Dinitrogen tetroxide (N2O4) Lewis structure, molecular
geometry or shape, electron geometry, resonance structure, polar or nonpolar, hybridization
N2O4 is the chemical formula for dinitrogen tetroxide, also known as nitrogen tetraoxide. It exists as a reddish-brown liquid at r.t.p, containing two nitro (NO2) functional groups chemically bonded to each other.
The catalytic oxidation of ammonia produces N2O4, which is frequently used as a powerful oxidizer itself.
In this article, we will discuss how to draw the Lewis dot structure of N2O4, what is its molecular geometry or shape, electron geometry, bond angle, hybridization, formal charges, polarity and much more.
The Lewis dot structure of dinitrogen tetroxide (N2O4) consists of two nitrogen (N) atoms at the center, single covalently bonded to each other.
Both the N-atoms are bonded to two oxygen (O) atoms, each via single and double covalent bonds, respectively. Lone pairs of electrons are present on the terminal O-atoms; however, there is no lone pair on either of the two N-atoms.
You can easily learn to draw the Lewis structure of N2O4 by following the simple steps given below.
Steps for drawing the Lewis dot structure of N2O4
1. Count the total valence electrons present in N2O4
N2O4 consists of two distinct elements, i.e., nitrogen and oxygen.
Nitrogen (N) is present in Group V A (or 15) of the Periodic Table, containing 5 valence electrons in each atom. In contrast, oxygen (O) is present in Group VI A (or 16), possessing a total of 6 valence electrons.
Total number of valence electrons in nitrogen = 5
Total number of valence electrons in oxygen = 6
The N2O4 molecule comprises 2 N-atoms and 4 O-atoms.
∴ Therefore, the total valence electrons available for drawing the Lewis dot structure of N2O4 = 2(5) + 4(6) = 34 valence electrons.
2. Choose the central atom
By convention, the least electronegative atom out of all those available is chosen as the central atom while drawing the Lewis structure of a molecule.
The least electronegative atom can easily form covalent bonds with other atoms by sharing its electrons.
In N2O4, oxygen (E.N = 3.44) is more electronegative than nitrogen (E.N = 3.04), so the O-atoms cannot be selected as the central atoms.
Both the N-atoms are identical, so either of the two N-atoms can be considered a central atom in the N2O4 Lewis structure.
Conversely, the 4 O-atoms occupy terminal positions, with 2 O-atoms surrounding each nitrogen, as shown below.
3. Connect the outer atoms with the central atoms
In this step, each of the two O-atoms is joined to their adjacent N-atom via a single straight line. Also, the two N-atoms at the center are joined to each other.
A straight line represents a single covalent bond, i.e., a bond pair containing 2 electrons.
In the above structure, there are 5 single bonds, i.e., 5(2) = 10 valence electrons are already consumed out of the 34 initially available.
Now let’s see where we can place the remaining valence electrons.
4. Complete the octet of the outer atoms
An O-atom needs a total of 8 valence electrons in order to achieve a stable octet electronic configuration.
In the above Lewis structure, each O-atom is surrounded by only 2 valence electrons as per an N-O single bond. Therefore, to complete their octets, 3 lone pairs of electrons are placed around each O-atom, as shown below.
5. Complete the octet of the central atoms and convert lone pairs into covalent bonds if necessary
Total valence electrons used till step 4 = 5 single bonds + 4 (electrons placed around each O-atom, shown as dots) = 5(2) + 4(6) = 34 valence electrons.
Total valence electrons – electrons used till step 4 = 34 – 34 = 0 valence electrons.
As all the valence electrons initially available are already consumed so there is no lone pair on either of the N-atom at the center of the N2O4 Lewis structure. However, none of the two N-atoms have a complete octet in the structure drawn till yet.
Both the N-atoms are surrounded by a total of 6 valence electrons which hints at a deficiency of 2 electrons.
To overcome this deficiency, a lone pair from an O-atom adjacent to the N-atom is converted into an additional covalent bond. The same process is repeated for the other nitrogen atom.
In this way, both the central N-atom and all the terminal O-atoms obtain completely stable octet electronic configurations in N2O4.
As a final step, we just need to check the stability of the Lewis structure drawn above by applying the formal charge concept.
6. Check the stability of Lewis’s 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 charges can be calculated using the formula given below.
Formal charge = [valence electrons- nonbonding electrons- ½ (bonding electrons)].
Now let us use this formula and the Lewis structure obtained in step 5 to determine the formal charges present on the N2O4 bonded atoms.
