NO3- lewis structure, molecular geometry, bond angle, hybridization (2024)

The Lewis structure of a nitrate [NO₃]^–ion consists of a nitrogen (N) atom and three oxygen (O) atoms. The nitrogen (N) atom is present at the center of the molecular ion, while three oxygen (O) atoms occupy terminal positions, one on each side.

There are a total of 3 electron density regions around the central N-atom in the Lewis structure of [NO₃]^–. All 3 electron density regions or electron domains are constituted of N=O and N-O bond pairs. Thus, there is no lone pair on the central N-atom in NO₃^– Lewis dot structure.

Drawing NO₃^–Lewis structure is not that difficult at all.

You just need to grab a piece of paper and a pencil and follow the simple instructions given below, and you will learn to draw the Lewis dot structure of nitrate within no time.

Steps for drawing the Lewis dot structure of [NO₃]^–

1. Count the total valence electrons in [NO₃]^–

The very first step while drawing the Lewis structure of [NO₃]^–is to calculate the total valence electrons present in the concerned elemental atoms.

There are two different elemental atoms present in the nitrate [NO₃]^– ion, i.e., a nitrogen (N) atom, and an oxygen (O) atom. The nitrogen atom contains a total of 5 valence electrons, while 6 valence electrons are present in each atom of oxygen.

• Total number ofvalence electrons in Nitrogen = 5
• Total number ofvalence electrons in Oxygen= 6

The [NO₃]^– ion consists of 1 N-atom, 3 O-atoms, and it also carries a negative one (-1) charge, which means 1 extra valence electron. Thus, the valence electrons in the Lewis structure of [NO₃]^– = 1(5) + 3(6) + 1 = 24 valence electrons.

2. Choose the central atom

Electronegativity is defined as the ability of an atom to attract a shared pair of electrons from a covalent chemical bond. So, the atom which is least electronegative or most electropositive is placed at the center of a Lewis structure. This is because this atom is most likely to share its electrons with the more electronegative atoms surrounding it.

As nitrogen (N) is less electronegative than oxygen (O) so, an N-atom is placed at the center of the [NO₃]^–Lewis structure while the three O-atoms are spread around it, as shown below.

3. Connect outer atoms with the central atom

At this step of drawing the Lewis structure of a molecule or molecular ion, we need to connect the outer atoms with the central atom using single straight lines.

As the O-atoms are the outer atoms in the [NO₃]^–, ion, so all 3 oxygen atoms are joined to the central N-atom using straight lines, as shown below.

Each straight line represents a single covalent bond containing 2 electrons.

Now, if we count the total valence electrons used till this step out of the 24 available initially, there are a total of 3 single bonds in the structure above. Thus, 3(2) = 6 valence electrons are used tillstep 3.

• Total valence electrons available – electrons used tillstep 3 = 24 – 6 = 18 valence electrons.
• This means we still have 18 valence electrons to be accommodated in the Lewis dot structure of [NO₃]^–.

4. Complete the octet of outer atoms

There are three O-atoms present as outer atoms in [NO₃]^–. Each O-atom needs a total of 8 valence electrons in order to achieve a stable octet electronic configuration.

Each N-O bond already represents 2 electrons; therefore, all three O-atoms require 6 more electrons each to complete their octet. Thus, these 6 valence electrons are placed as 3 lone pairs around each O-atom, as shown below.

5. Complete the octet of the central atom

• Total valence electrons used tillstep 4= 3 single bonds + 3 (electrons placed around O-atom, shown as dots) =3(2) +3(6) = 24 valence electrons.
• Total valence electrons available – electrons used till step 4 =24 – 24 = 0 valence electrons.

As all the valence electrons initially available for drawing NO₃^– Lewis structure are already used up thus, there is no lone pair on the central N-atom in this structure.

However, a problem here is that this central N-atom only has 3 single bonds around it, which denotes 3(2) = 6 valence electrons and thus an incomplete octet.

So, to solve this issue, a lone pair present on any one of the three outer O-atoms is converted into an additional covalent chemical bond between the central N-atom and the respective O-atom, as shown below.

In this way, the central N-atom now has a complete octet with 2 single bonds + 1 double bond. Also, the octet of each outer O-atom is complete with 1 double + 2 lone pairs and 1 single bond + 3 lone pairs, respectively, in NO₃^– Lewis structure.

Finally, we need to check the stability of the NO₃^– Lewis structure using the formal charge concept.

6. Check the stability of the NO₃^–Lewis structure using the formal charge concept

The less the formal charge on the atoms of a molecule or molecular ion, 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 instep 5to determine the formal charges on the nitrate [NO₃]^–ion.

For nitrogen atom

• Valence electrons of nitrogen = 5
• Bonding electrons =1 double bond + 2 single bonds = 4 + 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 single-bonded oxygen atoms

• 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 double-bonded oxygen atom

• Valence electrons of oxygen = 6
• Bonding electrons = 1 double bond = 4 electrons
• Non-bonding electrons = 2 lone pairs = 2(2) = 4 electrons
• Formal charge = 6 – 4 – 4/2 = 6 – 4 – 2 = 6 – 6 = 0

The above calculation shows that a zero formal charge is present on the N=O double-bonded oxygen atom. However, a +1 formal charge is present on the central N-atom, while a -1 formal charge is present on each of the two N-O single-bonded oxygen atoms.

