How to Draw Lewis Structures with Examples

Lewis structures

We have already seen that valence electrons are represented by dots in the Lewis description of atoms; in the same way, lone pairs of electrons in molecules and the shared electron pairs of covalent bonds may also be represented by pairs of dots.

However, if all the molecular valence electrons are explicitly shown by dots, larger molecules are tedious to draw and too complicated to view. Consequently, from now on we shall follow the simpler convention that the shared electron pair of a covalent bond is represented by a short line, and that only lone pairs and unpaired valence electrons are shown as dots.

bond radius is determined from the atomic separations in the metal. The ionic radius of an element is determined from measurements on crystals of an ionic compound which contain the element as monoatomic ions (e.g. Na+, Ca2+, Cl−, and O2–).

Notice from the values in the table that the sizes (radii) of atoms tend to become smaller on going from left to the right across a period, and to become larger on going down a group. These tendencies result first from the increasing positive charge of the nucleus and its consequent hold on electrons on going from left to right across a period, and secondly from the expanding valence electron shell of the atom upon descending a group.

 

It is very important to be able to draw correct (simplified) Lewis structures in organic chemistry, especially when we look at how organic reactions occur (since this involves the redistribution of valence electrons). Note that all the electrons in almost all stable molecules and ions are spin paired; a chemical species with  an unpaired electron is called a radical and is usually very reactive.

 

 

Example 1

For each of the following compounds, show the connectivity of the atoms and all valence electrons, then draw the (simplified) Lewis structure representing covalent bonds (if there are any) by lines.

(a) NH3

(b) BH3

(c) NaCl

Solution

The neutral N and B atoms have five and three valence electrons, respectively. The N in ammonia has a full valence shell (an octet) which includes one lone pair. Boron with just three valence electrons can form no more than three simple covalent bonds, e.g. with three H atoms to give the molecule shown, borane. Even then, it still has only six valence electrons, i.e. an incomplete valence shell. NaCl is an ionic compound.

 

 

 

Guide for drawing correct Lewis structures

The following is a guide for drawing correct Lewis structures of more complex molecules, with bonding pairs represented by short lines.

 

1.      First, arrange the atoms of the molecule or ion according to how they are bonded together; add dots around each atom corresponding to the number of valence electrons of the neutral atom. Then, add extra dots according to the negative charge for an anion, or delete dots according to the positive charge for a cation.

2.     As examples, consider Lewis structures of the series water (H2O), hydroxide ion (HO−), and oxonium ion (H3O+). First, put the atoms in place with their valence electrons. Then  identify each bond  between  the O  atom  and an  H with a (shared) pair of electrons—one from the O and one from an H. Thirdly, add an extra electron to the O of the anion, and remove one from the O of the cation.

          

3.     Replace pairs of dots representing shared electron pairs (covalent bonds) by short lines, and confirm that each non-hydrogen atom has a valence octet. Here, up to three bonds are possible, and the number of bonds plus lone pairs is four (i.e. eight valence electrons around the O in each case).

 

4.     Determine the formal charge for each atom and add the + or – value as a superscript to its symbol. The formal charge on an atom is the number of valence electrons in the neutral atom (before bonding), minus the number of unshared electrons around the bonded atom, minus the number of bonding pairs; in abbreviated form, this formula is given by:

 

Formalcharge = (valenceelectrons) − (unsharedelectrons) − (bondingspairs)

 

 

Example 2

Calculation of the formal charge on the oxygen atom in H2O, HO−, and H3O+:

 

 

Methanol CH3OH

(i) Arrange all the constituent atoms and  put in dots for valence electrons (for the sake of clarity, numbers of valence electrons are summarized on the left); (ii) identify pairs of electrons which correspond to covalent bonds and replace each by a line (this gives the Lewis structure); confirm that all non-hydrogen atoms have valence octets;

(iii) calculate the formal charges (equal to zero).

 

 

Methanal CH2O

Proceeding as above, we find that making one bond between C and O leaves one unpaired electron on each of C and O. These electrons are paired to form a second bond between C and O to give the C=O double bond; both C and O now have full valence octets. Formal charges are all zero.

 

 

Boron trifluoride BF3

This molecule contains only three bonds since B has only three valence electrons. It follows that B in BF3 has only six electrons whereas all three F atoms have full valence octets.

 

Ammonium ion NH+4

This ion has a positive charge so, according to step (1) in the protocol for drawing Lewis structures described in Subsection 1.3.1, one electron is deleted from the normal number of  valence electrons  for  N. Calculation  of  the formal charge  on  the N shows that it is +1 even though the N has a complete valence octet.

 

 

Methyl anion CH3−

This ion has a negative charge so one extra electron is put onto the C. Bond lines are drawn and the formal charge is calculated: with an unshared electron pair on the C, the formal charge is –1.

 

Ethanoate anion CH3CO−2

One extra electron is added to one of the O atoms for this anion. In the structure with only single bonds, the O carrying the extra electron has an octet, but the other O has only seven electrons. The C to which it is bonded also has an unpaired electron, however, so a C=O double bond can be formed and both of these atoms then achieve complete octets. When the procedure for calculating formal charge is carried out on the singly bonded O atom with the extra electron, it turns out to be –1; all other formal charges are zero.

 

 

In the above, the extra electron was arbitrarily put on one of the O atoms, and we obtained Lewis structure 1a below. We could equally well have chosen the other O, in which case we would have finished up with Lewis structures 1b; both 1a and 1b are legitimate Lewis structures for ethanoate anion. The significance of this will be discussed in the next section.

 

 

Nitromethane CH3NO2

Finally, we construct the Lewis structure of nitromethane. The connection of atoms in the nitro group (NO2) may not be familiar yet, but it is shown below. Each of the two O atoms in the singly bonded structure has seven electrons, so each has an unpaired electron (all other atoms have complete valence shells). A bond between the two O atoms would satisfy the valence requirements but would give a three-membered ring which would be very strained and hence unstable.

Alternatively, if we tried to make two N=O (double) bonds using another two electrons on N, the result would be an impossible structure with 10 valence electrons on the N. Thirdly, if one electron is transferred from the N lone pair to one O, the unpaired electron left on the N can form a second bond to the other O. By this procedure, the valence requirements of all atoms are satisfied and the outcome is the reasonable Lewis structure shown; it simply needs the addition of formal charges.

 

The nitrogen has four bonds and no lone pairs, so its formal charge by eqn 1.10 is (5 − 4) =+1. The value for the singly-bonded O is (6 – 6 – 1) = –1 since it has 6 unshared electrons (three lone pairs) and one bond. The other (double-bonded) O has four unshared electrons and two bonds so the formal charge on this atom is 6 − 4 − 2 = 0. This Lewis structure, therefore, involves charge separation even though the molecule is neutral overall.

The bonding shown for nitromethane is similar to that within the ethanoate ion described above. One of the two O atoms is double bonded and the singly bonded one bears a formal negative charge, as shown in structure 2a. And just as in the case of ethanoate, the O atoms could be switched around to give an alternative equivalent Lewis structure, 2b.