BF4- Lewis structure

BF₄^– (tetrafluoroborate) has one boron atom and four fluorine atoms.

In BF₄^– Lewis structure, there are four single bonds around the boron atom, with four fluorine atoms attached to it, and on each fluorine atom, there are three lone pairs.

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

Alternative method: Lewis structure of BF₄^–

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 BF₄^– has one boron atom and four fluorine atoms, so…

Valence electrons of one boron atom = 3 × 1 = 3
Valence electrons of four fluorine atoms = 7 × 4 = 28

Now the BF₄^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 3 + 28 + 1 = 32

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

• Second, find the total electron pairs

We have a total of 32 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 = 32 ÷ 2 = 16

• 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 a total of 16 electron pairs. And four B — F bonds are already marked. So we have to only mark the remaining twelve electron pairs 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 zero lone pair because all twelve 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 boron atom, formal charge = 3 – 0 – ½ (8) = -1

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

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

Final structure

The final structure of BF₄^– involves a central boron atom linked to four fluorine atoms through single covalent bonds. In this arrangement, the boron atom satisfies the octet rule by sharing its valence electrons with the four surrounding halogens, while each fluorine atom completes its octet by maintaining three lone pairs. This specific configuration is the most stable because it results in a formal charge of -1 on the central boron atom and zero on each of the fluorine atoms, representing the most energetically favorable distribution for the ion. Consequently, this electronic pattern serves as the definitive and most accurate Lewis representation for the tetrafluoroborate anion.

To complete the representation, draw square brackets around the entire Lewis structure and place a “-” or “-1” sign as a superscript outside the upper right bracket. This notation signifies that the negative charge is a property of the whole ion.

Next: P₂ Lewis structure

External video

• BF4- Lewis Structure – How to Draw the Lewis Structure for BF4- – YouTube • Wayne Breslyn

External links

• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/BF4-lewis-structure.html
• https://www.quora.com/How-can-you-determine-the-Lewis-dot-structure-for-BF4
• https://homework.study.com/explanation/draw-the-lewis-dot-structure-for-bf4-and-provide-the-following-information-a-number-of-valence-electrons-b-hybridization-c-electron-geometry-d-molecular-geometry-e-polarity.html