Lewis Dot Structures
During chemical bonding it is the valence electrons which move amongst different atoms. In order to keep track of the valence electrons for each atom and how they may be shared in bonding we use the Lewis Dot Structure for atoms and molecules. In this approach we represent the valence electrons as dots around the element symbol. For example, oxygen has 6 valence electrons, so we write the symbol O for oxygen and surround it with 6 dots:
The unpaired electrons are represented as single dots, and the paired electrons as double dots. The placement of the single or double dots around the symbol is not critical. Alternatively, we can represent the paired electrons as a line. That is, we replace the double dots as shown below:
Let's consider other examples. A sodium atom has 11 electrons, but only one is a valence electron. The other 10 are inside a closed shell with a Neon electron configuration. Thus, we draw the Lewis structure for a sodium atom as the symbol Na with a single dot:
A chlorine atom has 17 electrons, but only 7 of these are valence electrons. Thus, we draw the Lewis structure as:
In Ionic Bonds valence electrons are completely transferred (not shared). Thus, we write the Lewis structure for NaCl as:
As you can see Chlorine is now surrounded by 8 electrons in the n=3 shell and Sodium has lost its one valence electron in the n=3 shell. Of course, Sodium, is still surrounded by the 8 electrons of the n=2 shell, but we do not show electrons in the inner closed shells.
For period 2 elements, where all the valence electrons of an atom are in s and p orbitals, we find that the Lewis dot structure of molecules will often follow the Octet Rule:
Octet Rule - Atoms tend to gain, lose, or share electrons until they are surrounded by eight electrons (4 electron pairs).
Using Lewis dot structures and the octet rule, we can predict and represent the electronic structure of covalently bonded molecules. For example, when two chlorine atoms, each with 7 valence electrons, come together to form a diatomic chlorine molecule, the Lewis structure shows that there will be a sharing of two electrons between the two chlorine atoms which allows both chlorine to be surrounded by 8 electrons.
Of course, hydrogen is a period 1 element, with only has a 1s orbital, so it has a maximum of two electrons allowed in its valence shell. When two hydrogen atoms come together into a diatomic H2 molecule the Lewis structure shows that there will be a sharing of two electrons between the two hydrogen, allowing both hydrogen to be surrounded by a closed n=1 shell of 2 electrons:
We can represent the electronic structure and reaction of hydrogen and chlorine atoms to form HCl with Lewis structures:
For diatomic oxygen, the Lewis dot structure predicts a double bond.
While the Lewis diagram correctly predict that there is a double bond between O atoms, it incorrectly predicts that all the valence electrons are paired (i.e., it predicts that each valence electron is in an orbital with another electron of opposite spin). Later we will examine a more advanced theoretical approach called Molecular Orbital Theory, which correctly predicts both the double bond of O2 and its unpaired valence electrons. Generally, Lewis-dot structures have the advantage that they are simple to work with, and often present a good picture of the electronic structure. Let's consider another example. For diatomic nitrogen, the Lewis-dot structure correctly predicts that there will be a triple bond between nitrogen atoms:
This triple bond is very strong. The strength of the triple bond makes the N2 molecule very stable against chemical change, and, in fact, N2 is considered to be a chemically inert gas.There is a relationship between the number of shared electron pairs and the bond length.
| Bond | Bond Length |
|---|---|
| N-N | 1.47 Å |
| N=N | 1.24 Å |
| N≡N | 1.10 Å |
The distance between bonded atoms decrease as the number of shared electron pairs increase.
- Ball and Stick Models of NH2-NH2, NH=NH, N≡N Structures
Rules for drawing Lewis dot structures
-
Count the number of valence e- each
atom brings into the molecule. For ions, the charge
must be taken into account.
How many valence electrons in BeCl2?
How many valence electrons in NO2- and NO2+?
-
Put electron pairs about each atom such that
there are 8 electrons around each atom (octet rule),
with the exception of H, which is only surrounded by 2
electrons. Sometimes it's necessary to form double and
triple bonds. Only C, N, O, P and S (rarely Cl) will
form multiple bonds.
Draw the Lewis dot structure for CF4.
