How to Draw Lewis Dot Structure : A step-by-step Guide
Lewis dot structure is the classical bonding model in which only valence electrons of the atoms are used. Lewis structure is very important in chemistry, because they are used in many important concepts of general chemistry such as chemical bonding, resonance, valence shell electron pair repulsion theory, prediction of the polarity of the molecules and understanding of reaction mechanisms. Hence it is very important to learn how to draw Lewis Dot structure correctly for an atom, ion, molecule, polyatomic ion and an ionic compound.
We can learn to make accurate Lewis dot structures in 4 simple steps. These steps are easy to understand and implement. Do not skip or try to rearrange any step during your learning process, as it is important to understand and implement each step to correctly design these structures. Once you master these, you can draw Lewis structure of any chemical entity quickly. In these steps, you will come across some terms like valence electrons ,electronegativity, stable electronic configuration, formal charges, bonding pair and lone pair, single ,double and triple bonds .If you do not know the meaning of any of these terms, do not worry as all terms will be explained in the explanation of each step .
STEP 1 : COUNT THE TOTAL VALENCE ELECTRONS.
In Lewis dot structure, only valence electrons are used for making of the structure. Valence electrons are the electrons present in the outermost shell of the electronic configuration of an atom. The example below should shed some light on this.
If you are not good at writing electronic configurations, then there is another easy way of predicting the valence electrons by using the periodic table. Valence electrons are equal to the group number of the element in the periodic table. You can work some examples on the periodic table right now:
O belongs to group number 6 and its valence electrons are also 6.
Be belongs to group number 2 and its valence electrons are also 2.
Well, that’s step one! Easy, isn’t it?
Lewis dot structure for an atom:
For neutral atoms only step one is required. Just use dots for valence electrons (outermost shell electrons) and place them as paired and unpaired around the four sides of the symbol of the atom as presented in the electronic configuration of the element. For example
Nitrogen atom:
Electronic configuration:
[He]2s22p3
Valence shell is 2s22p3 with total 5 electrons.
Let’s do one more example:
Se atom
Electronic configuration:
[Ar]3d104s24p4
Valence shell is 4s24p4 with total 6 electrons.
Lewis dot structure of Monoatomic ions:
Ions are formed by gain or loss of electrons, so this will change the total number of valence electrons in the ion for the Lewis dot structure .If an atom has a negative charge it means it has gained electrons equal to the charge present on that ion, and in case of a positive charge, it has lost electrons .No of electrons lost or gained are subtracted or added from the valence electrons of the neutral atom.
For an example, let’s find the Lewis dot structure of a nitride ion ( N3-).Three negative charges means nitrogen atom has gained three electrons so its electronic configuration is with 10 electrons (instead of 7).
[He]2s22p6
Valence electrons are 8 (2 in 2s and 6 in 2p)
Now let us try Lewis dot structure of Sulfide ion ( S2-).Two negative charges means sulfur atom has gained two electrons so its electronic configuration is with 18 electrons (instead of 16).
[Ne]4s24p6
Valence electrons are 8 (2 in 3s and 6 in 3p)
Lewis dot structure will have 4 paired dots around Sulfur atom.For atoms and monoatomic ions, step one is sufficient to get the correct Lewis structure.
Lewis dot structures for Polyatomic ions and molecules :
However for molecules and polyatomic ions we need to consider many more factors before drawing a correct Lewis dot structure. Let’s practice step one “count the total valence electrons’ on molecules and polyatomic ions.
Molecule:
SO2 (Sulfur dioxide)
S is in the 6th group and O is also in the same group in the periodic table.
Total valence electrons = 6(S) + 2*6(2O) = 6+12=18
Ion:
NO3– (nitrate ion)
Total valence electrons = 5(N) + 3*6(3O) +1 (-1 charge) = 5+18+1=24
STEP 2 : MAKE A SKELETON OF THE STRUCTURE
SELECT LEAST ELECTRO-NEGATIVE (EN) ATOM AS THE CENTRAL ATOM AND MAKE A SKELETON OF THE STRUCTURE WITH REST OF THE ATOMS AROUND IT
For selecting the central atom we should have a good knowledge of the electronegativity and electronegativity trends along the period and down the group.
Electronegativity (EN) is the tendency of an atom to pull a shared pair electrons which results in the polarity (charge separation) in the bond.
In a periodic table, EN decreases down the group (as the size of the atom increases) and increases along the period (as the size of the atom decreases). As the size of the atom increases bonded electrons move away from the nucleus of the atom and hence nucleus of atom will have less pull on the electrons.
Here is a table that depicts electronegativity trends in the periodic table
Now let us select least EN atom as the central atom in our molecule SO2.You can use the periodic table while deciding about it. S is placed below O in the periodic table and hence it is bigger in size and less EN than O.
