The **Lewis structure of N _{2}H_{4}** contains five single bonds, with two nitrogens in the center, and each nitrogen is attached with two hydrogens. There is one lone pair on each nitrogen atom, and the hydrogen atom does not have any lone pair.

## Steps

By using the following steps, you can easily draw the Lewis structure of N_{2}H_{4}:

#1 Draw skeleton

#2 Show chemical bond

#3 Mark lone pairs

#4 Calculate formal charge and check stability (if octet is already completed on central atom)

Letâ€™s one by one discuss each step in detail.

### #1 Draw skeleton

In this step, first calculate the total number of valence electrons. And then, decide the central atom.

- Letâ€™s calculate the total number of valence electrons

We know thatâ€¦ nitrogen is a group 15 element and hydrogen is a group 1 element. Hence, nitrogen has **five** valence electrons and hydrogen has **one** valence electron.

Now N_{2}H_{4} has two nitrogen atoms and four hydrogen atoms.

So the total number of valence electrons = (valence electrons of nitrogen atom Ã— 2) + (valence electrons of hydrogen atom Ã— 4)

Therefore, the **total number of valence electrons** = 10 + 4 = 14

- Now decide the central atom

We can not assume hydrogen as 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.

Therefore, choose nitrogen as the central atom.

But there are two nitrogen atoms in N_{2}H_{4}. Hence, we can assume any one as the central atom. Letâ€™s assume that **left nitrogen is the central atom**.

So now, put two nitrogens in the center and four hydrogens on either side. And draw the rough skeleton structure for the Lewis structure of N_{2}H_{4} something like this:

**Also read:** How to draw Lewis structure of HBr (4 steps)

### #2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since two nitrogens are surrounded by four hydrogens, use ten electrons to show **five chemical bonds** as follows:

**Also read:** How to draw Lewis structure of H_{2}SO_{4} (5 steps)

### #3 Mark lone pairs

As calculated earlier, we have a total of 14 valence electrons. And in the above structure, we have already used ten valence electrons. Hence, four valence electrons are remaining.

Two valence electrons represent one lone pair. So four valence electrons = **two lone pairs**.

Note that nitrogen is a period 2 element, so it can not keep more than 8 electrons in its last shell. And hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell.

Also, make sure that you start marking these lone pairs on outside atoms first. And then, on the central atom.

The outside atoms are hydrogens and right nitrogen. But hydrogen can not keep more than 2 electrons in its last shell. Hence, donâ€™t mark the lone pairs on hydrogen.

So right nitrogen will get one lone pair. And the central atom (left nitrogen) will also get one lone pair.

Now draw the Lewis structure of N_{2}H_{4} something like this:

In the above structure, you can see that the octet is completed on the central atom (left nitrogen), and also on the outside atoms. Therefore, the octet rule is satisfied.

After completing the octet, one last thing we need to do is, calculate the formal charge and check the stability of the above structure.

**Also read:** How to draw Lewis structure of OH^{–} (4 steps)

### #4 Calculate formal charge and check stability

The following formula is used to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – Â½ bonding electrons

Collect the data from the above structure and then, write it down below as follows:

- For
**each nitrogen**atom

Valence electrons = 5

Nonbonding electrons = 2

Bonding electrons = 6

Formal charge = 5 – 2 – Â½ (6) = 0

- For
**each hydrogen**atom

Valence electrons = 1

Nonbonding electrons = 0

Bonding electrons = 2

Formal charge = 1 – 0 – Â½ (2) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of N_{2}H_{4} looks something like this:

In the above structure, you can see that the formal charges of both (nitrogen and hydrogen) are zero. Therefore, this is the **stable Lewis structure of N**_{2}**H**** _{4}**.

And each horizontal line drawn in the above structure represents a pair of bonding valence electrons.

## Related

- Lewis structure of HBr
- Lewis structure of H
_{2}SO_{4} - Lewis structure of OH
^{–} - Lewis structure of SiH
_{4} - Lewis structure of N
_{2}H_{2}

## External video

- N2H4 Lewis Structure – How to Draw the Lewis Structure for N2H4 – Wayne Breslyn

## External links

- Chemical Bonding: N2H4 Lewis Structure – The Geoexchange
- N2H4 Lewis Structure, Geometry, Hybridization, and Polarity – Techiescientist
- N2H4 lewis structure, molecular geometry, polarity, hybridization, angle – Topblogtenz
- What is the Lewis structure for N2H4? – Homework.Study.com
- N2H4 Lewis Structure, Molecular Structure, Hybridization, Bond Angle, and Shape – Geometry of Molecules
- N2H4 Lewis Structure in 6 Steps (With Images) – Pediabay
- What is the Lewis dot structure for N2H4? How is it made? – Quora
- N2H4 Lewis Structure (Dinitrogen Tetrahydride) – Pinterest
- N2H4 Lewis Structure – Wyzant
- Drawing the Lewis Structure for N2H4 – Pearson
- Hydrazine Lewis Structure – The Student Doctor Network
- Lewis Dot of Hydrazine N2H4 – Kent’s Chemistry
- Draw the Lewis structures of N2H4, N2H2, and N2 – Chegg
- Draw the Lewis structures of N2H4, N2H2, and N2 – Brainly

Deep

Rootmemory.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.