BF2 Lewis structure

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.

Alternative method: Lewis structure of BF₂

Rough sketch

• 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

Lone pair

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:

Formal charge

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.

Final structure

The final structure of BF₂ contains a central boron atom linked to two fluorine atoms through single covalent bonds. In this layout, the boron atom does not complete a full octet but instead maintains five valence electrons, which comprise two bonding pairs and one lone electron (a radical state). Within this arrangement, each fluorine atom successfully satisfies the octet rule by retaining three lone pairs alongside its single shared bond. This configuration represents the most stable state for the neutral radical because it results in a formal charge of zero for every atom involved. Therefore, this specific electronic distribution serves as the definitive and most accurate Lewis representation of BF₂.

Next: HCP Lewis structure