For each nitrogen atom
Valence electrons of nitrogen = 5
Bonding electrons = 2 single bonds + 1 double bond = 2(2) + 4 = 8 electrons
Non-bonding electrons = no lone pair = 0 electrons
As per the above calculation, the double-bonded O-atoms carry no formal charges, while a -1 and +1 formal charge is present on each of the single-bonded O-atoms and the central N-atoms, respectively.
+1 charge of each N-atom cancels with -1 formal charge of an adjacent O-atom. Put another way, the +2 formal charge of two N-atoms cancels completely with the -2 charge of two O-atoms.
Thus, there is zero or no overall formal charge on the N2O4 Lewis dot structure, proving that we have drawn a stable and correct Lewis representation of dinitrogen tetraoxide.
However, you may note that the following resonance structures are possible for representing the N2O4 molecule.
Each resonance form is a way of representing the Lewis structure of a molecule. The unbonded and pi-bonded electrons keep changing their positions. This is known as electron delocalization.
The actual N2O4 structure is a hybrid of all the above resonance forms, known as the resonance hybrid.
Now that we know everything about the Lewis structure of N2O4, let’s move ahead and discuss its molecular geometry or shape and electron geometry.
What are the electron and molecular geometry of N2O4?
The molecular geometry or shape of N2O4 is identical to its ideal electronic geometry w.r.t each N-atom, i.e., trigonal planar. There is no lone pair of electrons on either of the nitrogen atoms present at the center of N2O4, hence no distortion is witnessed in its molecular shape or geometry.
Molecular geometry of N2O4
The molecular geometry or shape of N2O4 w.r.t each N-atom is trigonal planar.
In N2O4, to an N-atom at the center, one other N-atom and two O-atoms are directly attached, like three corners of an equilateral triangle. There is no lone pair of electrons on either of the N-atoms; thus, negligible lone pair-lone pair and lone pair-bond pair electronic repulsions exist in the molecule.
Hence, it occupies a symmetrical trigonal planar shape, as shown below.
Electron geometry of N2O4
According to the valence shell electron pair repulsion (VSEPR) theory of chemical bonding, the ideal electron geometry of a molecule containing a total of 3 electron density regions around the central atom is trigonal planar.
In N2O4, each of the two central N-atoms is surrounded by 3 bond pairs (N-N, N-O, and N=O bonds) and no lone pair, making a total of 3 electron density regions. Hence, the ideal electron pair geometry of the N2O4 molecule is trigonal planar.
An easy trick to finding a molecule’s electron and molecular geometry is using the AXN method.
AXN is a simple formula representing the number of bonded atoms and lone pairs on the central atom.
It is used to predict the shape and geometry of a molecule using the VSEPR concept.
AXN notation for N2O4 molecule
A in the AXN formula represents the central atom. In the N2O4 molecule, a nitrogen (N) atom is present at the center, so A = N.
X denotes the atoms bonded to the central atom. In N2O4, 2 O-atoms and another N-atom are directly bonded to the central N-atom. So, X = 3 for N2O4.
N stands for the lone pairs present on the central atom. As per the Lewis structure of N2O4, the central N-atom has no lone pairs of electrons. Thus, N = 0 for N2O4.
As a result, the AXN generic formula for N2O4 is AX3N0 or simply AX3.
Now, you may have a look at the VSEPR chart below.
The VSEPR chart confirms that the molecular geometry or shape of a molecule with an AX3 generic formula is identical to its electron geometry, i.e., trigonal planar, as we already noted down for the dinitrogen tetroxide (N2O4) molecule.
Hybridization of N2O4
Both the N-atoms are sp2 hybridized in N2O4.
The electronic configuration of a nitrogen (N) atom is 1s2 2s2 2p3.
During chemical bonding in N2O4, the 2s atomic orbital of nitrogen hybridizes with two of its three half-filled 2p orbitals. This produces three sp2 hybrid orbitals.
Each sp2 hybrid orbital possesses a 33.3 % s-character and a 66.7 % p-character.
There is a +1 charge on nitrogen which indicates a deficiency of 1 electron. Therefore, all the sp2 hybrid orbitals of nitrogen are equivalent in N2O4, i.e., each contains 1 electron only.
The central N-atom uses these hybrid orbitals to form N-N, N-O and N=O sigma bonds, as shown below.
On the other hand, the unhybridized half-filled p-orbital of nitrogen forms the N=O pi bond.
A shortcut to finding the hybridization present in a molecule is using its steric number against the table shown below. The steric number of the N-atom in N2O4 is 3, so it has sp2 hybridization.
