Hydrazine (N2H4) lewis dot structure, molecular geometry, polarity, hybridization
Hydrazine is highly toxic composed of two nitrogen and four hydrogens having the chemical formula N2H4. It has an odor similar to ammonia and appears colorless. It is also known as nitrogen hydride or diazane. It is corrosive to tissue and used in various rocket fuels.
In this article, we will discuss N2H4 lewis structure, molecular geometry, hybridization, bond angle, polarity, etc.
Hydrazine is toxic by inhalation and by skin absorption. Long-term exposure to hydrazine can cause burning, nausea, shortness of breath, dizziness, and many more health-related problems.
Properties of Hydrazine
- It is miscible in water.
- It appears as a colorless and oily liquid.
- It has a boiling point of 114 °C and a melting point of 2 °C.
- It is a strong base and has a conjugate acid(Hydrazinium).
- The dipole moment of N2H4 is 1.85 D.
|Name of Molecule
|Molecular geometry of N2H4
|Electron geometry of N2H4
|107° – 109.5º
|Total Valence electron in N2H4 lewis structure
How to draw N2H4 lewis structure?
N2H4 lewis structure is made up of two nitrogen (N) and four hydrogens (H) having two lone pairs on the nitrogen atoms(one lone pair on each nitrogen) and containing a total of 10 shared electrons.
There are three types of bonds present in the N2H4 lewis structure, one N-N, and two H-N-H.
Let’s start the construction of the lewis structure of N2H4 step by step-
So, here we go!
Follow some steps for drawing the Lewis dot structure of N2H4
1. Count total valence electron in N2H4
As we know, lewis’s structure is a representation of the valence electron in a molecule. So, in the first step, we have to count how many valence electrons are available for N2H4.
I assume that you definitely know how to find the valence electron of an atom. The valence electron of an atom is equal to the periodic group number of that atom.
So, nitrogen belongs to the 15th periodic group, and hydrogen to the 1st group. Therefore, the valence electron for nitrogen is 5 and for hydrogen, it is 1.
⇒ Total number of the valence electron in Nitrogen = 5
⇒ Total number of the valence electrons in hydrogen = 1
∴ Total number of valence electron available for the N2H4 lewis structure = 5(2) + 1(4) = 14 valence electrons [∴two nitrogen and four hydrogen]
2. Find the least electronegative atom and placed it at center
Now it’s time to find the central atom of the N2H4 molecule. It doesn’t matter which atom is more or less electronegative, if hydrogen atoms are there in a molecule then it always goes outside in the lewis diagram.
Because hydrogen only needs two-electron or one single bond to complete the outer shell. So, for N2H4, put away hydrogen outside and nitrogen as a central atom in the lewis diagram.
3. Connect outer atoms to central atom with a single bond
In this step, we need to connect every outer atom(hydrogen) to the central atom(nitrogen) with the help of a single bond. And make sure you must connect both nitrogens with a single bond also.
Now count the total number of valence electrons we used till now in the above structure. A single bond contains two-electron and as we see in the above structure, 5 single bonds are used, hence we used 10 valence electrons till now.
We had 14 total valence electrons available for drawing the N2H4 lewis structure and from them, we used 10 valence electrons.
∴ (14 – 10) = 4 valence electrons
So, we are left with 4 valence electrons more.
4. Place remaining valence electrons starting from outer atom first
Now we have to place the remaining valence electron around the outer atom first, in order to complete their octet. Lewis’s structure is all about the octet rule.
Octet rule said that each elements tend to bond in such a way that each atom has eight electrons in its valence shell
Always remember, hydrogen is an exception to the octet rule as it needs only two electrons to complete the outer shell. As hydrogen has only one shell and in one shell, there can be only two electrons.
So, as you see in the 3rd step structure, all hydrogen atoms complete their octet as they already share two electrons with the help of a single bond.
Now head over to the next step-
5. Complete central atom octet and make covalent bond if necessary
As hydrogen atom already completed their octet, we have to look at the central atom(nitrogen) in order to complete its octet. Nitrogen needs 8 electrons in its outer shell to gain stability, hence achieving octet.
