General Chemistry at Penn State: Tutorials

Lewis structure allow us to account for connectivity of atoms in molecules and keep track of all bonding and lone-pair electrons. They are also a starting point to understanding of molecular structure and bonding on the orbital level. The procedure to draw Lewis structure of any covalently bonded molecule is outlined below with four examples.

Step 1. Based on molecular formula, sum the valence electrons for all atoms participating in bonding.

The number of valence electrons for each atom can be obtained from the periodic table. If the species is an ion correct for the overall charge: for an anion add an electron for each negative charge, and for the cation subtract an electron for each positive charge. Write down the total number of electrons.

CH₂O	BF₃	NCS^-	CO₃²^-
formaldehyde	boron trifluoride	thiocyanate ion	carbonate ion
C 4e 2H 2e O 6e Total: 12e	B 3e 3F 21e Total: 24e	N 5e C 4e S 6e charge 1e Total: 16e	C 4e 3O 18e charge 2e Total: 24e

Step 2. Establish connectivity of atoms involved in bonding: draw single bonds between all connected atoms. The number of single bonds drawn must not exceed one for H (or He), four for any atom of the second row of the periodic table, and six for any other atom.

Usually the formula suggest connectivity; i.e. the atoms are listed in the order that they are connected in the molecule. Some other time, the central atom is listed first, followed by other atoms attached to it. In many cases the connectivity is indicated by some additional information provided (such as a name, for example). If no additional information is given, one must explore all possibilities: different connectivities represent different molecules (isomers). Often the "best" arrangements are found by placing the least electronegative element in the center.
You may use the common valence numbers (not to be confused with valence electrons) to guide you to some reasonable connectivities. The common valence numbers are H = 1, [F, Cl, Br, or I] = 1, [O or S] = 2, [N or P] = 3, C = 4 etc. Atoms from the third row and below (Cl, Br, I, S, P etc.) may have increased valence numbers (and often do). The initial connectivities should use number of bonds that are equal or less than the valence numbers. This method works well for organic compounds.
The real shapes of the molecules (both 2D and 3D) drawn at this stage are not established yet: any clear drawing will work.

CH₂O	BF₃	NCS^-	CO₃²^-

connectivity implied by the name and formula; excluding hydrogens (which cannot be central anyway), the most electropositive atom is in the center	connectivity implied by the name and formula; the most electropositive atom is in the center	connectivity implied by the name and formula; the most electropositive atom is in the center	connectivity implied by the name and formula; the most electropositive atom is in the center

*For other possible connectivities and their Lewis structures check a separate page (in preparation).

Step 3. Complete the octets on atoms bonded to the central atom. Remember H (and He) are limited to two electrons, and atoms of the second row are limited to an octet.

3 e-pairs for single bonds, 3 e-pairs for nonbonding electrons on oxygen: total 12 e was used

No electrons left; carbon does not have an octet

3 e-pairs for single bonds, 9 e-pairs for nonbonding electrons on fluorines: total 24 e was used

No electrons left; boron does not have an octet

2 e-pairs for single bonds, 6 e-pairs for nonbonding electrons on nitrogen and sulfur: total 16 e was used

No electrons left; carbon does not have an octet

3 e-pairs for single bonds, 9 e-pairs for nonbonding electrons on oxygens: total 24 e was used

No electrons left; carbon does not have an octet

Step 4. Place any remaining electrons on the central atom, even if more than an octet configuration results (only for atoms beyond the second row).

In our examples there were no electrons left. For examples where electrons are left over, see a separate page (in preparation).

Step 5. Check if all the atoms, and especially the central atom, have octets. If there are not enough electrons try to form multiple bonds.

Use unshared electrons to form double or triple bond in such a way as to give the central atom an octet. Formally move electrons from the lone-pairs to bonds.

lone electron pair on oxygen was moved to make a double bond to carbon

all atoms have now an octet (except hydrogens that have the required two electrons) .

lone electron pair on one fluorine was moved to make a double bond to boron

all atoms have now an octet; one fluorine exceeds its valence number and seems to be different than the others for no apparent reason

lone electron pairs on nitrogen and sulfur were moved to make double bonds to carbon

all atoms have now an octet

lone electron pair on one oxygen was moved to make a double bond to carbon

all atoms have now an octet, one oxygen seems to be different than the others for no apparent reason

Step 6. Assign formal charges to atoms in the Lewis structure.

The formal charge is the difference between the number of valence electrons of a given atom and the number of electrons assigned to it in a given Lewis structure. Each atom in the Lewis structure is assigned half of the electrons it shares, and all of its unshared electrons. For simplicity you may show only non-zero formal charges.
The overall charge of the species is shown outside of the square brackets, and the formal charges must add up to the overall charge of the species.


Formal charges: O: 6 - 6 = 0 C: 4 - 4 = 0 H: 1 - 1 = 0	Formal charges: F(d): 7 - 6 = 1 F(s): 7 - 7 = 0 B: 3 - 4 = -1 Net charge: 0	Formal charges: N: 5 - 6 = -1 C: 4 - 4 = 0 S: 6 - 6 = 0 Net charge: -1	Formal charges: O(d): 6 - 6 = 0 O(s): 6 - 7 = -1 C: 4 - 4 = 0 Net charge:-2

Step 7. Check if alternative Lewis structures are possible: find resonance structures.

