# BrCN Lewis structure

BrCN (cyanogen bromide) has one bromine atom, one carbon atom, and one nitrogen atom.

In the BrCN Lewis structure, there is a single bond between carbon and bromine atom, and a triple bond between carbon and nitrogen atom. The bromine atom has three lone pairs, and the nitrogen atom has one lone pair.

Contents

## Steps

Here’s how you can easily draw the BrCN 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
#4 Minimize formal charges by converting lone pairs of the atoms, and try to get a stable Lewis structure
#5 Repeat step 4 again if needed, until all charges are minimized

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, bromine lies in group 17, carbon lies in group 14, and nitrogen lies in group 15.

Hence, bromine has seven valence electrons, carbon has four valence electrons, and nitrogen has five valence electrons.

Since BrCN has one bromine atom, one carbon atom, and one nitrogen atom, so…

Valence electrons of one bromine atom = 7 × 1 = 7
Valence electrons of one carbon atom = 4 × 1 = 4
Valence electrons of one nitrogen atom = 5 × 1 = 5

And the total valence electrons = 7 + 4 + 5 = 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 carbon is less electronegative than bromine and nitrogen, assume that the central atom is carbon.

Therefore, place carbon in the center and bromine and nitrogen 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 bonds are already marked. So we have to only mark the remaining six 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 both (carbon and nitrogen) 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 bromine and nitrogen.

So for bromine and nitrogen, there are three lone pairs, and for carbon, 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 bromine atom, formal charge = 7 – 6 – ½ (2) = 0

For carbon atom, formal charge = 4 – 0 – ½ (4) = +2

For nitrogen atom, formal charge = 5 – 6 – ½ (2) = -2

Here, both carbon and nitrogen atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable Lewis structure because both carbon and nitrogen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

### #4 Minimize formal charges by converting lone pairs of the atoms

Convert a lone pair of the nitrogen atom to make a new C — N bond with the carbon atom as follows:

### #5 Since there are charges on atoms, repeat step 4 again

Since there are charges on carbon and nitrogen atoms, again convert a lone pair of the nitrogen atom to make a new C — N bond with the carbon atom as follows: