Lewis dot structures also called electron dot structures are diagrams that describe the chemical bonding between atoms in a molecule. They also display the total number of lone pairs present in each of the atoms that constitute the molecule. Lewis dot structures are commonly referred to as electron dot structures or Lewis structures. In 1916, ten years before the Schrodinger wave equation, G. Lewis suggested that a chemical bond involved sharing of electrons. He described what he called the cubical atom, because a cube has 8 corners, to represent the outer valence shell electrons which can be shared to create a bond.This was his octet rule. Rules for drawing Lewis dot structures.
- Lewis Electron Dot Structure Calculator Worksheet
- Lewis Electron Dot Structure Calculator Form
- Lewis Electron Dot Structure Calculator Solver
What is a Lewis Diagram?
Lewis diagrams, also called electron-dot diagrams, are used to represent paired and unpaired valence (outer shell) electrons in an atom. For example, the Lewis diagrams for hydrogen, helium, and carbon arewhere the symbol represents the element (in this case, hydrogen, helium, and carbon) and the dots represent the electrons in the outer shell (in this case, one, two, and four). These diagrams are based on the electron structures learned in the Atomic Structure and Periodic Table chapters.
What is a Lewis Structure?
The Lewis structure is used to represent the covalent bonding of a molecule or ion. Covalent bonds are a type of chemical bonding formed by the sharing of electrons in the valence shells of the atoms. Covalent bonds are stronger than the electrostatic interactions of ionic bonds, but keep in mind that we are not considering ionic compounds as we go through this chapter. Most bonding is not purely covalent, but is polar covalent (unequal sharing) based on electronegativity differences.
The atoms in a Lewis structure tend to share electrons so that each atom has eight electrons (the octet rule). The octet rule states that an atom in a molecule will be stable when there are eight electrons in its outer shell (with the exception of hydrogen, in which the outer shell is satisfied with two electrons). Lewis structures display the electrons of the outer shells because these are the ones that participate in making chemical bonds.
How to Build a Lewis Structure?
For simple molecules, the most effective way to get the correct Lewis structure is to write the Lewis diagrams for all the atoms involved in the bonding and adding up the total number of valence electrons that are available for bonding. For example, oxygen has 6 electrons in the outer shell, which are the pattern of two lone pairs and two singles. If the electrons are not placed correctly, one could think that oxygen has three lone pairs (which would not leave any unshared electrons to form chemical bonds). After adding the four unshared electrons around element symbol, form electron pairs using the remaining two outer shell electrons.
| Incorrect Structure | Correct Structure |
One good example is the water molecule. Water has the chemical formula of H2O, which means there
We can now see that we have eight valence electrons (six from oxygen and one from each hydrogen). With few exceptions, hydrogen atoms are always placed on the outside of the molecule, and in this case the central atom would be oxygen. Each of the two unpaired electrons of the oxygen atom will form a bond with one of the unpaired electrons of the hydrogen atoms. The bonds formed by the shared electron pairs can be represented by either two closely places dots between two element symbols or more commonly by a straight line between element symbols:
Let us try another one.
Example: Write the Lewis structure for methane (CH4).
Answer:Hydrogen atoms are always placed on the outside of the molecule, so carbon should be the central atom.
After counting the valence electrons, we have a total of 8 [4 from carbon + 4(1 from each hydrogen] = 8.
Each hydrogen atom will be bonded to the carbon atom, using two electrons. The four bonds represent the eight valence electrons with all octets satisfied, so your structure is complete.
Example: Write the Lewis structure for carbon dioxide (CO2).
Answer:Carbon is the lesser electronegative atom and should be the central atom.
After counting the valence electrons, we have a total of 16 [4 from carbon + 2(6 from each oxygen)] = 16.
Each oxygen atom has two unshared electrons that can be used to form a bond with two unshared electrons of the carbon atom, forming a double bond between the two atoms. The remaining eight electrons will be place on the oxygen atoms, with two lone pairs on each.
