4.2: Drawing Lewis Structures

 Learning Objectives

• To draw Lewis structures for molecular compounds

Representing Valence Electrons as Bonds and Lone Pairs

Lewis structures are used to show the arrangement of valence electrons in a molecule.

Dots are used to show valence electrons that belong to only one atom. Lines are used to represent a pair of shared electrons. Core electrons (any electrons that are not valence electrons) are not shown. Electrons are almost always arranged in pairs, either as a pair of dots or as a line which is worth two electrons.  

Lewis structures are used to clearly show whether valence electrons are shared or unshared. They are optimized for a 2D paper or screen and do not clearly show the correct angles between atoms or the overall 3D shape of a molecule; 3D arrangements will be covered in a later section. 

Drawing Lewis Structures

For very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms. In the examples below individual atoms are shown with the correct number of valence electrons. Put one valence electron on each side of the symbol before putting any of the dots in pairs. The unpaired electrons from each atom can be shared to create a bond. 

For larger molecules the method above is tedious and it can be complicated to determine where multiple bonds (double or triple bonds) should be. For organic compounds an easy method to determine the Lewis structure is to start at the edges of the molecule and fill in the number of bonds and lone pairs for each atom so that they follow the patterns shown in 

Typical number of bonds and lone pairs for nonmetals

Location of nonmetal on the Periodic Table	# of Covalent Bonds	# of Lone Pairs
Hydrogen	1	0
Group 14	4	0
Group 15	3	1
Group 16	2	2
Group 17	1	3

The nonmetals in Group 18 already have 8 valence electrons, so they do not usually form any bonds. The patterns for the Group 14 - 18 nonmetals result in an octet of electrons because each covalent bond consists of two electrons and each lone pair is also two electrons. Hydrogen is an exception; it is satisfied with only two electrons because it is so small. There are other exceptions to the octet rule, but they are not covered here. 

When given the molecular formula, which just states the number of atoms of each element such as H₂O or C₂H₄, then the element listed first goes in the middle, unless it is hydrogen. Hydrogen is never found in the middle of a molecule because it is not able to form more than one bond. Lay out the atoms. For small molecules they tend to be arranged symmetrically. 

Example \(\PageIndex{1}\)

Draw the Lewis structure for C₂H₄. 

Solution

1. Lay out the atoms. Carbon goes in the middle because it is listed first (and because hydrogen can never go in the middle). Atoms tend to be arranged symmetrically for small molecules, unless you are told otherwise. Therefore we give two hydrogen atoms to each carbon atom. All atoms can be arranged at 90 or 180 degree angles. It does not matter for Lewis structures if you chose to put the H atoms above, below, or to the sides of the carbon atoms. 
2. Work from the outside in. Hydrogen is satisfied with one bond and zero lone pairs. 
3.  Each carbon should have four bond (and zero lone pairs). The hydrogen atoms are full and bonds cannot point towards the empty space above the carbon atoms because the lines represent electrons that are shared by two atoms. Therefore, two bonds are needed between the two carbon atoms. 

The bonds represent shared electrons, so each atom in the Lewis structure has the correct bonding pattern: one bond and zero lone pairs for hydrogen atoms and four bonds and zero lone pairs for carbon atoms. 

Exercise \(\PageIndex{1}\)

Draw the Lewis structure for CH₂O. 

Answer

 The carbon atom goes in the middle because it is listed first. Each hydrogen atom gets one bond. The oxygen atom gets two bonds. They both have to go to the carbon atom (as a double bond) because the oxygen atom is only bonded to one other atom. The oxygen atom also has two lone pairs. 

For larger structures you may be provided with the layout of the atoms. Follow the same procedure: start from the outside filling in bonds and lone pairs according to the expected pattern for each element. 

Example \(\PageIndex{2}\)

Fill in the bonds and lone pairs to complete the Lewis structure. Note: The upper O atom is bonded only to C, not to H. 

Solution

Each H atom can only have one bond, so you can start by filling those in. 

Next do the N or either of the O atoms. Leave the C atoms for later since they are in the middle and more information about their neighboring atoms is needed before completing them. 

Nitrogen should have three bonds and one lone pair. Each oxygen atom should have two bonds and two lone pairs. The upper oxygen atom is only bonded to one other atom so it needs a double bond. The oxygen atom on the right is between two atoms so it needs two single bonds. Lone pairs must go on an empty side; they should never be drawn between two atoms.

Each carbon atom should have four bonds and zero lone pairs.

For polyatomic ions and compounds that have exceptions to the octet rule, it is helpful to follow the step-by-step procedure outlined here:

1. Determine the total number of valence (outer shell) electrons among all the atoms. For cations, subtract one electron for each positive charge. For anions, add one electron for each negative charge.
2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom. (Generally, the least electronegative element should be placed in the center.) Connect each atom to the central atom with a single bond (one electron pair).
3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen), completing an octet around each atom.
4. Place all remaining electrons on the central atom.
5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.

