
N2F4 (tetrafluorohydrazine) has two nitrogen atoms and four fluorine atoms.
In N2F4 Lewis structure, there is a single bond between the two nitrogen atoms, and each nitrogen is attached with two fluorine atoms. Each fluorine atom has three lone pairs, and each nitrogen atom has one lone pair.
Alternative method: Lewis structure of N2F4
Rough sketch
- First, determine the total number of valence electrons

In the periodic table, nitrogen lies in group 15, and fluorine lies in group 17.
Hence, nitrogen has five valence electrons and fluorine has seven valence electrons.
Since N2F4 has two nitrogen atoms and four fluorine atoms, so…
Valence electrons of two nitrogen atoms = 5 × 2 = 10
Valence electrons of four fluorine atoms = 7 × 4 = 28
And the total valence electrons = 10 + 28 = 38
Learn how to find: Nitrogen valence electrons and Fluorine valence electrons
- Second, find the total electron pairs
We have a total of 38 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 = 38 ÷ 2 = 19
- Third, determine the central atom
We have to place the least electronegative atom at the center.
Since nitrogen is less electronegative than fluorine, assume that the central atom is nitrogen.
Here, there are two nitrogen atoms, so we can assume any one as the central atom.
Let’s assume that the central atom is left nitrogen.
Therefore, place nitrogens in the center and fluorines on either side.
- And finally, draw the rough sketch

Lone pair
Here, we have a total of 19 electron pairs. And five bonds are already marked. So we have to only mark the remaining fourteen electron pairs as lone pairs on the sketch.
Also remember that both (nitrogen and fluorine) are the period 2 elements, so they can not keep more than 8 electrons in their last shell.
Always start to mark the lone pairs from outside atoms. Here, the outside atoms are fluorines and right nitrogen.
So for each fluorine, there are three lone pairs, and for each nitrogen, there is one lone pair.
Mark the lone pairs on the sketch as follows:

Formal charge
Use the following formula to calculate the formal charges on atoms:
Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons
For each nitrogen atom, formal charge = 5 – 2 – ½ (6) = 0
For each fluorine atom, formal charge = 7 – 6 – ½ (2) = 0
Here, the atoms do not have charges, so no need to mark the charges.
Final structure
The final structure of N2F4 consists of two nitrogen atoms connected to each other by a single covalent bond, with each nitrogen also bonded to two fluorine atoms. In this arrangement, both nitrogen atoms satisfy the octet rule by forming three bonds and retaining one lone pair. Each fluorine atom likewise fulfills its octet by maintaining three lone pairs of its own alongside the single shared bond. This configuration is the most stable because it results in formal charges of zero for all atoms involved, representing the most energetically favorable state for the molecule. Therefore, this specific electronic distribution serves as the definitive and most accurate Lewis representation of tetrafluorohydrazine.
Next: Cl2O2 Lewis structure
External video
- Lewis Dot Structure and Molecular Geometry for N2F4: Tetrafluorohydrazine – YouTube • Wayne Breslyn
External links
- https://lambdageeks.com/n2f4-lewis-structure/
- https://homework.study.com/explanation/draw-and-explain-the-lewis-structure-for-n2f4.html
- https://www.chegg.com/homework-help/questions-and-answers/n2f4-draw-lewis-dot-structure-n2f4-include-lone-pairs-electrons-pl-hon-sp-f-q50859053
- https://www.bartleby.com/questions-and-answers/for-the-molecule-n2f4-1-draw-the-lewis-structure-then-identify-the-hybrid-used.-2-draw-the-orbital-d/6fd2c4a4-06c0-4154-9fa0-e8d6d0bb5b42
Deep
Learnool.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.