The +1 charge on the central N-atom cancels with the -1 charge on one of the two outer single-bonded O-atoms. This leaves behind a -1 charge on only one O-atom, which is also the charge present on the nitrate (NO₃^–) ion overall.

This shows that it is a stable and thus correct Lewis representation of the nitrate (NO₃^–) ion. Consequently, the NO₃^– Lewis structure is enclosed in square brackets, and a -1 charge is placed at the top right corner, as shown below.

Another important point to remember is that three distinct resonance structures are possible for the nitrate (NO₃^– ) ion. This is because anyone O-atom out of the three available can form a double covalent bond with the central N-atom.

Each resonance structure is a way of representing the Lewis structure of a molecule or molecular ion. Non-bonded electrons keep revolving from one position to another, along with a consequent movement of the double bond and thus, the formal charges on the bonded atoms keep changing.

In accordance with that, the actual NO₃^– structure is a hybrid of the following three resonance structures.

Also check –

• How to draw a lewis structure?
• Formal charge calculator
• Lewis structure generator

The nitrate [NO₃]^– ion has an identical electron geometry and molecular geometry or shape, i.e., trigonal planar. There are a total of 3 electron density regions around the central N-atom in NO₃^–. However, no lone pair of electrons is present on the central N-atom; thus, no distortion is witnessed in its shape and/or geometry.

Molecular geometry of [NO₃]^–

The nitrate [NO₃]^– ion has a trigonal planar shape and molecular geometry. The three O-atoms arrange around the central N-atom as 3 vertices of an equilateral triangle. The bond pair-bond pair repulsions between N=O and N-O bonds keep the O-atoms at a maximum distance from one another. Therefore, the O-atoms occupy terminal positions.

However, there is no lone pair present on the central N-atom, so lone pair-bond pair or bond pair-bond pair repulsions do not exist, and the molecular ion maintains its symmetrical shape.

We should keep in mind that the molecular geometry or shape of a molecule or molecular ion gets influenced by the different bond pairs and lone pairs present on the central atom.

Contrarily, there is no such distinction between bonded and non-bonded (lone) electron pairs while considering the electron geometry of the molecule.

Rather, the ideal electronic geometry of a molecule or molecular ion depends on the total electron density regions around the central atom.

Electron geometry of [NO₃]^–

According to the valence shell electron pair repulsion (VSEPR) theory of chemical bonding, the ideal electron geometry of a molecule containing 3 regions of electron density around the central atom is trigonal planar.

In NO₃^–, a double bond around the central N-atom is considered one region of electron density. This N=O double bond, along with two N-O single bonds, makes a total of 3 electron density regions around the central N-atom, thus, the ideal electron pair geometry of NO₃^– isalso trigonal planar.

A simpler way of finding the electron and molecular geometry of NO₃^–is by using the AXN method.

AXN is a simple formula to represent the number of atoms bonded to the central atom in a molecule or molecular ion and the number of lone pairs present on it.

It is used to predict the shape and geometry of a molecule or molecular ion based on the VSEPR concept.

AXN notation for [NO₃]^–molecular ion

• A in the AXN formula represents the central atom. In [NO₃]^–, a nitrogen (N) atom is present at the center, so A =N for [NO₃]^–.
• X denotes the atoms bonded to the central atom. In [NO₃]^–, a total of three oxygen (O) atoms are bonded to the central nitrogen atom, so X = 3.
• N stands for the lone pairs present on the central atom. As per the Lewis structure of [NO₃]^–there is no lone pair of electrons on the central nitrogen atom, so N = 0.

Hence, the AXN generic formula for the nitrate [NO₃]^–ion isAX₃.

Now, you may have a look at the VSEPR chart given below.

The VSEPR chart confirms that the molecular geometry or shape of a molecule or molecular ion with an AX₃ generic formula is identical to its electron pair geometry, i.e., trigonal planar, as we already noted for the nitrate [NO₃]^– ion.

Hybridization of [NO₃]^–

The central N-atom has sp²hybridization in NO₃^–.

The electronic configuration of nitrogen (N) is 1s²2s²2p³. During chemical bonding, one of the paired 3s electrons gets unpaired and shifts to a 3p atomic orbital of the nitrogen atom.

Consequently, the half-filled 3s orbital mixes with two half-filled 3p orbitals to produce three sp²hybrid orbitals of equal energy, each containing a single electron only.

Two of these sp² hybrid orbitals overlap with the p orbitals of the oxygen atoms to form N-O sigma (σ) bonds by sp²-p overlap.

The third sp² hybrid orbital overlap with the sp² hybrid orbital of the third oxygen atom to form a sigma bond in the N=O double bond. The unhybridized p orbital of the central N-atom, however, overlaps with the unhybridized p-orbital of this outer O-atom to form the pi (π) bond in N=O double bond, by p-p side by side overlap.

A shortcut to finding the hybridization present in a molecule or a molecular ion is by using its steric number against the table given below.

The steric number of central N-atom in [NO₃]^–is 3, so it hassp² hybridization.

The [NO₃]^–bond angle

The nitrate (NO₃^–) ionhas an ideal trigonal planar electronic and molecular geometry or shape, so each O-N-O bond angle is 120°.

Conversely, it is due to the resonance present in the NO₃^– ion that each N-O bond length is also equal in NO₃^– i.e., 124 pm, as opposed to the expectation of one shorter N=O bond and two longer N-O bonds.

Also check:-How to find bond angle?