SO2 S is the central atom because S is less EN then O
In the skeleton of the molecule two oxygen atoms making single bonds with S
NO3– N is the central atom because N is less EN then O .In the skeleton of the ion three O atoms making three single bonds with central atom N.
Remember:
1. Central atom must be able to make more than 1 single bond around it.
2. First group elements (H and He) cannot have more than 2 electrons, since they have only 1s orbitals in their configurations.
Second period elements (C,N,O,F) cannot have more than 8 electrons around the central atom. This is due to the lack of empty d orbitals and hence these elements can not have expanded octet.
Elements from the third period onwards can have an expanded octet due to the introduction of d orbitals in these periods.
3. H and F can never be the central atom as they need only one electron to complete their respective duplet and octet. These elements make only single bonds with other elements.
STEP 3 : COMPLETE THE OCTET.
COMPLETE THE OCTET OF THE MOST ELECTRONEGATIVE ATOM WITH MINIMUM FORMAL CHARGES
Formal charge is the charge assigned to an atom in a molecule or ion on the basis of the difference in valence electrons and electrons used by the atom in the Lewis dot structure. It is defined as the valence electrons of the atom minus electrons used by atom in making bonds and as lone pairs. An atom is supposed to use all electrons of its valence shell, but if it uses more or less than the number of electrons in its valence shell, then it gets a formal charge. For every covalent bond, an atom gives one electron so number of bonds around each atom will give the number of electrons used in making covalent bonds. Similarly for every lone pair it uses a pair of electrons.
Hence formal charge = valence electrons – electrons used (for bonding and as lone pair) in the Lewis dot structure
Formal charge (FC) = Valence electrons – ½ electrons as bond pairs – electrons as lone pairs
As we know, valence electrons are equal to the group number, number of bonds is equal to the number of electrons used in making covalent bonds and each lone pair means two electrons. So, the equation can be re-written as:
FC = Group No – No of bonds – 2*No of lone pairs.
If an atom has more electrons than the valence electrons around it in Lewis dot structure, then it will acquire a formal negative charge. If the electrons are less than the valence electrons, then it will acquire a formal positive charge.
Example:
Oxygen (O)
It has 6 valence electrons so it is very happy with two bonds and two lone pairs in the Lewis dot structures
Valence electrons of O = 6
No of bonds = 2
Lone pairs = 2
FC = 6-2-(2*2) =0
However if Oxygen has one bond with three lone pairs in Lewis dot structure, then
Valence electrons of O = 6
No of bonds = 1
Lone pairs = 3
FC = 6-2-(2*3) =-1
Another example:
Nitrogen (N)
It has 5 valence electrons so it is very happy with three bonds and one lone pair in the Lewis dot structures
Valence electrons of N = 5
No of bonds = 3
Lone pairs = 1
FC = 5-3-(2*1) =0
However if there are 4 bonds around N which we generally see in many ammonium compounds than it will acquire a formal positive charge
Valence electrons of N = 5
No of bonds = 4
Lone pairs = 0
FC = 5-4-(2*0) =+1
Yet another example:
Carbon (C)
It has 4 valence electrons so it is very happy with four bonds and no lone pairs in the Lewis dot structures.
Valence electrons of C = 4
No of bonds = 4
Lone pairs = 0
FC = 4-4-(2*0) =0
The atoms discussed above are in the second period of the periodic table and hence cannot have more than 8 electrons in the outermost shell (no expanded octet due to lack of d orbitals).
Now, let’s take an element which can have an expanded octet.
Example:
Sulfur/Sulphur (S)
It has 6 valence electrons. So, like oxygen it is also very happy with zero formal charge on it. However, unlike oxygen it has more different combinations to get a zero formal charge. One of the combinations is just like oxygen atom (two bonds and two lone pairs)
Valence electrons of S = 6
No of bonds = 2
Lone pairs = 2
FC = 6-2-(2*2) =0
Second combination is four bonds and one lone pair .Here Sulfur has 10 electrons around it .(expanded octet and extra electrons are accommodated in the empty 3d orbitals of Sulfur).
Valence electrons of S = 6
No of bonds = 4
Lone pairs = 1
FC = 6-4-(2*1) = o
Third combination is 6 bonds and no lone pair . Here Sulfur has 12 electrons around it (expanded octet and extra electrons are accommodated in the empty 3d orbitals of sulfur)
Valence electrons of S = 6
No of bonds = 6
Lone pairs = 0
FC = 6-6-(2*0) =0
Phosphorus has 5 valence electrons so like nitrogen it is also very happy with zero formal charge on it. However, unlike nitrogen it has more different combinations to get zero formal charge .One feasible is just like nitrogen atom three bonds and one lone pair.
