CBr2F2 Lewis structure

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CBr₂F₂ (dibromodifluoromethane) has one carbon atom, two bromine atoms, and two fluorine atoms.

In CBr₂F₂ Lewis structure, there are four single bonds around the carbon atom, with two bromine atoms and two fluorine atoms attached to it, and on each bromine and fluorine atom, there are three lone pairs.

Steps

Use these steps to correctly draw the CBr₂F₂ Lewis structure:

#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required

Let’s discuss each step in more detail.

#1 First draw a rough sketch

• First, determine the total number of valence electrons

In the periodic table, carbon lies in group 14, and both bromine and fluorine lie in group 17.

Hence, carbon has four valence electrons, both bromine and fluorine have seven valence electrons.

Since CBr₂F₂ has one carbon atom, two bromine atoms, and two fluorine atoms, so…

Valence electrons of one carbon atom = 4 × 1 = 4
Valence electrons of two bromine atoms = 7 × 2 = 14
Valence electrons of two fluorine atoms = 7 × 2 = 14

And the total valence electrons = 4 + 14 + 14 = 32

Learn how to find: Carbon valence electrons, Bromine 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 carbon is less electronegative than bromine and fluorine, assume that the central atom is carbon.

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

• And finally, draw the rough sketch

#2 Mark lone pairs on the atoms

Here, we have a total of 16 electron pairs. And four 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 (carbon and fluorine) are the period 2 elements, so they can not keep more than 8 electrons in their last shell. And bromine is a period 4 element, so it can keep more than 8 electrons in its last shell.

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

So for each bromine and fluorine, there are three lone pairs, and for carbon, there is zero lone pair because all twelve electron pairs are over.

Mark the lone pairs on the sketch as follows:

#3 Calculate and mark formal charges on the atoms, if required

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

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

For carbon atom, formal charge = 4 – 0 – ½ (8) = 0

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

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

Here, the atoms do not have charges, so no need to mark the charges.

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

Therefore, this structure is the stable Lewis structure of CBr₂F₂.

Next: SiH₃^– Lewis structure

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External links

• https://www.bartleby.com/questions-and-answers/draw-the-lewis-dot-structure-for-cbr2f2-.-choose-the-answer-below-that-gives-in-this-exact-order-the/6902624e-d633-402a-aa58-492cba47fbeb
• https://brainly.com/question/13633531