Nitric acid (HNO3) Lewis structure, molecular geometry or shape, resonance structure, bond angle, hybridization, polar or nonpolar
Wouldn’t it be a bit surprising if a chemistry student is unfamiliar with nitric acid (HNO3)? Well, it would definitely be if you said so, considering the immense importance of nitric acid in the chemistry laboratory and in industrial synthesis.
Nitric acid is represented by the chemical formula HNO3. It is a colorless, fuming liquid, completely miscible with water.
But do you know how to draw its Lewis structure? Let’s learn that through this article, along with other interesting facts about HNO3, including its molecular geometry or shape, electron geometry, bond angle, hybridization, formal charges, polarity, etc.
|Name of Molecule||Nitric acid|
|Molecular geometry of HNO3||Trigonal planar|
|Electron geometry of HNO3||Trigonal planar|
∠ O=N-OH (130.0°), ∠ O=N-O (115.9 °),
∠ O-N-O (113.9°), ∠ N-O-H (102.2°)
|Total Valence electron in HNO3||24|
|Overall Formal charge in HNO3||Zero|
How to draw lewis structure of HNO3?
The Lewis structure of nitric acid (HNO3) consists of three different elemental atoms. A nitrogen (N) atom is present at the center of the molecule. It is directly bonded to two oxygen (O) atoms and a hydroxyl (OH) functional group.
In this way, there are a total of 3 electron density regions around the central N-atom in the Lewis structure of HNO3. All three electron density regions are constituted of bond pairs; thus, there is no lone pair of electrons on the central N-atom in HNO3.
You can easily draw the Lewis dot structure of HNO3, and to do so, you just need to grab a piece of paper and a pencil and follow the simple steps given below.
Steps for drawing the Lewis dot structure of HNO3
1. Count the total valence electrons in HNO3
The very first step while drawing the Lewis structure of HNO3 is to calculate the total valence electrons present in its concerned elemental atoms.
In HNO3, there are atoms from three different elements of the Periodic Table. So you need to look for the group numbers of these elements in the Periodic Table.
Nitrogen (N) belongs to Group V A (or 15), so it has a total of 5 valence electrons. Oxygen (O) is present in Group VI A (or 16), so it has 6 valence electrons, while hydrogen (H) lies at the top of the Periodic Table containing a single valence electron only.
- Total number of valence electrons in hydrogen = 1
- Total number of valence electrons in nitrogen = 5
- Total number of valence electrons in oxygen =6
∴ The HNO3 molecule consists of 1 N-atom, 1 H-atom, and 3 O-atoms. Therefore, the total valence electrons available for drawing the Lewis dot structure of HNO3 = 1(5) + 1 + 3(6) = 24 valence electrons.
2. Choose the central atom
In this second step, usually the least electronegative atom out of all the concerned atoms is chosen as the central atom.
This is because the least electronegative atom is the one that is most likely to share its electrons with the atoms spread around it.
Oxygen is more electronegative than both nitrogen and hydrogen, so it cannot be chosen as the central atom. Hydrogen is less electronegative than nitrogen, but it can also not be chosen as the central atom because a hydrogen (H) atom can accommodate only 2 electrons; hence it can form a bond with a single adjacent atom only. This denotes that H is always placed as an outer atom in a Lewis structure.
As a result, the nitrogen (N) atom is chosen as the central atom in the Lewis dot structure of HNO3, while one H and three O atoms are placed around it as outer atoms, as shown in the figure below.
3. Connect outer atoms with the central atom
In this step, all the outer atoms are joined to the central atom using single straight lines. But as we discussed already, hydrogen can form a single bond with one adjacent atom only.
So H is first connected to its adjacent O-atom. Contrarily, all the other outer atoms are directly joined to the central N-atom using straight lines, as shown below.
Each straight line represents a single covalent bond, i.e., a bond pair containing 2 electrons. There are a total of 4 single bonds in the above diagram.
As 4(2) = 8, that means 8 valence electrons are already consumed out of the 24 initially available. But we still have 24 – 8 = 16 valence electrons to be accommodated in the Lewis dot structure of HNO3.
4. Complete the duplet and/or octet of the outer atoms
As we already identified, the hydrogen and oxygen atoms are the outer atoms in the Lewis dot structure of HNO3.
Each hydrogen (H) atom requires a total of 2 valence electrons in order to achieve a stable duplet electronic configuration.
The O-H single bond represents 2 valence electrons around the H-atom. This means it already has a complete duplet, and we do not need to make any changes in the Lewis structure obtained so far with regard to the hydrogen atom.
In contrast to that, an O -atom needs a total of 8 valence electrons in order to achieve a stable octet electronic configuration.
An O-H bond and an N-O bond represent a total of 2 + 2 = 4 electrons around this oxygen atom. This denotes it is still deficient in 4 more electrons to complete its octet. So these 4 electrons are placed as 2 lone pairs around this O-atom.
The other two oxygen atoms only form an N-O bond which denotes 2 electrons each. Thus, these are deficient in 6 more electrons to complete their octet. Consequently, these 6 electrons are placed as 3 lone pairs around each of the other two O-atoms, as shown below.
5. Complete the octet of the central atom and make a covalent bond if necessary
- Total valence electrons used till step 4 = 4 single bonds + electrons placed around the O-atom forming O-H bond + electrons placed around other two O-atoms, shown as dots = 4(2) + 4 + 2(6) = 24 valence electrons.
- Total valence electrons – electrons used till step 4 = 24 – 24 = 0 valence electrons.
As all the 24 valence electrons initially available are now consumed, so there is no lone pair on the central N-atom in the HNO3 Lewis structure.
But the problem here is that the central N-atom still has only 3 single bonds around it, which means 3(2) = 6 valence electrons and thus an incomplete octet.
But don’t worry because we can easily solve this problem by converting a lone pair present on an outer O-atom (not from O-H bonded O-atom) into a covalent bond between the central N and the concerned O-atom. Refer to the figure drawn below.
The figure below illustrates that the central N-atom now has a complete octet (2 single bonds + 1 double bond) in addition to the complete octet of each O-atom and a complete duplet of the H-atom.
As a final step, we just need to check the stability of the above Lewis structure, and we can do so by using the formal charge concept.
6. Check the stability of Lewis’s structure using the formal charge concept
The fewer formal charges present 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’s use this formula and the Lewis structure obtained in step 5 to determine the formal charges present on HNO3 atoms.
For nitrogen atom
- Valence electrons of nitrogen = 5
- Bonding electrons = 1 double bond + 2 single bonds = 2(2) + 2 + 2= 8 electrons
- Non-bonding electrons = no lone pair = 0 electrons
- Formal charge = 5-0-8/2 = 5-0-4 = 5-4 = +1
For O-H bonded oxygen atom
- Valence electrons of oxygen = 6
- Bonding electrons =2 single bonds = 2 + 2 = 4 electrons
- Non-bonding electrons = 2 lone pairs = 2(2) = 4 electrons
- Formal charge = 6-4-4/2 = 6-4-2 = 6-6 = 0
For N-O single-bonded oxygen atom
- Valence electrons of oxygen = 6
- Bonding electrons = 1 single bond = 2 electrons
- Non-bonding electrons = 3 lone pairs = 3(2) = 6 electrons
- Formal charge = 6-6-2/2 =6-6-1 = 6-7 = -1
For N=O double-bonded oxygen atom
- Valence electrons of oxygen = 6
- Bonding electrons = 1 double bond = 2(2) = 4 electrons
- Non-bonding electrons = 2 lone pairs = 2(2) = 4 electrons
- Formal charge = 6-4-4/2 =6-4-2 = 6-6 = 0
For hydrogen atom
- Valence electrons of hydrogen =1
- Bonding electrons = 1 single bond = 2 electrons
- Non-bonding electrons = no lone pairs = 0 electrons
- Formal charge = 1-0-2/2 =1-0-1 = 1-1= 0
The above calculation shows that zero formal charges are present on the N=O double-bonded oxygen atom as well as the O-H bonded oxygen and hydrogen atoms, respectively. However, a +1 formal charge is present on the central N-atom. Similarly, a -1 formal charge is present on the N-O single bonded O-atom.
The important point is that a +1 formal charge cancels with -1; thus, there is no overall charge present on the HNO3 molecule, which accounts for its extraordinary stability.
Rather, the +1 and the -1 charges keep circulating from one position to another on the molecule with a concomitant movement of the double bond. This results in three different resonance structures of HNO3.
Each resonance structure is a way of representing the Lewis structure of a molecule. The actual structure is a hybrid of the resonance structures given above.
Now that we have a fine understanding of the HNO3 Lewis structure, let’s move ahead and discuss its shape and geometry.
Also check –
What are the electron and molecular geometry of HNO3?
