58070130
- https://brainly.com/question/29591973
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.