Steric number
Hybridization
2
sp
3
sp2
4
sp3
5
sp3d
6
sp3d2
The bond angles of N2O4
There are multiple bond angles (∠ (O-N-N) = 105.1°, ∠ (N-N=O) = 112.7°, ∠ (N-N-O) = 117.5°, ∠ (O-N=O) = 129.8°) present in N2O4.
The multiple bond angles and bond lengths present in the dinitrogen tetroxide molecule are shown in the figure drawn below.
As per Pauling’s electronegativity scale, a polar covalent bond is formed between two dissimilar atoms with an electronegativity difference between 0.4 and 1.6 units.
The two main types of covalent chemical bonds present in N2O4 are N-N and N-O (or N=O).
The N-N bond is purely non-polar as it is formed between two identical nitrogen atoms having no electronegativity difference. In contrast, the N-O (or N=O) bond is polar as an electronegativity difference of 0.40 units is present between an oxygen (E.N = 3.44) and a nitrogen (E.N = 3.04) atom.
Oxygen being more electronegative, attracts the N-O or N=O electron cloud largely towards itself. Thus, all four O-atoms gain partial negative (δ–) charges in N2O4 while both the N-atoms obtain partial positive (δ+) charges.
The dipole moment of a polar N-O or N=O bond points from Nδ+ to Oδ–.
However, it is due to the symmetrical trigonal planar shape of the N2O4 molecule that the N-O and N=O dipole moments get canceled equally in opposite directions to yield an overall non-polar N2O4 molecule (net µ = 0).
The Lewis dot structure of N2O4 displays a total of 34 valence electrons i.e., 34/2 = 17 electron pairs.
Out of the 17 electron pairs, there are 10 bond pairs and 7 lone pairs of electrons.
Two N-atoms are single covalently bonded to each other in the N2O4 Lewis structure. Each N-atom, in turn, is bonded to 2 O-atoms at the sides.
The N=O double-bonded oxygen atoms carry 2 lone pairs of electrons, while 3 lone pairs are present on the N-O single-bonded oxygen atoms.
Is there a formal charge present on the N2O4 Lewis structure?
No. There is no overall formal charge on the N2O4 Lewis structure. Both the central N-atoms carry a +1 formal charge each, while a -1 formal charge is present on the two O-atoms in N2O4.
However, +2 cancels with -2 to yield a net zero formal charge on dinitrogen tetroxide.
What is the shape of N2O4?
The molecular shape of N2O4 w.r.t each nitrogen atom at the center is trigonal planar.
Is the molecular shape of N2O4 the same as its electron geometry? Why?
Yes. The dinitrogen tetroxide (N2O4) molecule possesses an identical electron and molecular geometry or shape, i.e., trigonal planar.
This is because there is no lone pair of electrons present on either N-atom at the center of N2O4 thus, no lone pair –lone pair or lone pair-bond pair electronic repulsions exist at the center of the molecule. Consequently, no distortion is witnessed in its shape and/or geometry.
How are the shapes of the following nitrogen compounds the same or different? NO2, NO3, N2O4 and HNO3.
The shape of NO2 is different from the other three nitrogen compounds mentioned above, which possess similar shapes and molecular geometries.
In NO2, a nitrogen (N) atom is present at the center. It is bonded to 2 O-atoms at the sides which implies that there is a lone pair of electrons on the central N-atom in NO2.
This lone pair leads to strong lone pair-bond pair electronic repulsions. The shape of the molecule gets distorted. It thus occupies a bent, angular or V-shape and molecular geometry.
The shapes of NO3, N2O4, and HNO3 are trigonal planar w.r.t central N-atom as there is no lone pair of electrons on the central atoms. In contrast, the bond pairs occupy 3 corners of an equilateral triangle.
The total number of valence electrons available for drawing the N2O4 Lewis structure is 34.
The molecular geometry or shape of N2O4 w.r.t each N-atom is identical to its ideal electronic geometry, i.e., trigonal planar.
Both the N-atoms are sp2 hybridized in N2O4.
There are multiple bond angles (∠ (O-N-N) = 105.1°, ∠ (N-N=O) = 112.7°, ∠ (N-N-O) = 117.5°, ∠ (O-N=O) = 129.8°) and bond lengths (N=O = 120.2 pm, N-O = 121.7 pm and N-N = 186.4 pm) present in N2O4.
N2O4 is a non-polar molecule (net µ = 0) as the dipole moments of individually polar N-O and N=O bonds get canceled equally in opposite directions.
The +2 formal charge on nitrogen atoms cancels out with an overall charge of -2 on the oxygen atoms of N2O4. Thus, there is no net charge on N2O4 which ensures that the Lewis structure drawn in the article is stable and absolutely correct.
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