Let us determine the Lewis structures of OF₂ and HCN as examples in following this procedure:

1. Determine the total number of valence (outer shell) electrons in the molecule or ion. For a molecule, we add the number of valence electrons (use the main group number) on each atom in the molecule. This is the total number of electrons that must be used in the Lewis structure.

O + 2 (F) = OF₂

6e^- + (2 x 7e^-) = 20e^-

H + C + N = HCN

1e^-+ 4e^-+ 5e^-= 10e^-

2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom and connecting each atom to the central atom with a single (one electron pair) bond. Note that H and F can only form one bond, and are always on the periphery rather than the central atom.

3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen) to complete their valence shells with an octet of electrons.

• In OF_2, six electrons are placed on each F.
• In HCN, six electrons placed on N

4. Place all remaining electrons on the central atom.

• In OF₂, 4 electrons are placed on O.
• In HCN: no electrons remain (the total valence of 10e^-is reached) so nothing changes.

5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.

• In OF₂, each atom has an octet as drawn, so nothing changes.
• In HCN, form two more C–N bonds

H dash C dash N with 6 dots around the N. 4 of the dots are pointing to the dash connecting the C and the N. This turns the formula into H dash C and three lines connecting the C and N together, with two dots on the outside of the N.

Finally, check to see if the total number of valence electrons are present in the Lewis structure. And then, inspect if the H atom has 2 electrons surrounding it and if each of the main group atoms is surrounded by 8 electrons.

MULTIPLE BONDS

In many molecules, the octet rule would not be satisfied if each pair of bonded atoms shares only two electrons. Review HCN in Step 5 above. Another example is carbon dioxide (CO₂). CO₂ has a total valence of 4e^- + (2 x 6e^-) = 16e^-. Following steps 1 to 4, we draw the following:

This does not give the carbon atom a complete octet; only four electrons are in its valence shell. This arrangement of shared electrons is far from satisfactory.

In this case, more than one pair of electrons must be shared between two atoms for both atoms to have an octet. A second electron pair from each oxygen atom must be shared with the central carbon atom shown by the arrows above. A lone pair from each O must be converted into a bonding pair of electrons.

In this arrangement, the carbon atom shares four electrons (two pairs) with the oxygen atom on the left and four electrons with the oxygen atom on the right. There are now eight electrons around each atom. Two pairs of electrons shared between two atoms make a double bond between the atoms, which is represented by a double dash:

Some molecules contain triple bonds (like HCN, shown above). Triple bonds are covalent bonds in which three pairs of electrons are shared by two atoms. Another compound that has a triple bond is acetylene (C₂H₂), whose Lewis diagram is as follows:

Example \(\PageIndex{1}\)

Draw the Lewis diagram for each molecule.

1. \(\ce{N2}\)
2. \(\ce{CH2O}\) (The carbon atom is the central atom.) One application of CH₂O, also called formaldehyde, is the preservation of biological specimens. Aqueous solutions of CH₂O are called formalin and have a sharp, characteristic (pungent) odor.

Solution

1. The total number of electrons is 2 x 5 = 10 electrons. The bond between the two nitrogen atoms is a triple bond. The Lewis diagram for N₂is as follows:

2. The total number of electrons is 4 x 2(1) + 6 = 12 electrons. In CH₂O, the central atom is surrounded by two different types of atoms. The Lewis diagram that fills each atom’s valence electron shell is as follows:

   

Exercise \(\PageIndex{1}\)

Draw the Lewis diagram for each molecule.

1. \(\ce{O2}\)
2. \(\ce{C2H4}\)

Answer a:

or

Answer b:

or or

 

Example \(\PageIndex{2}\)

Which is the correct Lewis structure for N₂H₂?

A.

B.

C.

Solution

Lewis structure A is the correct answer. It has a total of (2 x 5e^-) + (2 x 1e^-) = 12e^-. Each of the N atoms satisfy the octet requirement and the H atoms follow the duet rule.

Structure B is electron deficient. It has only 10e^- instead of 12.

Structure C has 14 (2 extra) electrons. The N atoms do not satisfy the octet.

Exercise \(\PageIndex{2}\)

Which is the correct Lewis structure for NOCl?

A.

B.

C.

Answer

Structure A violates the octet rule; N is surrounded by only 6e^-.

Structure B violates the octet rule; Cl has 10e^- around it. Furthermore, there are a total of 20e^- instead of 18e^-.

Structure C is the correct structure. It has a total of 6e^- + 5e^- + 7e^- = 18e^-. Each atom is surrounded by 8 electrons (octet rule).

Key Takeaways

• A Lewis structure shows the bonding and nonbonding electrons around individual atoms in a molecule.
• Some molecules must have multiple covalent bonds between atoms to satisfy the octet rule.
• A double bond contains four electrons and a triple bond contains six electrons.