Lewis structures of an atom denote the valence electrons of an atom by placing dots around the symbol of an atom. The Lewis structures of the compound involves the representation of symbols of the elements surrounded by dots, which indicates the electrons taking part in the bond formation as well as the non-bonding electrons.
The hydrogen has one dot representing its 1s electron, carbon has four dots (2s2 2p2), nitrogen has five dots (2s2 2p4), oxygen has six dots (2s2 2p4), fluorine has seven dots (2s2 2p5) and so on.
Rules to Draw the Lewis Structures
- To select the central atom, it should be in least number or least electronegative or largest size or highest atomic number. Hydrogen and Fluorine cannot be central atom. Carbon, Nitrogen, Oxygen, Sulphur, Phosphorus, Silicon, Chlorine, Bromine can be the central atoms.
- Octet of the corner atom must be complete. If Hydrogen is present, it should complete the duet (2e–).
- The central atom can have eight or more electrons.
- Central atom tries to remain in maximum covalency. The maximum covalency of Oxygen = 2, Sulphur = 6, Nitrogen = 5 (can form maximum 4 bond), Phosphorus = 5, Carbon= 4, Silicon =4, Chlorine = 7. The oxygen can show covalency (1,2), Sulphur (2,4,6), Nitrogen (3,4), Phosphorus (3,5), Carbon (4), Silicon (4), Cl (1,3,5,7).
- If positive charge is present (H3O+, NH4+), it will be on central atom.
- If negative charge is present (NO3–, SO42-, CO32-), it will be on corner atom.
Calculations to Draw Lewis Structures
Q = Valence electrons of all the atoms + (any negative charge) – (any positive charge)
Where Q = Total Electrons
Bond Pair Electrons = 2 x No. of Bonds
Lone Pair Electrons = Q – Bond Pair Electrons
Lone pair electrons can be on lone pair of double and triple bond electrons.
Examples of Lewis Structures
Lewis Structure of H2
Q = 2
Bond Pair Electrons = 2
Lone Pair e- = Q – Bond Pair e- = 2 – 2 = 0
In this H2 molecule Lewis structure, both the atoms have duet.
Lewis Structure of O2
Q = 6 x2 = 12
Bond Pair Electrons = 2 e-
Lone Pair e- = Q- Bond Pair e- = 12 – 2 = 10
In this first step, O – O will be formed and then assign the lone pair electrons to complete the octet of corner atoms first. There will be two less electron of one oxygen atom to complete the octet, so it suggest that there will one pi- bond between two oxygen atoms. The Lewis structure O2 is:
Lewis Structure of H3O+
Q= 6 + 3 x 1 – 1 = 8
Bonding Pair e- = 6 e-
Lone Pair e- = Q – Bonding Pair e- = 8 – 6 = 2
First, draw the connections of the atoms according to the specified rules, selection of central atom. The oxygen is the least atom and also electronegative, so it is central atom and then connects three hydrogen atoms with single bond.
After that count the electrons and assign remaining lone pair electrons and the positive charge will be assigned over central atom. The Lewis structure of H3O+ is:
Lewis Structure of NH4+
Q = 5 + 4 x 1 – 1 = 8
Bond Pair e- = 8
Lone Pair e- = 8 – 8 = 0
In the structure of NH4+, nitrogen is central atom because it is in least number and more electronegative atom. Nitrogen has covalency five and it was stated in rules that Nitrogen can only make maximum four bonds. The positive charge will always be assigned on central atoms according to rules. The Lewis structure of NH4+ is:
Lewis Structure for NO31-
Q = 5 + 3 x 6 +1 = 24
Bond Pair e- = 6
Lone Pair e- = Q – Bond Pair e- = 24 – 6 = 18
Here, the central atom is nitrogen because it is in least number than oxygen. First of all, complete the octet of the corner atoms.
Now the octet of the corner atoms is complete but nitrogen has six electrons, it is mentioned in the rules that central atom must have compete octet. One of the lone pair will be used to make pi bond to complete the octet of nitrogen.
All the atoms in the structure have complete octet. We know that negative charge will be assigned on corner atom but there are three corner atoms. To decide the position of negative atom, formal charge must be calculated.
What is Formal Charge?
The formal charge is the equal charge on atom within a larger molecule or polyatomic ion. The sum of the formal charges on any molecule or ion results in the net overall charge.
FC = Formal Charge
V = Valence Electrons
N = Non Bonding Electrons
B = Bonding electrons
Calculate the formal charge of NO31- atoms
Assign the numbering on the atoms and then calculate the formal charge of each atom to know exactly which atom possess the charge.
FC = V – N – B/2
IO FC = 6 – 4 – 4/2 = 6 – 4 – 2 = 0
IIO FC = 6 – 6 – 2/2 = 6 – 4 – 1 = – 1
IIIO FC = 6 – 6 – 2/2 = 6 – 4 – 1 = – 1
IVN FC = 5 – 0 – 8/2 = 5 – 4 = +1
Lewis structure of NO31- after assigning the charge, the net charge will be -1. The structure of NO31- is:
Lewis Structure of NO21-
Q = 5 + 2 x 6 + 1 = 18
Bone Pair e- = 4
Lone Pair e- = Q – Bond Pair e- = 18 – 4 = 14
In this structure, the central atom is nitrogen because it is in least number. Make the bonds between central atom and corner atoms. Assign the lone pairs to complete the octet of the corner atoms first and then remaining one lone pair over central atom.
Nitrogen has six electrons and it needs eight to complete the octet. One of the lone pair from one oxygen atom will be utilized to from pi bond between nitrogen and oxygen.
Assign the negative charge; we know that negative charge is always present on corner atoms. When oxygen is doubly bonded, it does not possess any charge so other oxygen atom will possess negative charge.
FC = V – N – B/2
IO FC = 6 – 6 – 2/2 = 6 – 4 – 1 = – 1
IIO FC = 6 – 4 – 4/2 = 6 – 4 – 2 = 0
IIIN FC = 5 – 2 – 6/2 = 5 – 2 – 2 = 0
The Lewis structure of NO21- is: