ClF_{2}^{–} has **one chlorine** atom and **two fluorine** atoms.

In ClF_{2}^{–} Lewis structure, there are two single bonds around the chlorine atom, with two fluorine atoms attached to it, and each atom has three lone pairs.

Also, there is a negative (-1) charge on the chlorine atom.

## Steps

Use these steps to correctly draw the ClF_{2}^{–} 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, both chlorine and fluorine lie in group 17.

Hence, both chlorine and fluorine have **seven** valence electrons.

Since ClF_{2}^{–} has one chlorine atom and two fluorine atoms, so…

Valence electrons of one chlorine atom = 7 × 1 = 7

Valence electrons of two fluorine atoms = 7 × 2 = 14

Now the ClF_{2}^{–} has a negative (-1) charge, so we have to add one more electron.

So the **total valence electrons** = 7 + 14 + 1 = 22

Learn how to find: Chlorine valence electrons and Fluorine 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

We have to place the least electronegative atom at the center.

Since chlorine is less electronegative than fluorine, assume that the **central atom is chlorine**.

Therefore, place chlorine in the center and fluorines 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 two Cl — F bonds are already marked. So we have to only mark the remaining nine electron pairs as lone pairs on the sketch.

Also remember that chlorine is a period 3 element, so it can keep more than 8 electrons in its 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 fluorines.

So for each atom, there are **three** lone pairs.

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 **chlorine** atom, formal charge = 7 – 6 – ½ (4) = -1

For **each fluorine** atom, formal charge = 7 – 6 – ½ (2) = 0

Here, the chlorine atom has a charge, so mark it on the sketch as follows:

In the above structure, you can see that the central atom (chlorine) forms an octet. And the outside atoms (fluorines) also form an octet. Hence, the octet rule is satisfied.

Now there is still a negative (-1) charge on the chlorine atom.

This is not okay, right? Because the structure with a negative charge on the most electronegative atom is the best Lewis structure. And in this case, the most electronegative element is fluorine.

But if we convert a lone pair of the chlorine atom to make a new Cl — F bond with the fluorine atom, and calculate the formal charge, then we do not get the formal charges on atoms closer to zero.

And the structure with the formal charges on atoms closer to zero is the best Lewis structure.

Therefore, this structure is the **most stable Lewis structure** of ClF_{2}^{–}.

And since the ClF_{2}^{–} has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

**Next:** ClF_{4}^{–} Lewis structure

## External links

- https://lambdageeks.com/clf2-lewis-structure/
- https://homework.study.com/explanation/draw-the-lewis-structure-for-clf2.html
- https://www.chegg.com/homework-help/sketch-lewis-structures-clf2-clf2-electron-pair-molecular-g-chapter-10-problem-24gq-solution-9780495112990-exc

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.