Azide [N3]- ion Lewis structure, molecular geometry or shape, resonance structure, polar or non-polar, hybridization, bond angle
N_{3}^{–} is the chemical formula for the azide ion, also known as hydrazoate. It is an anion composed of three nitrogen (N) atoms. It is the conjugate base of hydrazoic acid/ hydrogen azide (HN_{3}).
The azide [N_{3}]^{–} ion forms extremely toxic chemical agents, which may lead to mitochondrial respiratory chain inhibition in the human body.
In this article, we will discuss how to draw the Lewis dot structure of N_{3}^{–}, what is its molecular geometry or shape, electron geometry, bond angle, hybridization, formal charges, polarity, resonance structure, and much more.
So, stay connected and continue reading!
Name of Molecular ion | Azide |
Chemical formula | [N_{3}]^{–} |
Molecular geometry of [N_{3}]^{–} | Linear |
Electron geometry of [N_{3}]^{–} | Linear |
Hybridization | Sp |
Bond angle(O=Cl-O) | 180° |
Total Valence electron in [N_{3}]^{–} | 16 |
Overall Formal charge in [N_{3}]^{–} | -1 |
How to draw lewis structure of N3-?
The Lewis structure of an azide [N_{3}]^{–} ion consists of three identical nitrogen (N) atoms. One N-atom acts as the central atom, while the other two nitrogen atoms act as outer atoms in [N_{3}]^{–} Lewis structure.
There are a total of 2 electron density regions around the central N-atom in this structure. Both the electron density regions or electron domains are constituted of bond pairs which means there is no lone pair of electrons on the central N-atom in N_{3}^{–}.
In the step-by-step guide given below, we will teach you how to draw the Lewis dot structure of azide [N_{3}]^{–} ion. So grab a paper and pencil and draw this Lewis structure with us.
Steps for drawing the Lewis dot structure of [N_{3}]^{–}
1. Count the total valence electrons in [N_{3}]^{–}
The very first step while drawing the Lewis structure of [N_{3}]^{–} is to count the total valence electrons present in the concerned elemental atoms.
As the azide N_{3}^{–} ion is made up of three identical nitrogen (N) atoms, so we just need to look for the position of nitrogen in the Periodic Table. Nitrogen is present in Group V A (or 15) of the Periodic Table. Thus each atom of nitrogen has a total of 5 valence electrons.
The [N_{3}]^{–} ion consists of 3 N-atoms and a negative (-1) charge, which means 1 extra valence electron is gained.
∴ Therefore, the total number of valence electrons available for drawing [N_{3}]^{–} Lewis structure = 3(5) + 1 = 16 valence electrons.
2. Choose the central atom
In this second step, the least electronegative atom out of all the concerned atoms is chosen and placed as the central atom.
But in the case of N_{3}^{–}, the situation seems uncomplicated because it involves three identical nitrogen atoms. Hence any one N atom can be chosen as the central atom while the other two N atoms are placed in its surroundings, as shown in the figure below.
3. Connect outer atoms with the central atom
In this step, the outer atoms are joined to the central atom using single straight lines. So, the outer N-atoms are joined to the central nitrogen atom in the structure of N_{3}^{–} using straight lines.
Each straight line represents a single covalent bond, i.e., a bonded electron pair containing 2 electrons. There are a total of 2 single bonds in the above diagram; thus, 2(2) = 4 valence electrons.
- Total valence electrons available – electrons used till step 3 = 16 – 4 = 12 valence electrons.
- This means 12 valence electrons are still available to be accommodated in the Lewis dot structure of N_{3}^{–}.
4. Complete the octet of outer atoms
In this step, we need to complete the octet of the two N-atoms bonded to the central nitrogen atom. Each N-atom requires a total of 8 valence electrons to achieve a stable octet electronic configuration.
Both the outer N-atoms are bonded to the central nitrogen atom using N-N single bonds. That means each outer N-atom already has 2 valence electrons. It is thus short of 6 more electrons that are required to complete its octet.
Therefore, 6 valence electrons are placed around each outer N-atom as 3 lone pairs, respectively. Refer to the figure below.
5. Complete the octet of the central atom
- Total valence electrons used till step 4 = 2 single bonds + 2 (electrons placed around each outer N-atom, shown as dots) = 2(2) + 2(6) = 16 valence electrons.
- Total valence electrons – electrons used till step 4 = 16 – 16 = 0 valence electrons.
As all the valence electrons initially available for drawing N_{3}^{–} Lewis structure are already consumed thus, there is no lone pair on the central N-atom in this Lewis structure.
But the problem here is that the central N-atom only has 2 single bonds, i.e., 4 valence electrons in this structure. This denotes that it still has an incomplete octet, and it needs 4 more electrons to achieve a stable octet electronic configuration.
To solve this problem, we need to convert 2 lone pairs, one from each outer N-atom, into two new covalent chemical bonds, as shown below.
In this way, the central N-atom now has a complete octet with 2 double bonds, i.e., 2(4) = 8 valence electrons. Rest assured, each outer N-atom also has a complete octet in the above Lewis structure with 1 double bond and 2 lone pairs of electrons, respectively.
