Diazene (N2H2) Lewis dot structure, molecular geometry or
shape, electron geometry,
hybridization, polar or nonpolar, bond angle
N2H2 is the chemical formula for diazene, also known as diimide. It is a yellowish color gas that is popularly used as a reagent in organic synthesis.
In this article, we have talked about some important concepts related to the chemistry of diazene including information about its Lewis structure, molecular geometry or shape, electron geometry, bond angle, hybridization, and any formal charges present on N2H2 atoms.
To gain all this insightful information and much more, continue reading!
The Lewis structure of N2H2 consists of two nitrogen (N) atoms at the center which are bonded to two atoms of hydrogen (H), one on each side. The N-atoms are bonded to each other via a double covalent bond. There is a lone pair of electrons present on each N-atom in the N2H2 lewis dot structure.
Drawing the Lewis structure of N2H2 can be a bit tricky sometimes. However, to make things seem easier, we have split the task into the following simple steps that can you follow one by one and draw the Lewis structure of diazene with us. So, let’s start.
Steps for drawing the Lewis dot structure of N2H2
1. Count the total valence electrons in N2H2
The Lewis dot structure of a molecule is referred to as a simplified representation of all the valence electrons present in it. Therefore, the very first step while drawing the Lewis structure of N2H2 is to count the total valence electrons present in the concerned elemental atoms.
Nitrogen (N) is situated in Group VA of the Periodic Table of elements, so it has a total of 5 valence electrons. On the other hand, hydrogen (H) is present at the top of the Periodic Table with a single valence electron only.
∴ The diazene (N2H2) molecule consists of two atoms of nitrogen and two hydrogen atoms. Hence, the total valence electrons available for drawing the Lewis dot structure of N2H2 are 2(5) + 2(1) = 12 valence electrons.
2. Choose the central atom
The second step while drawing the Lewis structure of a molecule is to determine the least electronegative atom and place it at the center. All the other atoms are spread in its surroundings. This is because the least electronegative is most likely to share its electrons with the atoms in its surroundings.
But here is an exception. Although hydrogen (H) is less electronegative than nitrogen (N) but it cannot be selected as the central atom in any Lewis structure.
This is because H can only form a single bond with its adjacent atom, as it has a maximum capacity of accommodating 2 valence electrons only.
Keeping this fact in mind, the nitrogen (N) atom is chosen as the central atom in the Lewis diagram of N2H2. As there are 2 identical N atoms in N2H2 therefore to keep things fair, both the N atoms are placed at the center. The H atoms are placed as outer atoms at the terminals of the molecule, as shown in the figure below.
3. Connect outer atoms with the central atom
In this step, the outer H atoms are joined to the central N atoms using single straight lines. Both the central N atoms are also joined to each other using straight lines, as shown below.
Each straight line represents a single covalent bond i.e., a bond pair containing 2 electrons.
Now if we count the total valence electrons used so far out of the 12 available, there are a total of 3 single bonds in the structure above. Thus, 4(2) = 8 valence electrons are used till step 3.
Total valence electrons available – electrons used till step 3 = 12-6 = 6 valence electrons.
This means we still have 6 valence electrons to be accommodated in the Lewis dot structure of N2H2.
4. Complete the duplet and/or octet of the outer atoms
As already noted, the two hydrogens (H) atoms are the outer atoms in the Lewis structure of N2H2. Each H atom requires a total of 2 valence electrons only in order to achieve a stable duplet electronic configuration.
Each H atom is bonded to a central N atom via a single bond which means it already has a complete duplet. So, we need not worry about the H atoms. Consequently, there is no lone pair on any H atom.
Total valence electrons available – electrons used till step 4 = 12- 6 = 6 valence electrons .
There are still 6 electrons which means 6/2 = 3 lone pairs of electrons are available.
Each N atom requires a total of 8 valence electrons to complete its octet shell. If one of the two N-atoms is considered an outer atom at this stage the situation looks as shown in the figure below.
This ‘outer’ N atom is bonded to the central N via a single bond and also to an H atom via another single bond. It means it has a total of 4 valence electrons. It needs 4 more electrons to complete its octet. So, these 4 electrons are placed as 2 lone pairs around this outer N atom.
5. Place the remaining valence electrons as a lone pair on the central atom
Electrons used till step 5 = 3 single bonds + electrons placed around an N atom, shown as dots = 3(2) + 4 = 10 valence electrons.
Total valence electrons available – electrons used till step 5 = 12-10 = 2 valence electrons.
Thus, these 2 valence electrons are placed as a lone pair on the central N atom.