The number of valence electrons is 4 + 4 ( 7 ) = 32 electrons.
So, we obtain:
Draw the Lewis dot structure for CO.
The number of valence electrons is 4 + 6 = 10 electrons or 5 pairs. Since both C and O allow multiple bonds we can still follow the octet and write:
-
If there is not enough electrons to follow the
octet rule, then the least electronegative atom is left
short of electrons.
Draw the Lewis dot structure for BeF2.
In BeF2 number of valence e- = 2+ 2(7) = 16 e- or 8 pairs. Since neither Be or F form multiple bonds readily and Be is least electronegative we obtain:
-
If there are too many electrons to follow the
octet rule, then the extra electrons are placed on the
central atom.
Draw the Lewis dot structure for SF4.
In SF4 the number of valence electrons is 6 + 4 ( 7 ) = 34 electrons or 17 pairs. Placing the extra electrons on S we obtain:
How can the octet rule be violated in this last example? The octet rule arises because the s and p orbitals can take on up to 8 electrons. However, once we reach the third row of elements in the periodic table we also have d-orbitals, and these orbitals help take the extra electrons. Note that you still need to know how the atoms are connected in a polyatomic molecule before using the Lewis-Dot structure rules.
Resonance Forms
The Lewis structure for certain molecules or ions can be drawn in more than one way. For example, for NO2- the number of valence eletrons is 5 + 2 (6) + 1 = 18 e- (or 9 pairs), and we find that there are two equally valid Lewis structures that can be drawn:
Which one is correct? Well, you would expect that the doubly bonded oxygen-nitrogen distance to be slightly less than the singly bonded distance. Actually, what is found experimentally is that both N-O distances are equivalent. The true structure of the molecule is a combination of the two. Anytime you have more than one valid structure for a molecule or ion, you have what are known as resonance structures. So in this case, both resonances structures contribute equally to the final structure of the molecule. Sometimes you will have multiple resonance structures which do not contribute equally to the final structure of the molecule. In these cases it can be helpful to know which structure has the greatest contribution to the final structure. If you have many possible resonance forms, you choose the most likely resonance form by calculating the formal charge on each atom in each resonance form. In these situations it is helpful to calculate the formal charge on each atom in each possible resonance structure, and use the formal charges to determine the most representative structure.
Formal charge = Group number - number of nonbonding e-- (number of bonding e-) / 2.
In the example below, we calculate the formal charge on each atom in a Lewis structure.
What are the formal charges on each atom in NO2-?
The sum of the formal charges must equal that of the compund or ion. So, 0 + 0 - 1 = -1 as expected for NO2-.
To use the formal charge to determine most representative resonance forms we follow:
Rules for determining most representative resonance form
- The resonance form(s) with the least number of atoms with formal charge is (are) the most preferred.
- Resonance forms with low formal charges are favored over high formal charge. (e.g., ±1 is favored over ±2).
- Resonance forms with negative formal charge or most electronegative atoms are favored.
- Resonance forms with the same charge on adjacent atoms are not favored.
For example, N2O has number of 2 ( 5 ) + 6 = 16 valence electrons or 8 pairs. We can draw the three valid Lewis structures below, labeled A, B, and C:
For each structure we can calculate the formal charges below on each atom:
| A | B | C | |
|---|---|---|---|
| N1 | 5 - 2 - ( 6 ) / 2 = 0 | 5 - 4 - ( 4 ) / 2 = - 1 | 5 - 6 - ( 2 ) / 2 = -2 |
| N2 | 5 - 0 - ( 8 ) / 2 = + 1 | 5 - 0 - ( 8 ) / 2 = + 1 | 5 - 0 - ( 8 ) / 2 = + 1 |
| O | 6 - 6 - ( 2 ) / 2 = - 1 | 6 - 4 - ( 4 ) / 2 = 0 | 6 - 2 - ( 6 ) / 2 = + 1 |
Examining the formal charges above we see that Formula C is less representative because it has a -2 charge, and formula B is less representative because it has a -1 charge on N and 0 charge on O. Oxygen is more electronegative and should get -1 charge, therefore Formula A is most representative.