Finally, we just need to check the stability of the N_{3}^{–} Lewis structure by calculating the formal charges present on all three bonded atoms.
6. Check the stability of the N_{3}^{–} 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 in step 5 to determine the formal charges present on three N-atoms in the azide [N_{3}]^{–} ion.
For the central nitrogen atom
- Valence electrons of nitrogen = 5
- Bonding electrons = 2 double bonds = 2 (4) = 8 electrons
- Non-bonding electrons = no lone pairs = 0 electrons
- Formal charge = 5 – 0 – 8/2 = 5 – 0 – 4 = 5 – 4= +1
For outer nitrogen atoms
- Valence electrons of nitrogen = 5
- Bonding electrons = 1 double bond = 4 electrons
- Non-bonding electrons = 2 lone pairs = 2(2) = 4 electrons
- Formal charge = 5 – 4 – 4/2 = 5 – 4 – 2= 5 – 6 = -1
The above calculation shows that a +1 formal charge is present on the central N-atom while a -1 formal charge is present on each of the outer N-atoms in N_{3}^{–} Lewis structure.
+1 formal charge on the central N-atom cancels with -1 formal charge on one of the two outer N-atoms. This leaves behind a -1 formal charge which is the charge present on the azide ion overall.
The above Lewis structure is finally enclosed in square brackets, and a -1 charge is placed at the top right corner, as shown below.
An important point to remember is that the azide (N_{3}^{–}) ion consists of three resonance structures. Each of the following resonance structures is a way of representing the Lewis structure of the azide ion.
The non-bonded electrons keep revolving from one position to another on the ion with a consequent movement of the covalent chemical bonds. However, resonance structures 1 and 3 are relatively unstable due to higher formal charges on the bonded atoms.
In contrast to that, resonance structure 2 is the best possible Lewis representation of the azide ion, as the formal charges present on the bonded atoms are minimized in this structure.
Now that we have discussed everything about N_{3}^{– }Lewis structure, we will discuss about the shape and geometry of the azide ion in the next section.
Also check –
What are the electron and molecular geometry of N3-?
The azide [N_{3}]^{–} ion has an identical electron geometry and molecular geometry or shape, i.e., linear. There are a total of 2 electron density regions around the central N-atom in N_{3}^{–}. 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 [N_{3}]^{–}
The azide [N_{3}]^{–} ion has a linear shape or molecular geometry. Two nitrogen (N) atoms are bonded to the central N-atom in a perfectly symmetrical linear arrangement.
There is no lone pair of electrons on the central N-atom in N_{3}^{–}. Therefore, no lone pair-lone pair or lone pair-bond pair electronic repulsions exist in the molecular ion.
Only a bond pair-bond pair repulsive effect exists that pushes the bonded N-atoms away so that they occupy the opposite corners of a linear shape, as shown below.
Electron geometry of [N_{3}]^{–}
According to the valence shell electron pair repulsion (VSEPR) theory of chemical bonding, the ideal electron geometry of a molecule or a molecular ion containing a total of 2 electron density regions around the central atom is linear.
In N_{3}^{–} ion, two N-atoms are bonded to the central N-atom, which makes a total of 2 electron density regions or electron domains around the central atom. Thus, the electron geometry of the azide [N_{3}]^{–} ion is also linear.
A shortcut to finding the electron and the molecular geometry of a molecule or a molecular ion 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 [N_{3}]^{–}
- A in the AXN formula represents the central atom. In [N_{3}]^{–}, a nitrogen (N) atom is present at the center, so A =N for [N_{3}]^{–}.
- X denotes the atoms bonded to the central atom. In [N_{3}]^{–}, two nitrogen (N) atoms are bonded to the central nitrogen atom, so X = 2.
- N stands for the lone pairs present on the central atom. As per the Lewis structure of [N_{3}]^{– }there is no lone pair of electrons on the central nitrogen atom, so N = 0.
Hence, the AXN generic formula for the azide [N_{3}]^{–} ion is simply AX_{2}.
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 AX_{2} generic formula is identical to its electron pair geometry, i.e., linear, as we already noted for azide [N_{3}]^{–} ion.
Hybridization of [N_{3}]^{–}
The central nitrogen atom has sp hybridization in the azide [N_{3}]^{–} ion.
The electronic configuration of nitrogen (N) is 1s^{2} 2s^{2} 2p^{3}.
During chemical bonding, one 2s electron of nitrogen shifts to a 2p atomic orbital. The half-filled 2s orbital of nitrogen consequently hybridizes with one half-filled 2p orbital to yield two equivalent sp hybrid orbitals.
Each sp hybrid orbital of nitrogen contains a 50% s-character and a 50% p-character, and each contains a single valence electron only. The sp hybrid orbital overlaps with the sp^{2} hybridized orbital of an outer nitrogen atom to form the N-N sigma (σ) bond.
However, the unhybridized p-orbitals of central nitrogen overlap with the unhybridized p-orbitals of outer nitrogen atoms to form the required pi (π) bonds in N=N double bonds, one on each side of the molecular ion.
