1. Lewis Dot Structure Practice Worksheet
2. Lewis Electron Dot Structure Calculator Worksheet
3. Lewis Electron Dot Structure Calculator

What is a Lewis Diagram?

A Lewis electron dot diagram (or electron dot diagram or a Lewis diagram or a Lewis structure) is a representation of the valence electrons of an atom that uses dots around the symbol of the element. The number of dots equals the number of valence electrons in the atom. These dots are arranged to the right and left and above and below the. Given descriptions, diagrams, scenarios, or chemical symbols, students will model covalent bonds using electron dot formula (Lewis structures).

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 are

where 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

are two hydrogen atoms and one oxygen atom. The Lewis structure of each of these atoms would be as follows:

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^-).

Answer: Since there are only two atoms, we can begin with just a bond between the two atoms.

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:

1. Hydronium ion (H₃O⁺)
2. Hypochlorite ion (ClO^-)
3. Carbonate ion (CO₃^-2)
4. Ammonia (NH₃)
5. Hydrogen fluoride (HF)
6. Ozone (O₃)
7. Xenon difluoride (XeF₂)

Using Lewis Dot Symbols to Describe Covalent Bonding

This sharing of electrons allowing atoms to “stick” together is the basis of covalent bonding. There is some intermediate distant, generally a bit longer than 0.1 nm, or if you prefer 100 pm, at which the attractive forces significantly outweigh the repulsive forces and a bond will be formed if both atoms can achieve a completen s²np⁶ configuration. It is this behavior that Lewis captured in his octet rule. The valence electron configurations of the constituent atoms of a covalent compound are important factors in determining its structure, stoichiometry, and properties. For example, chlorine, with seven valence electrons, is one electron short of an octet. If two chlorine atoms share their unpaired electrons by making a covalent bond and forming Cl₂, they can each complete their valence shell:

Each chlorine atom now has an octet. The electron pair being shared by the atoms is called a bonding pair ; the other three pairs of electrons on each chlorine atom are called lone pairs. Lone pairs are not involved in covalent bonding. If both electrons in a covalent bond come from the same atom, the bond is called a coordinate covalent bond.

We can illustrate the formation of a water molecule from two hydrogen atoms and an oxygen atom using Lewis dot symbols:

The structure on the right is the Lewis electron structure, or Lewis structure, for H₂O. With two bonding pairs and two lone pairs, the oxygen atom has now completed its octet. Moreover, by sharing a bonding pair with oxygen, each hydrogen atom now has a full valence shell of two electrons. Chemists usually indicate a bonding pair by a single line, as shown here for our two examples:

The following procedure can be used to construct Lewis electron structures for more complex molecules and ions:

1. Arrange the atoms to show specific connections. When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl₄ and CO₃²⁻, which both have C as the central atom), which is another clue to the compound’s structure. Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central.

Note:

The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal.

1. Determine the total number of valence electrons in the molecule or ion. Add together the valence electrons from each atom. (Recall from Chapter 2 that the number of valence electrons is indicated by the position of the element in the periodic table.) If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion. For CO₃²⁻, for example, we add two electrons to the total because of the −2 charge.
2. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. In H₂O, for example, there is a bonding pair of electrons between oxygen and each hydrogen.
3. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). These electrons will usually be lone pairs.
4. If any electrons are left over, place them on the central atom. Some atoms are able to accommodate more than eight electrons.
5. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. This will not change the number of electrons on the terminal atoms.

Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed.

H₂O

1. Because H atoms are almost always terminal, the arrangement within the molecule must be HOH.

2. Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons

3. Placing one bonding pair of electrons between the O atom and each H atom gives H:O:H, with 4 electrons left over.

4. Each H atom has a full valence shell of 2 electrons.

5. Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure:

This is the Lewis structure we drew earlier. Because it gives oxygen an octet and each hydrogen two electrons, we do not need to use step 6.

1. With only two atoms in the molecule, there is no central atom.

2. Oxygen (group 16) has 6 valence electrons, and chlorine (group 17) has 7 valence electrons; we must add one more for the negative charge on the ion, giving a total of 14 valence electrons.

3. Placing a bonding pair of electrons between O and Cl gives O:Cl, with 12 electrons left over.

4. If we place six electrons (as three lone pairs) on each atom, we obtain the following structure:

Each atom now has an octet of electrons, so steps 5 and 6 are not needed. The Lewis electron structure is drawn within brackets as is customary for an ion, with the overall charge indicated outside the brackets, and the bonding pair of electrons is indicated by a solid line. OCl⁻ is the hypochlorite ion, the active ingredient in chlorine laundry bleach and swimming pool disinfectant.

1. Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. One possible arrangement is as follows:

2. Each hydrogen atom (group 1) has one valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons.

3. Placing a bonding pair of electrons between each pair of bonded atoms gives the following:

Six electrons are used, and 6 are left over.

4. Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following:

Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons.

5. There are no electrons left to place on the central atom.

6. To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond:

Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid.

An alternative structure can be drawn with one H bonded to O. Formal charges, discussed later in this section, suggest that such a structure is less stable than that shown previously.

Example

Write the Lewis electron structure for each species.

1. NCl₃
2. S₂²⁻
3. NOCl

Given: chemical species

Asked for: Lewis electron structures

Strategy:

Use the six-step procedure to write the Lewis electron structure for each species.

Show Answer

Nitrogen trichloride is an unstable oily liquid once used to bleach flour; this use is now prohibited in the United States.

1. In a diatomic molecule or ion, we do not need to worry about a central atom. Each sulfur atom (group 16) contains 6 valence electrons, and we need to add 2 electrons for the −2 charge, giving a total of 14 valence electrons. Using 2 electrons for the S–S bond, we arrange the remaining 12 electrons as three lone pairs on each sulfur, giving each S atom an octet of electrons:

2. Because nitrogen is less electronegative than oxygen or chlorine, it is the central atom. The N atom (group 15) has 5 valence electrons, the O atom (group 16) has 6 valence electrons, and the Cl atom (group 17) has 7 valence electrons, giving a total of 18 valence electrons. Placing one bonding pair of electrons between each pair of bonded atoms uses 4 electrons and gives the following:

   Adding three lone pairs each to oxygen and to chlorine uses 12 more electrons, leaving 2 electrons to place as a lone pair on nitrogen:

3. Because this Lewis structure has only 6 electrons around the central nitrogen, a lone pair of electrons on a terminal atom must be used to form a bonding pair. We could use a lone pair on either O or Cl. Because we have seen many structures in which O forms a double bond but none with a double bond to Cl, it is reasonable to select a lone pair from O to give the following:

   All atoms now have octet configurations. This is the Lewis electron structure of nitrosyl chloride, a highly corrosive, reddish-orange gas.

Example

Write Lewis electron structures for CO₂ and SCl₂, a vile-smelling, unstable red liquid that is used in the manufacture of rubber.

It is sometimes possible to write more than one Lewis structure for a substance that does not violate the octet rule, as we saw for CH₂O, but not every Lewis structure may be equally reasonable. In these situations, we can choose the most stable Lewis structure by considering the formal charge on the atoms, which is the difference between the number of valence electrons in the free atom and the number assigned to it in the Lewis electron structure. The formal charge is a way of computing the charge distribution within a Lewis structure; the sum of the formal charges on the atoms within a molecule or an ion must equal the overall charge on the molecule or ion. A formal charge does not represent a true charge on an atom in a covalent bond but is simply used to predict the most likely structure when a compound has more than one valid Lewis structure.

To calculate formal charges, we assign electrons in the molecule to individual atoms according to these rules:

• Nonbonding electrons are assigned to the atom on which they are located.
• Bonding electrons are divided equally between the bonded atoms.

For each atom, we then compute a formal charge:

formal charge = valence e⁻−(free atom)(non−bonding e⁻ + bonding e⁻/2)

To illustrate this method, let’s calculate the formal charge on the atoms in ammonia (NH₃) whose Lewis electron structure is as follows:

A neutral nitrogen atom has five valence electrons (it is in group 15). From its Lewis electron structure, the nitrogen atom in ammonia has one lone pair and shares three bonding pairs with hydrogen atoms, so nitrogen itself is assigned a total of five electrons [2 nonbonding e⁻ + (6 bonding e⁻/2)]. Substituting into the below equation, we obtain:

formal charge(N)= 5 valence e^–−(2non−bonding e⁻ + 6 bonding e⁻/2)=0

A neutral hydrogen atom has one valence electron. Each hydrogen atom in the molecule shares one pair of bonding electrons and is therefore assigned one electron [0 nonbonding e⁻ + (2 bonding e⁻/2)]. Using the below equation to calculate the formal charge on hydrogen, we obtain:

formal charge(H)= 1 valence e^–−(0 non−bonding e⁻ + 2 bonding e⁻/2)=0

The hydrogen atoms in ammonia have the same number of electrons as neutral hydrogen atoms, and so their formal charge is also zero. Adding together the formal charges should give us the overall charge on the molecule or ion. In this example, the nitrogen and each hydrogen has a formal charge of zero. When summed the overall charge is zero, which is consistent with the overall charge on the NH₃ molecule.

Lewis Dot Structure Practice Worksheet

Typically, the structure with the most charges on the atoms closest to zero is the more stable Lewis structure. In cases where there are positive or negative formal charges on various atoms, stable structures generally have negative formal charges on the more electronegative atoms and positive formal charges on the less electronegative atoms. The next example further demonstrates how to calculate formal charges.