Lewis Structures of Polyatomic Ions
Building the Lewis Structure for a polyatomic ion can be done in the same way as with other simple molecules, but we have to consider that we will need to adjust the total number of electrons for the charge on the polyatomic ion. If the ion has a negative charge, the number of electrons that is equal to the charge on the ion should be added to the total number of valence electrons. If the ion has a positive charge, the number of electrons that is equal to the charge should be subtracted from the total number of valence electrons. After writing the structure, the entire structure should then be placed in brackets with the charge on the outside of the brackets at the upper right corner.
Example: Write the Lewis structure for the ammonium ion (NH4+).
Answer: Hydrogen atoms are always placed on the outside of the molecule, so nitrogen should be the central atom.
After counting the valence electrons, we have a total of 9 [5 from nitrogen + 4(1 from each hydrogen)] = 9. The charge of +1 means an electron should be subtracted, bringing the total electron count to 8.
Each hydrogen atom will be bonded to the nitrogen atom, using two electrons. The four bonds represent the eight valence electrons with all octets satisfied, so your structure is complete. (Do not forget your brackets and to put your charge on the outside of the brackets)
Example: Write the Lewis structure for the hydroxide ion (OH-).
Lewis Electron Dot Structure Calculator Worksheet
After counting the valence electrons, we have a total of 7 [6 from Oxygen + 1 from each Hydrogen)] = 7. The charge of -1 indicates an extra electron, bringing the total electron count to 8.
Oxygen will be bonded to the hydrogen, using two electrons. Place the remaining six electrons as three lone pairs on the oxygen atom. All octets are satisfied, so your structure is complete. (Do not forget your brackets and to put your charge on the outside of the brackets)
Lewis Structures for Resonance Structures
The existence of some molecules often involves two or more structures that are equivalent. Resonance can be shown using Lewis structures to represent the multiple forms that a molecule can exist. The molecule is not switching between these forms, but is rather an average of the multiple forms. This can be seen when multiple atoms of the same type surround the central atom. When all lone pairs are placed on the structure, all the atoms may still not have an octet of electrons. To deal with this problem, the atoms (primarily in a C, N, or O formula) form double or triple bonds by moving lone pairs to form a second or third bond between two atoms. The atom that originally had the lone pair does not lose its octet because it is sharing its lone pair. Double-headed arrows are placed between the multiple structures of the molecule or ion to show resonance. Let us look at how to build a nitrate ion (NO3-).
Nitrogen is the least electronegative atom and should be the central atom.
After counting the valence electrons, we have a total of 23[5 from nitrogen + 3(6 from each oxygen)] = 23. The charge of -1 indicates an extra electron, bringing the total electron count to 24.
Each oxygen atom will be bonded to the nitrogen atom, using a total of six electrons. We then place the remaining 18 electrons initially as 9 lone pairs on the oxygen atoms (3 pairs around each atom).
Although all 24 electrons are represented in the structure (two electrons for each of the three bonds and 18 for each of the nine lone pairs), the octet for the nitrogen atom is not satisfied. To satisfy the octet rule for the nitrogen atom, a double bond needs to be made between the nitrogen and one of the oxygen atoms. Because of the symmetry of the molecule, it does not matter which oxygen atoms is chosen. Because there are three different oxygen atoms that could form the double bond, there will be three different resonance structures showing each oxygen atom with a double bond to the nitrogen atom. Double-headed arrows will be placed between these three structures. (Do not forget your brackets and to put your charge on the outside of the brackets)
Example: What is the Lewis structure for the nitrite ion (NO2−)?
Answer: Nitrogen is the least electronegative atom and should be the central atom.
After counting the valence electrons, we have a total of 17 [5 from nitrogen + 2(6 from each oxygen)] = 17. The charge of -1 indicates an extra electron, bringing the total electron count to 18.
Each oxygen will be bonded to the nitrogen, using two electrons. Place the remaining 16 electrons initially as nine lone pairs on the oxygen atoms (3 pairs around each atom) and the nitrogen (one pair).