6. Complete the octet of the central N atom and make a covalent bond if necessary
All the 12 valence electrons are now used but the central N atom has a total of 6 valence electrons only. The problem is that it still needs 2 more electrons to complete its octet. An easy solution to this problem is to convert a lone pair present on the nitrogen adjacent to the central N atom into a covalent bond.
Consequently, the 2 N atoms have a double covalent bond between each other, a single bond, and 1 lone pair is present on each. Both the N atoms now have a complete octet, and the duplet of each H atom is also complete.
The final step is to check the stability of this Lewis structure. We can do that by using the formal charge concept.
7. Check the stability of the N2H2 Lewis structure using the formal charge concept
The less the formal charge on the atoms of a molecule, the better the stability of its Lewis structure.
The formal charge can be calculated using the formula given below.
Now let us use this formula and the Lewis structure of diazene (N2H2) obtained in step 6 to calculate the formal charges present on the bonded N and H atoms.
For nitrogen atom
Valence electrons of nitrogen = 5
Bonding electrons = 1 single bond + 1 double bond = 2 + 4 = 6 electrons
Non-bonding electrons =1 lone pair = 2 electrons
Formal charge = 5-2-6/2 = 5-2-3 = 5-5 = 0
For hydrogen atom
Valence electrons of hydrogen = 1
Bonding electrons = 1 single bond = 2 electrons
Non-bonding electrons = no lone pair = 0 electrons
Formal charge = 1-0-2/2 = 1-0-1 = 1-1 = 0
Zero formal charges present on all the N and H atoms in the N2H2 molecule ensure the stability of its Lewis structure and that we have drawn it correctly.
What are the electron and molecular geometry of N2H2?
The ideal electron geometry of the diazene (N2H2) molecule is trigonal planar while its molecular geometry or shape is bent. While considering the shape and geometry of N2H2 any one N atom (left or right) can be considered a central atom.
The shape and geometry will stay the same in either case because both the N atoms are bonded to 2 identical chemical entities i.e., an H and an N-H at the sides. Both the N atoms also have a lone pair that distorts the symmetry of the molecule and makes it occupy a different shape from the ideal electron pair geometry.
Molecular geometry of N2H2
N2H2 has a bent shape and molecular geometry. The bent shape is also known as angular or V-shape. Consider the figure below and try to understand this concept according to the left N atom as the central atom. This central N atom has a lone pair of electrons.
Therefore, lone pair-bond pair repulsions exist in addition to bond pair-bond pair repulsions. The electron repulsive effect distorts the geometry of the molecule and makes it attain a bent or V-shape. The central N atom lies at the center of the inverted V while the other N-H and H atoms occupy the terminal positions.
An important point to remember is that the molecular geometry or shape of a molecule depends on the distinction between the bond pairs and lone pairs of electrons present on the central atom in the molecule. Contrarily, the ideal electron geometry depends on the total number of electron density regions around the central atom in the molecule.
In N2H2, the N-H bond next to the central N atom is considered one region of electron density. The central N atom having an N-H bond on one side, an H atom on the other side and a lone pair at the top has a total of 3 electron density regions.
Electron geometry of N2H2
According to the valence shell electron pair repulsion (VSEPR) theory of chemical bonding, the ideal electron geometry of a molecule having a total of 3 electron density regions around the central atom is trigonal planar. Thus, the electron geometry of N2H2 is trigonal planar.
An easy way to find the shape and geometry of the molecule is to use the AXN method.
AXN is a simple formula to represent the number of atoms bonded to the central atom in a molecule and the number of lone pairs present on it.
It is used to predict the geometry or shape of a molecule using the VSEPR concept.
AXN notation for the N2H2 molecule is :
A in the AXN formula represents the central atom of the molecule. Nitrogen (N) acts as a central atom in N2H2 so A=N.
X represents the bond pairs around the central atom. In N2H2 there are 2 bond pairs around the central N atom so X=2.
N denotes the number of lone pairs present on the central atom. As per the Lewis structure of N2H2, there is 1 lone pair on the central N atom so N=1.
Thus, the AXN generic formula for the N2H2 molecule is AX2N1.
Now, you can use the VSEPR chart given below and identify the electron and molecular geometries assigned against AX2N1.
The VSEPR chart clearly indicates that the electron geometry of a molecule having AX2N1 generic formula is trigonal planar while its molecular geometry is bent or V-shaped, as we already noted down for the N2H2 molecule.
Hybridization of N2H2
Both the N atoms lying at the center of the N2H2 molecule are sp2 hybridized.
The electronic configuration of nitrogen (N) is 1s2 2s2 2p3.
During chemical bond formation, the 2s atomic orbital of nitrogen hybridizes with two of the three half-filled 2p orbitals to yield three sp2 hybrid orbitals. One of the three sp2 hybrid orbitals contains paired electrons which are situated as a lone pair on the N atom.
