# Lewis Structure of N2H4 (With 6 Simple Steps to Draw!)

I’m super excited to teach you the lewis structure of N2H4 in just 6 simple steps.

Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of N2H4 molecule.

So, if you are ready to go with these 6 simple steps, then let’s dive right into it!

Lewis structure of N2H4 contains a single bond between the two Nitrogen (N) atoms as well as between Nitrogen (N) & Hydrogen (H) atoms. Both the nitrogen atoms have one lone pair on them.

Let’s draw and understand this lewis dot structure step by step.

(Note: Take a pen and paper with you and try to draw this lewis structure along with me. I am sure you will definitely learn how to draw lewis structure of N2H4).

## 6 Steps to Draw the Lewis Structure of N2H4

### Step #1: Calculate the total number of valence electrons

Here, the given molecule is N2H4. In order to draw the lewis structure of N2H4, first of all you have to find the total number of valence electrons present in the N2H4 molecule.
(Valence electrons are the number of electrons present in the outermost shell of an atom).

So, let’s calculate this first.

Calculation of valence electrons in N2H4

• For Nitrogen:

Nitrogen is a group 15 element on the periodic table.

Hence, the valence electrons present in nitrogen is 5 (see below image).

• For Hydrogen:

Hydrogen is a group 1 element on the periodic table.

Hence, the valence electron present in hydrogen is 1 (see below image).

Hence in a N2H4 molecule,

Valence electrons given by each Nitrogen (N) atom = 5
Valence electron given by each Hydrogen (H) atom = 1
So, total number of Valence electrons in N2H4 molecule = 5(2) + 1(4) = 14

### Step #2: Select the center atom (H is always outside)

While selecting the center atom, always put the least electronegative atom at the center.

(Remember: Fluorine is the most electronegative element on the periodic table and the electronegativity decreases as we move right to left in the periodic table as well as top to bottom in the periodic table).

Here in the N2H4 molecule, if we compare the nitrogen atom (N) and hydrogen atom (H), then hydrogen is less electronegative than nitrogen. But as per the rule, we have to keep hydrogen outside.

So, both the nitrogen atoms should be placed in the center and the remaining 4 hydrogen atoms will surround it.

### Step #3: Put two electrons between the atoms to represent a chemical bond

Now in the above sketch of N2H4 molecule, put the two electrons (i.e electron pair) between the nitrogen-nitrogen atoms and nitrogen-hydrogen atoms to represent a chemical bond between them.

These pairs of electrons present between the Nitrogen atoms as well as between the Nitrogen and Hydrogen atoms form a chemical bond, which bonds these atoms with each other in a N2H4 molecule.

### Step #4: Complete the octet (or duplet) on outside atoms. If the valence electrons are left, then put the valence electrons pair on the central atom

Don’t worry, I’ll explain!

In the Lewis structure of N2H4, the outer atoms are hydrogen atoms.

So now, you have to check whether these hydrogen atoms are forming a duplet or not! (because hydrogen requires only 2 electrons to have a complete outer shell).

You can see in the above image that all the hydrogen atoms form a duplet.

Also, only 10 valence electrons of N2H4 molecule are used in the above structure.

But there are total 14 valence electrons in N2H4 molecule (as calculated in step #1).

So the number of electrons left to be kept on the central nitrogen atoms = 14 – 10 = 4.

So let’s keep these four electrons (i.e 2 electron pairs) on the central nitrogen atoms.

Now, let’s move to the next step.

### Step #5: Check whether the central atom has octet or not

In this step, we have to check whether the central atom (i.e nitrogen atoms) has an octet or not.

In simple words, we have to check whether the central Nitrogen (N) atoms are having 8 electrons or not.

As you can see from the above image, both the nitrogen atoms have 8 electrons. So it fulfills the octet rule and the nitrogen atoms are stable.

### Step #6: Final step – Check the stability of lewis structure by calculating the formal charge on each atom

Now, you have come to the final step and here you have to check the formal charge on nitrogen atoms (N) as well as hydrogen atoms (H).

For that, you need to remember the formula of formal charge;

Formal charge = Valence electrons – Nonbonding electrons – (Bonding electrons)/2

• For Nitrogen:
Valence electrons = 5 (as it is in group 15)
Nonbonding electrons = 2
Bonding electrons = 6
• For Hydrogen:
Valence electron = 1 (as it is in group 1)
Nonbonding electrons = 0
Bonding electrons = 2

So you can see above that the formal charges on nitrogen as well as hydrogen are “zero”.

Hence, there will not be any change in the above structure and the above lewis structure of N2H4 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of N2H4 represents the single bond ( | ). So the above lewis dot structure of N2H4 can also be represented as shown below.

Related lewis structures for your practice:
Lewis structure of CH3NH2
Lewis structure of SiO2
Lewis structure of SiH4
Lewis structure of ClO4-
Lewis structure of ClO-