Although all 18 electrons are represented in the structure (2 electrons for each of the two bonds and 14 for each of the seven lone pairs), the octet for the nitrogen atom is not satisfied. To satisfy the octet rule for the nitrogen atom, a double bond needs to be made between the nitrogen atom and one of the oxygen atoms. Because of the symmetry of the molecule, it does not matter which oxygen is chosen. Because there are two different oxygen atoms that could form the double bond, there will be two different resonance structures showing each oxygen atom with a double bond to the nitrogen atom. A double-headed arrow will be placed between these structures. (Do not forget your brackets and to put your charge on the outside of the brackets)
Lewis Structures for Electron-rich Compounds
Elements with atomic number greater than 13 often form compounds or polyatomic ions in which there are “extra” electrons. For these compounds we proceed as above. Once all of the octets are satisfied, the extra electrons are assigned to the central atom either as lone pairs or an increase in the number of bonds. (Never use multiple bonds with these compounds—you already have too many electrons.)Example: Draw the Lewis structure for phosphorus pentafluoride, PF5.
Answer:The electronegativity of fluorine is greater than that of phosphorus—so the phosphorus atom is placed in the center of the molecule.The total number of electros is 40 [5(7 from each fluorine) + 5 from the phosphorus] = 40. Using a single bond between the phosphorus atom and each of the fluorine atoms and filling the remaining electrons to satisfy the octet rule for the fluorine atoms accounts for all 40 electrons. Note that there are five bonds around the central atom.
Lewis Structures for Electron-poor Compounds
There is another type of molecule or polyatomic ion in which there is an electron deficiency of one or more electrons needed to satisfy the octets of all the atoms. In these cases, the more electronegative atoms are assigned as many electrons to complete those octets first and then the deficiency is assigned to the central atom.Example: Draw the Lewis structure for boron trifluoride, BF3.
Answer:The electronegativity of fluorine is greater than that of boron—so the boron atom is placed in the center of the molecule.The total number of electron is 24 [3(7 from each fluorine) + 3 from boron] = 24. Using a single bond between the boron and each of the fluorine atoms and filling the remaining electron as lone pairs around the fluorine atoms to satisfy the octets accounts for all 24 electrons.
The boron atom is two electrons shy of its octet. You may ask about the formation of a double bond (and even resonance). But, fluorine and boron are not in the list that can form double bonds (C, N, O, P, S) and so the compound is electron poor.
Try It Out!
Draw the Lewis structure for the following:
- Hydronium ion (H3O+)
- Hypochlorite ion (ClO-)
- Carbonate ion (CO3-2)
- Ammonia (NH3)
- Hydrogen fluoride (HF)
- Ozone (O3)
- Xenon difluoride (XeF2)
| Writing Lewis Structures by Trial and Error | A Step-By-Step Approach to Writing Lewis Structures | Drawing Skeleton Structures |
| Molecules that Contain Too Many or Not Enough Electrons | Resonance Hybrids | Formal Charge |
Writing Lewis Structures by Trial and Error
The Lewis structure of a compound can be generated by trial and error. We start by writing symbols that contain the correct number of valence electrons for the atoms in the molecule. We then combine electrons to form covalent bonds until we come up with a Lewis structure in which all of the elements (with the exception of the hydrogen atoms) have an octet of valence electrons.
Example: Let's apply the trial and error approach to generating the Lewis structure of carbon dioxide, CO2. We start by determining the number of valence electrons on each atom from the electron configurations of the elements. Carbon has four valence electrons, and oxygen has six.
C: [He] 2s2 2p2
O: [He] 2s2 2p4
We can symbolize this information as shown at the top of the figure below. We now combine one electron from each atom to form covalent bonds between the atoms. When this is done, each oxygen atom has a total of seven valence electrons and the carbon atom has a total of six valence electrons. Because none of these atoms have an octet of valence electrons, we combine another electron on each atom to form two more bonds. The result is a Lewis structure in which each atom has an octet of valence electrons.
The trial-and-error method for writing Lewis structures can be time consuming. For all but the simplest molecules, the following step-by-step process is faster.