The remaining two sp2 hybrid orbitals containing 1 electron each overlap with the sp2 hybrid orbital of the other nitrogen atom on one side and with the s orbital of hydrogen on the other side.
This leads to the formation of N-N and N-H sigma (σ) bonds respectively. The unhybridized p orbitals of each N atom form the required pi (π) bond in N=N, as shown in the figure below.
A short trick for finding the hybridization present in a molecule is to use its steric number against the table given below. The steric number of central N atom in N2H2 is 3 (1 single bond +1 double bond + 1 lone pair) so N2H2 has sp2 hybridization.
Steric number
Hybridization
2
sp
3
sp2
4
sp3
5
sp3d
6
sp3d2
The N2H2 bond angle
The ideal bond angle in a trigonal planar molecule is 120°. But it is due to the presence of a lone pair on the central N atom in N2H2 that lone pair-bond pair repulsions exist in the molecule which distorts the geometry of the molecule and makes it occupy a bent shape.
The lone pair-lone pair repulsions between the lone pairs on the two N atoms lying at the center also contribute to this distortion. Consequently, the H-N=N bond angle decreases and becomes approx. 109.5°. The N=N bond length is 125 pm while the N-H bond lengths are close to 100 pm in the N2H2 molecule.
Considering how fascinating chemistry is as a subject, the polarity of diazene (N2H2) is also an interesting concept. Whether the N2H2 molecule is polar or not, depends upon its cis-trans conformation.
Cis and trans-diazene are the two possible geometric isomers of diazene. Geometric isomers are chemical compounds having the same molecular formula and functional groups or atoms, but the spatial arrangement of these atoms differs.
The central N=N bond in the N2H2 molecule is non-polar as no electronegativity difference exists between the identical nitrogen atoms. However, both the N-H bonds are polar owing to an electronegativity difference of 3.04 -2.20 = 0.84 units between the bonded nitrogen (E.N = 3.04) and hydrogen (E.N =2.20) atoms.
In cis-diazene, the molecule is arranged such that the dipole moments of individually polar N-H bonds add up. Hence the N2H2 molecule is polar (net μ > 0) in its cis-form.
Contrarily, in trans-diazene, a horizontal axis of symmetry exists, one H atom lies above the symmetrical plane while the other H-atom lies below the symmetrical plane. As a result, the dipole moments of individual N-H bonds get canceled in opposite directions thus the N2H2 molecule is non-polar (net μ = 0) in its trans-form.
The Lewis structure of diazene (N2H2) shows a total of 4 atoms i.e., 2 nitrogen (N) atoms and 2 hydrogens (H) atoms. There are a total of 12 valence electrons in this Lewis structure i.e., 12/2 = 6 electron pairs.
Out of these 6 electron pairs, there are 4 bond pairs and 2 lone pairs. One lone pair is present on each N-atom at the center of the molecule.
How are the shapes and geometry of N2H2 and N2H4 different?
In N2H2, 2 hydrogen atoms are bonded to the central N atoms, one on each side while in N2H4, a total of 4 hydrogen atoms are bonded to the central N atoms, two on each side.
The central N-atoms are bonded to each other via a double covalent bond in N2H2 in contrast to a single bond present between the central N-atoms in N2H4.
However, there is one lone pair on each N atom at the center in both N2H2 and N2H4
Considering these differences, the AXN notation for the N2H2 molecule is AX2N1 thus according to the VSEPR concept, it has a bent shape and molecular geometry while its ideal electron geometry is trigonal planar.
The AXN notation for the N2H4 molecule is AX3N1 consequently it has a trigonal pyramidal shape and molecular geometry while its electron geometry is tetrahedral.
The total valence electrons available for drawing diazene (N2H2) Lewis structure are 12.
The N2H2 molecule has a bent shape and molecular geometry.
The ideal electron pair geometry of the N2H2 molecule with respect to any of the two central N-atoms is trigonal planar.
It is due to the repulsive effect of the lone pair of electrons present in the N2H2 molecule that it adopts a different shape from its ideal electronic geometry and the bond angle decreases from the ideal 120° to 109.5°.
N2H2 has sp2
Cis-diazene is a polar molecule while trans-diazene is non-polar in nature due to the symmetry present in the trans conformation.
The absence of any formal charge on the bonded nitrogen and hydrogen atoms in the N2H2 molecule marks the extraordinary stability of its Lewis structure.
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/
Topblogtenz is a website dedicated to providing informative and engaging content related to the field of chemistry and science. We aim to make complex subjects, like chemistry, approachable and enjoyable for everyone.
