BrF4- Lewis structure

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BrF4- Lewis Structure
BrF4 Lewis structure

BrF4 has one bromine atom and four fluorine atoms.

In BrF4 Lewis structure, there are four single bonds around the bromine 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 bromine atom.

Steps

To properly draw the BrF4 Lewis structure, follow these steps:

#1 Draw a rough sketch of the structure
#2 Next, indicate lone pairs on the atoms
#3 Indicate formal charges on the atoms, if necessary

Let’s break down each step in more detail.

#1 Draw a rough sketch of the structure

  • First, determine the total number of valence electrons
Periodic table

In the periodic table, both bromine and fluorine lie in group 17.

Hence, both bromine and fluorine have seven valence electrons.

Since BrF4 has one bromine atom and four fluorine atoms, so…

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

Now the BrF4 has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 7 + 28 + 1 = 36

  • Second, find the total electron pairs

We have a total of 36 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 = 36 ÷ 2 = 18

  • Third, determine the central atom

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

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

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

  • And finally, draw the rough sketch
BrF4- Lewis Structure (Step 1)
Rough sketch of BrF4 Lewis structure

#2 Next, indicate lone pairs on the atoms

Here, we have a total of 18 electron pairs. And four Br — F bonds are already marked. So we have to only mark the remaining fourteen electron pairs as lone pairs on the sketch.

Also remember that bromine is a period 4 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 fluorine, there are three lone pairs, and for bromine, there are two lone pairs.

Mark the lone pairs on the sketch as follows:

BrF4- Lewis Structure (Step 2)
Lone pairs marked on BrF4 Lewis structure

#3 Indicate formal charges on the atoms, if necessary

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

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

For bromine atom, formal charge = 7 – 4 – ½ (8) = -1

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

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

BrF4- Lewis Structure (Step 3)
Formal charges marked, and got the most stable Lewis structure of BrF4

In the above structure, you can see that the central atom (bromine) 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 bromine 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 bromine atom to make a new Br — 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 BrF4.

And since the BrF4 has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

BrF4- Lewis Structure (Final)
BrF4 Lewis structure showing a negative (-1) charge

Next: AsF6 Lewis structure

External links

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.

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