SF5Cl Lewis structure

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.

Alternative method: Lewis structure of SF₅Cl

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

Lone pair

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:

Formal charge

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.

Final structure

The final structure of SF₅Cl comprises a central sulfur atom connected to five fluorine atoms and one chlorine atom through single covalent bonds. In this configuration, the sulfur atom utilizes an expanded valence shell to accommodate twelve electrons, forming six total bonds. Within this layout, each of the five fluorine atoms and the single chlorine atom satisfies the octet rule by maintaining three lone pairs alongside their respective single shared bonds. This octahedral arrangement represents the most stable state for the molecule because it results in a formal charge of zero for every atom involved, including the central sulfur and all six halogen atoms. Thus, this specific electronic distribution serves as the definitive and most accurate Lewis representation of sulfur chloride pentafluoride.

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