# NH2OH Lewis structure

NH2OH or H3NO (hydroxylamine) has three hydrogen atoms, one nitrogen atom, and one oxygen atom.

In the NH2OH Lewis structure, there are two N — H bonds, one N — O bond, and one O — H bond. The nitrogen atom has one lone pair, and the oxygen atom has two lone pairs.

Contents

## Steps

To properly draw the NH2OH 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

In the periodic table, nitrogen lies in group 15, hydrogen lies in group 1, and oxygen lies in group 16.

Hence, nitrogen has five valence electrons, hydrogen has one valence electron, and oxygen has six valence electrons.

Since NH2OH has one nitrogen atom, three hydrogen atoms, and one oxygen atom, so…

Valence electrons of one nitrogen atom = 5 × 1 = 5
Valence electrons of three hydrogen atoms = 1 × 3 = 3
Valence electrons of one oxygen atom = 6 × 1 = 6

And the total valence electrons = 5 + 3 + 6 = 14

• Second, find the total electron pairs

We have a total of 14 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 = 14 ÷ 2 = 7

• Third, determine the central atom

Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.

Now we have to choose the central atom from nitrogen and oxygen. Place the least electronegative atom at the center.

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

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

• And finally, draw the rough sketch

### #2 Next, indicate lone pairs on the atoms

Here, we have a total of 7 electron pairs. And four bonds are already marked. So we have to only mark the remaining three 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 hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are hydrogens and oxygen. But no need to mark on hydrogen, because each hydrogen has already two electrons.

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

Mark the lone pairs on the sketch as follows:

### #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 nitrogen atom, formal charge = 5 – 2 – ½ (6) = 0

For each hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

For oxygen atom, formal charge = 6 – 4 – ½ (4) = 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 (nitrogen) forms an octet. The outside atom (oxygen) also forms an octet, and all hydrogens form a duet. Hence, the octet rule and duet rule are satisfied.

Therefore, this structure is the stable Lewis structure of NH2OH.