BF4- Lewis structure

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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.

Steps

Here’s how you can easily draw the BF₄^– 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 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

#2 Mention lone pairs on the atoms

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:

#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 – 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:

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

Therefore, this structure is the most stable Lewis structure of BF₄^–.

And since the BF₄^– has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: P₂ Lewis structure

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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