Valence electrons of P = 5
No of bonds = 3
Lone pairs = 1
FC = 5-3-(2*1) =0
Second feasible combination to get zero formal charge is five bonds around P . Here phosphorus is with 10 electrons around it (expanded octet and extra electrons are accommodated in the empty 3d orbitals of Phosphorous)
Valence electrons of P = 5
No of bonds = 5
Lone pairs = 0
FC = 5-5-2*0=0
Now let us apply step 3 on our molecule
SO2
From step 2 skeleton of the molecule is
Now let us complete the octet of the most electronegative element O first with minimum formal charge. As you have seen that oxygen is happy with two bonds and two lone pairs so very safely we can put a double bond and two lone pairs on each oxygen atom.
Let’s take nitrate ion as the next example.
In the nitrate ion – NO3−
From step 2 skeleton of the molecule is
Now let us complete the octet of the most electronegative atom O with minimum formal charge. Oxygen being terminal is very happy with a double bond and two lone pairs
This structure is wrong because N cannot have more than 8 electrons around it .In the above structure we have made 6 bonds around Nitrogen means 6*2 (2 electrons in each bond) =12 electrons .Now we need to replace two of the double bonds of the oxygen atom with nitrogen into single bond .To complete the octet of these oxygen, we need to put one extra lone pair on each of them and in the structure you can see two singly bonded oxygen atoms with three lone pairs.
STEP 4 : COMPLETE THE STRUCTURE
COMPLETE THE STRUCTURE BY PLACING THE REMAINING VALENCE ELECTRONS FROM THE TOTAL VALENCE ELECTRONS AS LONE PAIRS ON THE CENTRAL ATOM
Let’s understand this using an example:
In SO2 molecule
Total valence electrons = 18 (from step 1)
Last step is to calculate the total bond pairs and lone pairs placed in the molecule and subtract it from total valence electrons calculated in step 1
Number of electron used up to step 3 are
4 bond pairs and 4 lone pairs hence total is 4*2(Bond pair) +4*2 (lone pair) =16
No of electrons left unused = Total valence electrons – electrons used in Lewis dot structure
= 18-16 =2
These left electrons pair is put on the S atom
Now let us calculate the formal charge on each atom in the lewis dot structure of SO2 molecule
Now let us check for NO3– (nitrate ion)
Total valence electrons = 24
Electrons used are as 4 bond pairs and 8 lone pairs =4*2+8*2=24
Hence all 24 valence electrons are used up .
Let us calculate formal charge on each atom using the equation
FC = Valence electrons – No of bonds – 2*Lone pairs
Final Lewis dot structure of NO3– (nitrate ion)
In brief we need to master 4 steps for making a correct Lewis dot structure
- Count total valence electrons in the molecule or ion.
- Select the central atom and make a skeleton of the molecule or ion.
- Complete the octet of the most electronegative atom with minimum formal charges.
Formal charge = Valence electrons – no of bonds – 2*Lone pairs
Or Formal charge = Group No – Bond pairs – 2*Lone pairs
- Complete the structure by placing unused electrons from the total valence electrons as lone pairs on the central atom.
Practice Examples on Lewis Dot Structure:
NH4+ (ammonium ion) Lewis Dot Structure
Step 1
Total valence electrons = 5(N) + 4*1 (4 H s)-1 (due to one positive charge) = 8
Step 2
Central atom is N because H can never be the central atom and N is more EN than H. (remember mentioned earlier also)
Skeleton of NH4+
Step 3 is already taken care of ,as N has 8 electrons around it and each H is with two electrons on it .
Step 4 :
Total electrons used are as 4 bond pairs = 4*2 = 8
Formal charge on N= Valence electrons – no of bonds – 2*Lone pairs
5-4-0 = +1
Formal charge on H = Valence electrons – no of bonds – 2*Lone pairs
= 1-1-0 = 0
Final correct Lewis dot structure of ammonium ion is:
ClO4– ion (Perchlorate ion) Lewis dot structure
Step 1
Total valence electrons = 7(Cl) + 4*6 (4 SO)+1 (due to one negative charge) = 32
Step 2
Central atom is Cl because O is more electronegative than Cl (check the periodic table)
Skeleton of ClO4– ion
Step 3
Complete the octet of oxygen with minimum formal charge .
Oxygen being terminal is very happy with a double bond and two lone pairs
Remember Cl can have maximum 7 bonds around it because it has 7 valence electrons. In the above structure Cl has 8 bonds around it which will give a negative formal charge to Cl. So this can be taken care if we replace one double bond of oxygen with a single bond and complete the octet of O with one lone pair.
(FC = Valence electrons – no of bonds – 2*Lone pairs)
Step 4:
Electrons used are as 7 bond pairs and 9 lone pairs = 7*2+9*2=32 electrons
Hence all valence electrons are used and no more electrons are left.
Final completed correct lewis dot structure of perchlorate ion is