HNO3 has an identical electron and molecular geometry or shape i.e., trigonal planar. There are a total of 3 electron density regions around the central N-atom comprised of 2 oxygen atoms and one hydroxyl functional group. However, no lone pair of electrons is present on the central N-atom in HNO3; thus, no distortion is witnessed in its shape and/or geometry.
Molecular geometry of HNO3
The molecular geometry or shape of the nitric acid (HNO3) molecule is trigonal planar.
The hydroxyl (O-H) functional group present next to the central N-atom is considered one region of electron density. This O-H group, in addition to two O-atoms, makes the molecule adopt a trigonal planar shape in which the bonded atoms lie along the three vertices of an equilateral triangle.
There is no lone pair of electrons on the central N-atom; thus, no lone pair-bond pair and lone pair-lone pair electronic repulsions exist in the molecule.
Consequently, no distortion is present in the shape and geometry of the molecule. Only a bond pair-bond pair repulsive effect exists that pushes the bonded atoms away.
The atoms occupy positions as far apart from one another as possible to minimize the electron-repulsive effect.
Electron geometry of HNO3
According to the valence shell electron pair repulsion (VSEPR) concept of chemical bonding, the ideal electronic geometry of a molecule containing 3 regions of electron density around the central atom is trigonal planar.
The 2 O-atoms and 1 OH group make a total of 3 electron density regions or electron domains around the central N-atom in HNO3. Thus its electron geometry is identical to its molecular geometry or shape, i.e., trigonal planar.
A more straightforward way of determining the shape and geometry of a 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 shape and geometry of a molecule based on the VSEPR concept.
AXN notation for the HNO3 molecule
- A in the AXN formula represents the central atom. In HNO3, nitrogen (N) acts as the central atom, so A= N.
- X denotes the electron domains bonded to the central atom.2 O-atoms and 1 OH group is directly bonded to the central N atom in HNO3; thus, X=3.
- N stands for the lone pairs present on the central atom. As per the Lewis dot structure of HNO3, there is no lone pair present on central nitrogen; thus, N=0.
So, the AXN generic formula for HNO3 is AX3.
Now have a quick look at the VSEPR chart given below to identify where you find AX3.
The VSEPR chart confirms that the molecular geometry or shape of a molecule with an AX3 generic formula is identical to its electron pair geometry, i.e., trigonal planar, as we already noted for HNO3.
Hybridization of HNO3
The central N atom is sp2 hybridized in the HNO3 molecule.
The electronic configuration of nitrogen (N) is 1s2 2s2 2p3.
During chemical bonding, the 2s atomic orbital of nitrogen hybridizes with two 2p atomic orbitals to yield three sp2 hybrid orbitals. Each sp2 hybrid orbital possesses a 33.3% s-character and a 67.7% p-character, and each contains a single electron only.
The sp2 hybrid orbitals of the central nitrogen atom overlap with the p atomic orbital and the sp2 and sp3 hybrid orbitals of respective oxygen atoms to form the N-O, N=O, and N-OH sigma (σ) bonds.
The unhybridized p-orbitals of nitrogen overlap with the p-orbital of the oxygen atom to form the required pi (π) bond in the N=O double bond in the HNO3 molecule, as shown below.
A short trick for finding the hybridization present in a molecule is to memorize the table given below. You can determine the steric number of a molecule and use that against this table to find its hybridization.
The steric number of central nitrogen (N) in HNO3 is 3, so it has sp2 hybridization.
The HNO3 bond angle
The ideal bond angle in a trigonal planar molecule is 120°. However, it is due to the different types of electron domains around the central N-atom in HNO3 that more than one bond angle is formed between the bonded atoms.
The N-O-H bond angle is 102.2°, the sideways O-N=O bond angles are 115.9° and 130.3° while the O-N-O bond angle is 113.9°. The 130.3 + 115.9 + 113.9 makes a total of 360°, i.e., one complete rotation around the center of a trigonal planar shape.
Similarly, there is more than one different bond length present in the HNO3 molecule. Refer to the figure below.
Also check:- How to find bond angle?
Is HNO3 polar or nonpolar?
Pauling’s electronegativity scale states that a covalent bond is polar if the bonded atoms possess an electronegativity difference between 0.5 to 1.6 units.
An electronegativity difference of 0.4 units exists between the bonded nitrogen (E.N = 3.04) and oxygen (E.N = 3.44) atoms in each of the N-O and N=O bonds in the HNO3 molecule. Thus both N-O and the N=O bonds are slightly polar in the HNO3 molecule.