Sometimes it is possible to draw several Lewis structures for a given atom connectivity. These structures differ only in electron distribution between atoms (and not in positioning of atoms). They are called resonance structures. When drawing these alternative structures, you may not move atoms, you may only assign electrons in alternative ways. The double-headed arrow (n) is used to show that the Lewis structures are resonance structures of each other.
When such a situation is encountered (i.e. when drawing more than one Lewis structure is possible for a given atomic arrangement), it means that the distribution of electrons in the molecule cannot be adequately presented with one picture. This is not because the molecule is unusual, but only because our pictures do not do an adequate job in representing molecules. The problem is that by drawing lines for electron pairs we rigidly assign them to predetermined spaces. In real molecules there are no such restrictions, electrons can be shared by more than two atoms, and can be shared to a varying degree.
When the molecule is described by using several (more than one) of such resonance structures, it does not mean that the molecule oscillates between the alternatives, but that the real structure is a superposition of these alternatives. There is only one structure with defined bond lengths and angles, but we need several drawings to represent it.
Molecules or ions that show resonance are more stable (because their electron sharing is better) than analogous molecules without the possibility of resonance.

	The resonance structures are obtained by assigning one electron pair of the double bond to oxygen or carbon. The second resonance structure leaves carbon without an octet. In the third resonance structure oxygen does not have an octet.		The resonance structures are obtained by moving one electron pair from a double bond to the "other" side, forming a triple bond. In all structures all atoms have octets. The structures are not equivalent.
	The top three resonance structures are obtained by forming double bonds to boron from fluorines (one at a time). They suggest that all fluorines are identical (they are). Top three structures are equivalent: they have octet for all atoms. In the bottom structure boron does not have an octet.		The top three resonance structures are obtained by forming double bonds to carbon from oxygens (one at a time). They show that all oxygens are identical. Top three structures are equivalent and have octet for all atoms. In the bottom structure carbon does not have an octet.

Step 8. Evaluate the relative importance of resonance structures.

The molecule may be described by several equivalent resonance structures, or by structures that are not equivalent. The equivalent structures contribute equally to the description of the molecule, but non-equivalent structures will have different contributions. Which structures are important? Which contribute to the description of the molecule more, and which are only a minor component? These questions are sometimes difficult to answer, but there are some general rules that help to sort the relative importance of various resonance structures.

Structures in which the second row atoms (B, C, N, O, F) have filled their valence shells (the octet rule) are generally more important than structures where the octet rule is not satisfied.

The non-octet structures are valid for third and lower row elements of the periodic table. The non-octet structures for the second row atoms may be valid, and even the most important, if in the corresponding octet-structures, the formal charges on atoms go against electronegativity of these atoms.

If alternative resonance structures have octets on all atoms, the structure with the smallest formal charges that are in agreement with the electronegativity trends are the most important.

Importance of the structure is decreased by an increase in charge separation. Structures containing formal charges of more than �2 on an atom usually contribute very little. Structures with two like charges on adjacent atoms are especially unfavorable.

Structures with negative charges assigned to a more electronegative atom are more important than those in which the charge is on a less electronegative atom. Similarly positive charges are best carried on atoms of low electronegativity.

Ordinarily, the structures with more covalent bonds are more important than those with fewer such bonds.

Structures with strongly distorted bond angles and lengths are not important (application of this rule requires knowledge of molecular geometry, a topic to be covered latter).

	The first structure is the most important (it contributes the most). It has octet for C and O and no formal charges. It also has one bond more than other structures. The second structure is less important. It has no octet on carbon, and has charge separation, but the formal charges are in agreement with the electronegativity of the elements (negative charge on oxygen). The third structure is the least important (it does not contribute to the description of the molecule). It has no octet on oxygen and has a charge separation that goes against electronegativity trend (positive formal charge on oxygen, negative on carbon).		In all structures all atoms have octets and equal number of bonds. The structure that will contribute the most is in the middle. It has the lowest possible charge residing on the most electronegative element (nitrogen). In the bottom structure the formal charge is on sulfur that is less electronegative. This structure would be second in importance. The top structure is the least important. It has the largest charge separation, and a positive charge on a relatively electronegative element (sulfur).
	Although the first three equivalent structures have octet for all the atoms, and one more bond than the bottom structure, they have charge separation that goes against the electronegativity of the elements involved. The most electronegative element (fluorine) is forced to have a formal positive charge. These structures do not contribute a lot to the description of the molecule. The last structure (without an octet on boron) is the most important.		The top three equivalent structures are the most important. They have octet for all atoms, minimal possible charges on the most electronegative element present, and one bond more than the bottom structure. The last structure also suffers from the larger charge separation and it does not have an octet on carbon. The last structure contributes less to the description of that ion than any of top structures.

Additional examples and special cases of Lewis structures are presented on a separate page (in preparation)