If you look at the structure in the 3rd step, each nitrogen has three single bonds around it. So, each nitrogen already shares 6 valence electrons(3 single bonds).
In order to complete the octet, we need two more electrons for each nitrogen. We have already 4 leftover valence electrons in our account.
So, put two and two on each nitrogen. Hurry up!
N2H4 lewis dot structure
Yes, we completed the octet of both atoms(nitrogen and hydrogen) and also used all available valence electrons.
I think we completed the lewis dot structure of N2H4? No, we need one more step to verify the stability of the above structure with the help of the formal charge concept.
6. Check the stability with the help of a formal charge concept
A formal charge is the charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms.
The structure with the formal charge close to zero or zero is the best and most stable lewis structure.
To calculate the formal charge on an atom. Use the formula given below-
⇒ Formal charge = (valence electrons – lone pair electrons – 1/2shared pair electrons)
We will calculate the formal charge on the individual atoms of the N2H4 lewis structure.
For nitrogen atom:
⇒ Valence electrons of nitrogen = 5
⇒ Lone pair electrons on nitrogen = 2
⇒ Shared pair electrons(3 single bond) = 6
∴ (5 – 2 – 6/2) = 0 formal charge on the nitrogen atom
For hydrogen atom
⇒ Valence electrons of hydrogen = 1
⇒ Lone pair on hydrogen = 0
⇒ Shared pair electrons(one single bond) = 2
∴ (1 – 0 – 2/2) = o formal charge on the hydrogen atom
Hence, the overall formal charge in the N2H4 lewis structure is zero.
Therefore, we got our best lewis diagram.
Also check –
What is the molecular geometry of N2H4?
The molecular geometry of N2H4 is trigonal pyramidal. Each nitrogen(left side or right side) has two hydrogen atoms. The lone pair electron present on nitrogen and shared pair electrons(around nitrogen) will repel each other. As a result, they will be pushed apart giving the trigonal pyramidal geometry on each nitrogen side.
Since both nitrogen sides are symmetrical in the N2H4 structure, hence there shape will also be the same.
This is the only overview of the N2H4 molecular geometry. Now we will learn, How to determine the shape of N2H4 through its lewis diagram?
We will use the AXN method to determine the geometry. Generally, AXN is the representation of electron pairs(Bond pairs + Lone pairs) around a central atom, and after that by applying the VSEPR theory, we will predict the shape of the geometry of the molecule.
That’s how the AXN notation follows as shown in the above picture.
Now we have to find the molecular geometry of N2H4 by using this method.
AXN notation for N2H4 molecule:
- A represents the central atom, so as per the N2H4 lewis structure, nitrogen is the central atom. A = Nitrogen
- X represents the bonded atoms, as we know, nitrogen is making three bonds(two with hydrogen and one with nitrogen also). Therefore, X = 3
- N represents the lone pair, nitrogen atom has one lone pair on it. Hence, N = 1
So, the AXN notation for the N2H4 molecule becomes AX3N1.
As per the VSEPR theory and its chart, if a molecule central atom is attached with three bonded atoms and has one lone pair then the molecular geometry of that molecule is trigonal pyramidal.
Hence, the molecular shape or geometry for N2H4 is trigonal pyramidal.
N2H4 molecular geometry
As you see in the molecular shape of N2H4, on the left side, nitrogen is attached to the two hydrogen atoms and both are below of plane of rotation and on the right side, one hydrogen is above and one is below in the plane.
The electron geometry for N2H4 is tetrahedral.
Hybridization of N2H4
To find the hybridization of an atom, we have to first determine its hybridization number.
Hybridization number is the addition of a total number of bonded atoms around a central atom and the lone pair present on it.
∴ Hybridization number of N2H4 = (Number of bonded atoms attached to nitrogen + Lone pair on nitrogen)
According to the N2H4 lewis dot structure, we have three bonded atoms attached to the nitrogen and one lone pair present on it.
∴ Hybridization number of N2H4 = (3 + 1) = 4
So, for a hybridization number of four, we get the Sp3 hybridization on each nitrogen atom in the N2H4 molecule.
