BF2 Lewis structure

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BF₂ has one boron atom and two fluorine atoms.

In BF₂ Lewis structure, there are two single bonds around the boron atom, with two fluorine atoms attached to it. Each fluorine atom has three lone pairs, and the boron atom has one unpaired electron.

Steps

Here’s how you can easily draw the BF₂ Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms

Now, let’s take a closer look at each step mentioned above.

#1 Draw a rough skeleton structure

• First, determine the total number of valence electrons

In the periodic table, boron lies in group 13, and fluorine lies in group 17.

Hence, boron has three valence electrons and fluorine has seven valence electrons.

Since BF₂ has one boron atom and two fluorine atoms, so…

Valence electrons of one boron atom = 3 × 1 = 3
Valence electrons of two fluorine atoms = 7 × 2 = 14

And the total valence electrons = 3 + 14 = 17

Learn how to find: Boron valence electrons and Fluorine valence electrons

• Second, find the total electron pairs

We have a total of 17 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

But 17 can not be divided by two. Hence, there are a total of 8 electron pairs and one unpaired electron.

• Third, determine the central atom

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

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

Therefore, place boron in the center and fluorines on either side.

• And finally, draw the rough sketch

#2 Mention lone pairs on the atoms

Here, we have 8 electron pairs and one unpaired electron. And two B — F bonds are already marked. So we have to only mark the remaining six electron pairs and one unpaired electron as lone pairs on the sketch.

Also remember that both (boron and fluorine) are the period 2 elements, so they can not keep more than 8 electrons in their last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are fluorines.

So for each fluorine, there are three lone pairs, and for boron, there is one unpaired electron.

Mark the lone pairs on the sketch as follows:

#3 If needed, mention formal charges on the atoms

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For boron atom, formal charge = 3 – 1 – ½ (4) = 0

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

Here, both boron and fluorine atoms do not have charges, so no need to mark the charges.

In the above structure, you can see that the central atom (boron) doesn’t form an octet. But, boron has an exception that it does not require eight electrons to form an octet. So no need to worry about the octet rule here.

Therefore, this structure is the stable Lewis structure of BF₂.

Next: HCP Lewis structure

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