Step 1: Determine the total number of valence electrons.
Step 2: Write the skeleton structure of the molecule.
Step 3: Use two valence electrons to form each bond in the skeleton structure.
Step 4: Try to satisfy the octets of the atoms by distributing the remaining valence electrons as nonbonding electrons.
The first step in this process involves calculating the number of valence electrons in the molecule or ion. For a neutral molecule this is nothing more than the sum of the valence electrons on each atom. If the molecule carries an electric charge, we add one electron for each negative charge or subtract an electron for each positive charge.
Example: Let's determine the number of valence electrons inthe chlorate (ClO3-) ion.
A chlorine atom (Group VIIA) has seven valence electrons and each oxygen atom (Group VIA) has six valence electrons. Because the chlorate ion has a charge of -1, this ion contains one more electron than a neutral ClO3 molecule. Thus, the ClO3- ion has a total of 26 valence electrons.
ClO3-: 7 + 3(6) + 1 = 26
The second step in this process involves deciding which atoms in the molecule are connected by covalent bonds. The formula of the compound often provides a hint as to the skeleton structure. The formula for the chlorate ion, for example, suggests the following skeleton structure.
The third step assumes that the skeleton structure of the molecule is held together by covalent bonds. The valence electrons are therefore divided into two categories: bonding electrons and nonbonding electrons. Because it takes two electrons to form a covalent bond, we can calculate the number of nonbonding electrons in the molecule by subtracting two electrons from the total number of valence electrons for each bond in the skeleton structure.
There are three covalent bonds in the most reasonable skeleton structure for the chlorate ion. As a result, six of the 26 valence electrons must be used as bonding electrons. This leaves 20 nonbonding electrons in the valence shell.
| 26 valence electrons |
| - 6 bonding electrons |
| 20 nonbonding electrons |
The nonbonding valence electrons are now used to satisfy the octets of the atoms in the molecule. Each oxygen atom in the ClO3- ion already has two electrons the electrons in the Cl-O covalent bond. Because each oxygen atom needs six nonbonding electrons to satisfy its octet, it takes 18 nonbonding electrons to satisfy the three oxygen atoms. This leaves one pair of nonbonding electrons, which can be used to fill the octet of the central atom.
The most difficult part of the four-step process in the previous section is writing the skeleton structure of the molecule. As a general rule, the less electronegative element is at the center of the molecule.
Example: The formulas of thionyl chloride (SOCl2) and sulfuryl chloride (SO2Cl2) can be translated into the following skeleton structures.
It is also useful to recognize that the formulas for complex molecules are often written in a way that hints at the skeleton structure of the molecule.
Example: Dimethyl ether is often written as CH3OCH3, which translates into the following skeleton structure.
Finally, it is useful to recognize that many compounds that are acids contain O-H bonds.
Example: The formula of acetic acid is often written as CH3CO2H, because this molecule contains the following skeleton structure.
Too Few Electrons
Occasionally we encounter a molecule that doesn't seem to have enough valence electrons. If we can't get a satisfactory Lewis structure by sharing a single pair of electrons, it may be possible to achieve this goal by sharing two or even three pairs of electrons.
Example: Consider formaldehyde (H2CO) which contains 12 valence electrons.
H2CO: 2(1) + 4 + 6 = 12
The formula of this molecule suggests the following skeleton structure.
There are three covalent bonds in this skeleton structure, which means that six valence electrons must be used as bonding electrons. This leaves six nonbonding electrons. It is impossible, however, to satisfy the octets of the atoms in this molecule with only six nonbonding electrons. When the nonbonding electrons are used to satisfy the octet of the oxygen atom, the carbon atom has a total of only six valence electrons.
Lewis Electron Dot Structure Calculator Form
We therefore assume that the carbon and oxygen atoms share two pairs of electrons. There are now four bonds in the skeleton structure, which leaves only four nonbonding electrons. This is enough, however, to satisfy the octets of the carbon and oxygen atoms.