Similarly, an electronegativity difference of 1.24 units exists between the bonded hydrogen (E.N = 2.20) and oxygen atoms (E.N = 3.44). So the O-H bond is also polar and possesses a specific dipole moment value (symbol µ).
The dipole moments of these polar bonds do not get canceled equally; thus, the HNO3 molecule overall is polar with a non-uniformly distributed electron cloud (net µ = 2.17 Debye).
The extreme solubility of HNO3 in polar solvents such as H2O also hints at the polarity in its nature. Like dissolves like. Polar water molecules attract polar HNO3 molecules using the oppositely charged partial positive and partial negative centers.
Hydrogen bonding in the presence of OH functional groups also endorses HNO3 polarity and, thus, its water miscibility.
Read in detail–
How is the Lewis dot structure for nitric acid (HNO3) determined?
It is bonded to 2 O-atoms via a single and a double covalent bond, respectively as well as to a hydroxyl (OH) functional group.
How many lone pairs are present in the Lewis structure of HNO3?
There are a total of 7 lone pairs in the Lewis structure of HNO3. Three lone pairs of electrons are present on the single-bonded O-atom in the N-O bond.
Two lone pairs are present on each of the N=O and N-OH oxygens. However, there is no lone pair of electrons on the central N-atom in the HNO3 Lewis structure.
Is there any formal charge present on the HNO3 Lewis structure?
A +1 formal charge is present on the central N-atom, and a -1 formal charge is present on the single bonded O-atom in the HNO3 Lewis structure.
However, the +1 charge cancels with -1. Thus, there is no overall charge present on the HNO3 Lewis structure.
Is the shape of HNO3 different from its ideal electronic geometry?
No. The nitric acid (HNO3) molecule possesses an identical electron and molecular geometry or shape, i.e., trigonal planar.
There are a total of 3 electron density regions or electron domains around the central N-atom in HNO3, and no lone pairs of electrons are present on this central atom.
Thus no lone pair-bond pair and lone pair-lone pair electronic repulsions exist in the molecule. As a result, there is no distortion in the HNO3 shape and geometry.
How is the shape of HNO3 different from that of HNO2?
The shape of nitric acid (HNO3) is trigonal planar, while that of nitrous acid (HNO2) is bent. In HNO2, in addition to an O-atom and an OH functional, a lone pair of electrons is present on the central N-atom.
Lone pair-bond pair repulsions distort the shape of the molecule. The molecule thus adopts a bent shape, different from its ideal electron geometry.
- IF5 lewis structure and its molecular geometry
- CH2Cl2 lewis structure and its molecular geometry
- CH3COOH lewis structure and its molecular geometry
- C2H2Cl2 lewis structure and its molecular geometry
- CHCl3 lewis structure and its molecular geometry
- CH3F lewis structure and its molecular geometry
- CF2Cl2 lewis structure and its molecular geometry
- CH3CN lewis structure and its molecular geometry
- CH2O lewis structure and its molecular geometry
- The total number of valence electrons available for drawing the nitric acid (HNO3) Lewis structure is 24.
- The molecular geometry or shape of HNO3 is identical to its ideal electron pair geometry, i.e., trigonal planar.
- The central N-atom is sp2 hybridized in HNO3.
- There are multiple bond lengths and angles present in the HNO3 molecule due to the different types of electron domains present around the central N-atom.
- HNO3 is a polar molecule overall (net µ = 2.17 D).
- Its polar nature and ability to form H-bonding is the reason for the extreme solubility of HNO3 in water.
- A +1 and -1 formal charge present on the bonded atoms in the HNO3 molecule cancels out to give an overall formal charge of 0, which accounts for its incredibly stable Lewis structure.
About the author
My name is Vishal Goyal and I am the founder of Topblogtenz. I hold a degree in B.tech (Chemical Engineering) and have a strong passion for the life sciences and chemistry. As a highly qualified and experienced chemistry tutor with 4 years of experience, I possess a deep understanding of the unique challenges that students often encounter when attempting self-study in the field of chemistry. I have created this website as a comprehensive resource for those who are seeking guidance and support in their chemistry studies. I have brought together a team of experts, including experienced researchers, professors, and educators, to provide our readers with accurate and engaging information on a wide range of chemistry and science topics. Our goal is to make complex subjects like chemistry accessible and understandable for all. I hope you find the information and resources on our site helpful in your studies. Let's connect through LinkedIn: https://www.linkedin.com/in/vishal-goyal-2926a122b/