# BeBr2 Lewis structure

BeBr2 (beryllium bromide) has one beryllium atom and two bromine atoms.

In the BeBr2 Lewis structure, there are two single bonds around the beryllium atom, with two bromine atoms attached to it, and on each bromine atom, there are three lone pairs.

Contents

## Steps

Here’s how you can easily draw the BeBr2 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, beryllium lies in group 2, and bromine lies in group 17.

Hence, beryllium has two valence electrons and bromine has seven valence electrons.

Since BeBr2 has one beryllium atom and two bromine atoms, so…

Valence electrons of one beryllium atom = 2 × 1 = 2
Valence electrons of two bromine atoms = 7 × 2 = 14

And the total valence electrons = 2 + 14 = 16

• Second, find the total electron pairs

We have a total of 16 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 = 16 ÷ 2 = 8

• Third, determine the central atom

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

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

Therefore, place beryllium in the center and bromines on either side.

• And finally, draw the rough sketch

### #2 Mention lone pairs on the atoms

Here, we have a total of 8 electron pairs. And two Be — Br bonds are already marked. So we have to only mark the remaining six electron pairs as lone pairs on the sketch.

Also remember that beryllium is a period 2 element, so it can not keep more than 8 electrons in its 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.

So for each bromine, there are three lone pairs, and for beryllium, there is zero lone pair because all six 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 beryllium atom, formal charge = 2 – 0 – ½ (4) = 0

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

Here, both beryllium and bromine atoms do not have charges, so no need to mark the charges.

In the above structure, you can see that the central atom (beryllium) doesn’t form an octet. But, beryllium has an exception that it does not require eight electrons to form an octet. So no need to worry about the octet rule here.

Therefore, this structure is the stable Lewis structure of BeBr2.

Next: CSe2 Lewis structure