Every once in a while, we encounter a molecule for which it is impossible to write a satisfactory Lewis structure.
Example: Consider boron trifluoride (BF3) which contains 24 valence electrons.
BF3: 3 + 3(7) = 24
There are three covalent bonds in the most reasonable skeleton structure for the molecule. Because it takes six electrons to form the skeleton structure, there are 18 nonbonding valence electrons. Each fluorine atom needs six nonbonding electrons to satisfy its octet. Thus, all of the nonbonding electrons are consumed by the three fluorine atoms. As a result, we run out of electrons while the boron atom has only six valence electrons.
The elements that form strong double or triple bonds are C, N, O, P, and S. Because neither boron nor fluorine falls in this category, we have to stop with what appears to be an unsatisfactory Lewis structure.
Too Many Electrons
It is also possible to encounter a molecule that seems to have too many valence electrons. When that happens, we expand the valence shell of the central atom.
Lewis Electron Dot Structure Calculator Solver
Example: Consider the Lewis structure for sulfur tetrafluoride (SF4) which contains 34 valence electrons.
SF4: 6 + 4(7) = 34
There are four covalent bonds in the skeleton structure for SF4. Because this requires using eight valence electrons to form the covalent bonds that hold the molecule together, there are 26 nonbonding valence electrons.
Each fluorine atom needs six nonbonding electrons to satisfy its octet. Because there are four of these atoms, so we need 24 nonbonding electrons for this purpose. But there are 26 nonbonding electrons in this molecule. We have already satisfied the octets for all five atoms, and we still have one more pair of valence electrons. We therefore expand the valence shell of the sulfur atom to hold more than eight electrons.
This raises an interesting question: How does the sulfur atom in SF4 hold 10 electrons in its valence shell? The electron configuration for a neutral sulfur atom seems to suggest that it takes eight electrons to fill the 3s and 3p orbitals in the valence shell of this atom. But let's look, once again, at the selection rules for atomic orbitals. According to these rules, the n = 3 shell of orbitals contains 3s, 3p, and 3d orbitals. Because the 3d orbitals on a neutral sulfur atom are all empty, one of these orbitals can be used to hold the extra pair of electrons on the sulfur atom in SF4.
S: [Ne] 3s2 3p4 3d0
| Practice Problem 3: Write the Lewis structure for xenon tetrafluoride (XeF4). |
Two Lewis structures can be written for sulfur dioxide.
The only difference between these Lewis structures is the identity of the oxygen atom to which the double bond is formed. As a result, they must be equally satisfactory representations of the molecule.
Interestingly enough, neither of these structures is correct. The two Lewis structures suggest that one of the sulfur-oxygen bonds is stronger than the other. There is no difference between the length of the two bonds in SO2, however, which suggests that the two sulfur-oxygen bonds are equally strong.
When we can write more than one satisfactory Lewis structure, the molecule is an average, or resonance hybrid, of these structures. The meaning of the term resonance can be best understood by an analogy. In music, the notes in a chord are often said to resonate they mix to give something that is more than the sum of its parts. In a similar sense, the two Lewis structures for the SO2 molecule are in resonance. They mix to give a hybrid that is more than the sum of its components. The fact that SO2 is a resonance hybrid of two Lewis structures is indicated by writing a double-headed arrow between these Lewis structures, as shown in the figure above.
| Practice Problem 4: Write the Lewis structures for the acetate ion, CH3CO2-. |
Formal Charge
It is sometimes useful to calculate the formal charge on each atom in a Lewis structure. The first step in this calculation involves dividing the electrons in each covalent bond between the atoms that form the bond. The number of valence electrons formally assigned to each atom is then compared with the number of valence electrons on a neutral atom of the element. If the atom has more valence electrons than a neutral atom, it is assumed to carry a formal negative charge. If it has fewer valence electrons it is assigned a formal positive charge.
| Practice Problem 5: The formula of the amino acid known as glycine is often written as H3N+CH2CO2-. Use the concept of formal charge to explain the meaning of the positive and negative signs in the following Lewis structure. |