# NOBr Lewis structure

NOBr (nitrosyl bromide) has one nitrogen atom, one oxygen atom, and one bromine atom.

In NOBr Lewis structure, there is a double bond between nitrogen and oxygen atom, and a single bond between nitrogen and bromine atom. The oxygen atom has two lone pairs, the nitrogen atom has one lone pair, and the bromine atom has three lone pairs.

Contents

## Steps

Use these steps to correctly draw the NOBr 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
#4 Convert lone pairs of the atoms, and minimize formal charges
#5 Repeat step 4 if needed, until all charges are minimized, to get a stable Lewis structure

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, nitrogen lies in group 15, oxygen lies in group 16, and bromine lies in group 17.

Hence, nitrogen has five valence electrons, oxygen has six valence electrons, and bromine has seven valence electrons.

Since NOBr has one nitrogen atom, one oxygen atom, and one bromine atom, so…

Valence electrons of one nitrogen atom = 5 × 1 = 5
Valence electrons of one oxygen atom = 6 × 1 = 6
Valence electrons of one bromine atom = 7 × 1 = 7

And the total valence electrons = 5 + 6 + 7 = 18

• Second, find the total electron pairs

We have a total of 18 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 = 18 ÷ 2 = 9

• Third, determine the central atom

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

Since nitrogen is less electronegative than oxygen and bromine, assume that the central atom is nitrogen.

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

• And finally, draw the rough sketch

### #2 Mark lone pairs on the atoms

Here, we have a total of 9 electron pairs. And two bonds are already marked. So we have to only mark the remaining seven electron pairs as lone pairs on the sketch.

Also remember that both (nitrogen and oxygen) 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 oxygen and bromine.

So for oxygen and bromine, there are three lone pairs, and for nitrogen, there is one lone pair.

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 nitrogen atom, formal charge = 5 – 2 – ½ (4) = +1

For oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

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

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

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

### #4 Convert lone pairs of the atoms, and minimize formal charges

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

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

Also, the above structure is more stable than the previous structures. Therefore, this structure is the stable Lewis structure of NOBr.

Next: BH4 Lewis structure