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 [N_{3}]^{–} is 2, so it has sp hybridization.
Steric number | Hybridization |
2 | sp |
3 | sp^{2} |
4 | sp^{3} |
5 | sp^{3}d |
6 | sp^{3}d^{2} |
The [N_{3}]^{–} bond angle
In N_{3}^{–}, the three nitrogen atoms ion lie in a planar arrangement, on a straight line, and form a mutual bond angle of 180°.
Each bond length is equal in N_{3}^{–} i.e., 115 pm.
Also check:- How to find bond angle?
Is N3- polar or nonpolar?
According to Pauling’s electronegativity scale, a covalent chemical bond is considered polar if the bonded atoms have an electronegativity difference between 0.5 to 1.6 units.
As we have already seen that the N_{3}^{–} ion consists of three identical nitrogen atoms. The electronegativity value of each nitrogen atom is 3.04 units. Therefore, a zero electronegativity difference exists between the bonded atoms in both N-N and N=N bonds.
Also, N_{3}^{–} has a planar linear shape. Thus, by definition, the azide [N_{3}]^{–} ion should be non-polar.
But some chemists report that it is due to the negative charge present on the ion overall in N_{3}^{–} that its symmetry gets disturbed and it exhibits some polar characteristics.
A particular evidence of this argument is that N_{3}^{–} is water soluble. Like dissolves like, and H_{2}O is definitely a polar molecule; therefore, the azide (N_{3}^{–}) ion may also exhibit some polar characteristics (net dipole moment µ > 0).
Read in detail–
FAQ
What is the Lewis structure for [N_{3}]^{–}? |
However, 2 lone pairs of electrons are present on each outer N-atom, as shown below. |
How many resonance structures are possible for N_{3}^{–}? |
The following resonance structures are possible for N_{3}^{–}. The actual structure of N_{3}^{–} is a hybrid of these three resonance structures. |
What is the most stable Lewis structure for N_{3}^{–}? |
The best possible and most stable Lewis representation for N_{3}^{–} is the one in which the formal charges present on the bonded atoms are minimized. In this structure, the +1 formal charge on the central N-atom cancels with the -1 charge of one of the two outer N-atoms. ∴ So the overall charge present on the ion is -1. |
What is the molecular geometry of N_{3}^{–}? |
The molecular geometry or shape of N_{3}^{–} is identical to its ideal electronic geometry, i.e., linear. A total of 2 electron domains are present around the central N-atom in N_{3}^{–}, and there is no lone pair of electrons present on it. |
How is the shape of azide [N_{3}]^{–} ion different from that of the azanide [NH_{2}]^{–}? |
The azide [N_{3}]^{–} ion has a linear molecular geometry or shape. The three bonded atoms lie in a straight line in a linear arrangement. The azanide [NH_{2}]^{–} ion has a bent or V-shape as opposed to its tetrahedral electron geometry. The central N-atom is bonded to two H-atoms via single bonds, and it has 2 lone pairs of electrons present on it. Lone pair-lone pair and lone pair-bond pair electronic repulsions distort the shape and geometry of the molecular ion. |
How is the shape of N_{2} similar to that of N_{3}^{–}? |
Both nitrogen (N_{2}) molecules and the azide [N_{3}]^{–} ion possesses a linear shape. In N_{2}, two N-atoms are triple covalently bonded to each other, and there is a lone pair of electrons on both N-atoms. So it adopts a linear shape. The azide [N_{3}]^{–} ion also has an identical linear electron and molecular geometry or shape, however, no lone pairs of electrons are present on the central N-atom in this case. |
Also Read:-
- ClO_{2}^{–} lewis structure and its molecular geometry
- ClO_{3}^{–} lewis structure and its molecular geometry
- CH_{2}Cl_{2} lewis structure and its molecular geometry
- CH_{3}COOH lewis structure and its molecular geometry
- C_{2}H_{2}Cl_{2} lewis structure and its molecular geometry
- CHCl_{3} lewis structure and its molecular geometry
- CH_{3}F lewis structure and its molecular geometry
- CF_{2}Cl_{2} lewis structure and its molecular geometry
- CH_{3}CN lewis structure and its molecular geometry
- CH_{2}O lewis structure and its molecular geometry
Summary
- The total number of valence electrons available for drawing azide [N_{3}]^{–} ion Lewis structure is 16.
- The negative 1 charge present on the ion accounts for 1 extra valence electron added in its Lewis structure.
- The [N_{3}]^{–} ion has an identical electronic and molecular geometry or shape, i.e., linear.
- The central N-atom is sp hybridized in N_{3}^{–}.
- The polarity of N_{3}^{–} ion is a debatable topic. It may be considered non-polar owing to its identical atoms and linear shape. However, the negative charge present on the ion and its water solubility accounts for some definite polar characteristics present in N_{3}^{–}.
- +1 formal charge present on the central N-atom cancels with a -1 charge on one of the two outer N-atoms.
- This leaves behind a -1 formal charge on the other N-atom that accounts for an overall negative charge on the monovalent anion, i.e., N_{3}^{–}.
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