Example

Calculate the formal charges on each atom in the NH₄⁺ ion.

Lewis Electron Dot Structure Calculator Worksheet

Given: chemical species

Asked for: formal charges

Strategy:

Identify the number of valence electrons in each atom in the NH₄⁺ ion. Use the Lewis electron structure of NH₄⁺ to identify the number of bonding and nonbonding electrons associated with each atom and then use the given formula to calculate the formal charge on each atom.

Show Answer

The Lewis electron structure for the NH₄⁺ ion is as follows:

The nitrogen atom shares four bonding pairs of electrons, and a neutral nitrogen atom has five valence electrons. Using the formula, the formal charge on the nitrogen atom is

formal charge(N)=5−(0+8/2)=1

Each hydrogen atom in has one bonding pair. The formal charge on each hydrogen atom is therefore

formal charge(H)=1−(0+2/2)=0

The formal charges on the atoms in the NH₄⁺ ion are thus

Adding together the formal charges on the atoms should give us the total charge on the molecule or ion. In this case, the sum of the formal charges is 0 + 1 + 0 + 0 + 0 = +1.

Example

Write the formal charges on all atoms in BH₄⁻.

Show Answer

If an atom in a molecule or ion has the number of bonds that is typical for that atom (e.g., four bonds for carbon), its formal charge is zero.

Using Formal Charges to Distinguish between Lewis Structures

As an example of how formal charges can be used to determine the most stable Lewis structure for a substance, we can compare two possible structures for CO₂. Both structures conform to the rules for Lewis electron structures.

1. C is less electronegative than O, so it is the central atom.

2. C has 4 valence electrons and each O has 6 valence electrons, for a total of 16 valence electrons.

3. Placing one electron pair between the C and each O gives O–C–O, with 12 electrons left over.

4. Dividing the remaining electrons between the O atoms gives three lone pairs on each atom:

This structure has an octet of electrons around each O atom but only 4 electrons around the C atom.

5. No electrons are left for the central atom.

6. To give the carbon atom an octet of electrons, we can convert two of the lone pairs on the oxygen atoms to bonding electron pairs. There are, however, two ways to do this. We can either take one electron pair from each oxygen to form a symmetrical structure or take both electron pairs from a single oxygen atom to give an asymmetrical structure:

Both Lewis electron structures give all three atoms an octet. How do we decide between these two possibilities? The formal charges for the two Lewis electron structures of CO₂ are as follows:

Both Lewis structures have a net formal charge of zero, but the structure on the right has a +1 charge on the more electronegative atom (O). Thus the symmetrical Lewis structure on the left is predicted to be more stable, and it is, in fact, the structure observed experimentally. Remember, though, that formal charges do not represent the actual charges on atoms in a molecule or ion. They are used simply as a bookkeeping method for predicting the most stable Lewis structure for a compound.

Note:

Lewis Electron Dot Structure Calculator

The Lewis structure with the set of formal charges closest to zero is usually the most stable.

Examples

The thiocyanate ion (SCN⁻), which is used in printing and as a corrosion inhibitor against acidic gases, has at least two possible Lewis electron structures. Draw two possible structures, assign formal charges on all atoms in both, and decide which is the preferred arrangement of electrons.

Given: chemical species

Asked for: Lewis electron structures, formal charges, and preferred arrangement

Strategy:

A Use the step-by-step procedure to write two plausible Lewis electron structures for SCN⁻.

B Calculate the formal charge on each atom using formal charge = valence e⁻−(free atom)(non−bonding e⁻ + bonding e⁻/2)

C Predict which structure is preferred based on the formal charge on each atom and its electronegativity relative to the other atoms present.

Show Answer

A Possible Lewis structures for the SCN⁻ ion are as follows:

B We must calculate the formal charges on each atom to identify the more stable structure. If we begin with carbon, we notice that the carbon atom in each of these structures shares four bonding pairs, the number of bonds typical for carbon, so it has a formal charge of zero. Continuing with sulfur, we observe that in (a) the sulfur atom shares one bonding pair and has three lone pairs and has a total of six valence electrons. The formal charge on the sulfur atom is therefore 6−(6+2/2)=−1 and 5−(4+4/2)=−1. In (c), nitrogen has a formal charge of −2.

C Which structure is preferred? Structure (b) is preferred because the negative charge is on the more electronegative atom (N), and it has lower formal charges on each atom as compared to structure (c): 0, −1 versus +1, −2.

Example

Salts containing the fulminate ion (CNO⁻) are used in explosive detonators. Draw three Lewis electron structures for CNO⁻ and use formal charges to predict which is more stable. (Note: N is the central atom.)

Show Answer

The second structure is predicted to be more stable.

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