Lewis structure of N2H4

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Lewis structure of N2H4
Lewis structure of N2H4 | Image: Root Memory

The Lewis structure of N2H4 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 N2H4:

#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)

Let’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.

  • Let’s calculate the total number of valence electrons

We know that… 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 N2H4 has two nitrogen atoms and four hydrogen atoms.

So the total number of valence electrons = (valence electrons of nitrogen atom × 2) + (valence electrons of hydrogen atom × 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 N2H4. Hence, we can assume any one as the central atom. Let’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 N2H4 something like this:

Skeleton structure for Lewis structure of N2H4 | Image: Root Memory

#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:

Five chemical bonds shown between atoms | Image: Root Memory

#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, don’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 N2H4 something like this:

Lone pairs marked on Lewis structure of N2H4 | Image: Root Memory

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.

#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 – ½ 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 – ½ (6) = 0

  • For each hydrogen atom

Valence electrons = 1
Nonbonding electrons = 0
Bonding electrons = 2

Formal charge = 1 – 0 – ½ (2) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of N2H4 looks something like this:

Formal charges are calculated, and got the stable Lewis structure of N2H4 | Image: Root Memory

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 N2H4.

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

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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.

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