The bond angle of N2H4
“A bond angle is the geometrical angle between two adjacent bonds”.
The bond angle of N2H4 is subtended by H-N-H and N-N-H will be between 107° – 109°. Normally, atoms that have Sp3 hybridization hold a bond angle of 109.5°. Since one lone pair is present on the nitrogen atom in N2H4, lower the bond angle to some extent.
Also check:- How to find bond angle?
Hydrazine polarity: is N2H4 polar or nonpolar
We know, there is one lone pair on each nitrogen in the N2H4 molecule, both nitrogens is Sp3 hybridized. One hybrid of each orbital forms an N-N bond. But the bond N-N is non-polar because of the same electronegativity and the N-H bond is polar because of the slight difference between the electronegativity of nitrogen and hydrogen.
As you see the molecular geometry of N2H4, on the left side and right side, there is the total number of four N-H bonds present. Three hydrogens are below their respective nitrogen and one is above.
Also, the presence of lone pair on each nitrogen distorted the shape of the molecule since the lone pair tries to repel with bonded pair. So, there is no point that they will cancel the dipole moment generated along with the bond.
As both sides in the N2H4 structure seem symmetrical to different planes i.e. left side symmetric to the vertical plane(both hydrogen below) and the right side symmetric to the horizontal plane(one hydrogen is below and one is above).
So, the resultant of four N-H bond moments and two lone electron pairs leads to the dipole moment of 1.85 D. hence, N2H4 is a polar molecule.
Shared pair electrons are also called the bonded pair electrons as they make the covalent between two atoms and share the electrons. Lone pair electrons are unshared electrons means they don’t take part in chemical bonding.
Shared pair electrons in N2H4 molecule = a total of 10 shared pair electrons(5 single bonds) are present in N2H4 molecule.
Lone pair electrons in N2H4 molecule = Both nitrogen central atom contains two lone pair. So, the lone pair of electrons in N2H4 equals, 2 ×(2) = 4 unshared electrons.
As you closely see the N2H4 lewis structure, hydrogen can occupy only two electrons in its outer shell, which means hydrogen can share only two electrons. So, there is no point we can use a double bond with hydrogen since a double bond contains a total of 4 electrons.
Now let’s talk about the N-N bond, each nitrogen has three single bonds and one lone pair. If we convert the lone pair into a covalent bond then nitrogen shares four bonds(two single and one double bond).
But the problem is if a double bond is present in the N2H4 dot structure, then it becomes unstable. As nitrogen atoms will get some formal charge.
“Lewis structure is most stable when the formal charge is close to zero”.
That’s why there is no need to make any double or triple bond as we already got our best and most stable N2H4 lewis structure with zero formal charges.
- NH2OH lewis structure and its molecular geometry
- C2H4 Lewis structure and its molecular geometry
- NH2Cl lewis structure and its molecular geometry
- CO2 Lewis structure and its molecular geometry
- NH3 Lewis structure and its molecular geometry
- The total valence electron available for the N2H4 lewis structure is 14.
- The hybridization of each nitrogen in the N2H4 molecule is Sp3.
- N2H4 is polar in nature and dipole moment of 1.85 D.
- The formal charge on nitrogen in N2H4 is zero.
- The molecular geometry or shape of N2H4 is trigonal pyramidal.
- Total 2 lone pairs and 5 bonded pairs are present in the N2H4 lewis dot structure.
- The electron geometry of N2H4 is tetrahedral.
About the author
Vishal Goyal is the founder of Topblogtenz, a comprehensive resource for students seeking guidance and support in their chemistry studies. He holds a degree in B.Tech (Chemical Engineering) and has four years of experience as a chemistry tutor. The team at Topblogtenz includes experts like experienced researchers, professors, and educators, with the goal of making complex subjects like chemistry accessible and understandable for all. A passion for sharing knowledge and a love for chemistry and science drives the team behind the website. Let’s connect through LinkedIn: https://www.linkedin.com/in/vishal-goyal-2926a122b/