How To Make A Lewis Dot Structure

The Lewis dot structure, also known as the Lewis structure or electron dot diagram, is a visual representation of the valence electrons of atoms within a molecule. It illustrates how these electrons are arranged around individual atoms in a molecule and how they contribute to the chemical bonds between the atoms. Understanding how to construct a Lewis dot structure is fundamental to grasping chemical bonding, molecular geometry, and reactivity.

Why Lewis Dot Structures Matter

Lewis dot structures provide a simplified method for visualizing the electronic structure of molecules. They offer insights into:

• Bonding: They show how atoms share electrons to form covalent bonds.
• Lone Pairs: They display the non-bonding electrons (lone pairs) that influence a molecule's properties.
• Molecular Shape: Although not directly, they hint at the molecular geometry which impacts chemical reactions and physical properties.
• Reactivity: They help predict how a molecule might react with other substances.

Prerequisites

Before diving into the steps of drawing Lewis dot structures, make sure you have a grasp on these basic concepts:

• Atoms and Elements: Understanding the basic building blocks of matter.
• Electrons: Knowing the role of electrons, especially valence electrons, in chemical bonding.
• Valence Electrons: Being able to determine the number of valence electrons for each element.
• Octet Rule: Understanding that atoms "want" to have eight valence electrons (except for hydrogen, which wants two).

Step-by-Step Guide to Drawing Lewis Dot Structures

Let's break down the process of drawing Lewis structures into clear, manageable steps. We'll use examples to illustrate each step.

Step 1: Determine the Total Number of Valence Electrons

This is the crucial first step. You need to know the total "electron budget" for the molecule.

• Identify each element in the molecule.
• Find the number of valence electrons for each element. This is usually the same as the element's group number on the periodic table (for main group elements).
• Multiply the number of valence electrons for each element by the number of atoms of that element in the molecule.
• Sum up the valence electrons from all the atoms.
• If you are dealing with a polyatomic ion, add electrons for negative charges and subtract electrons for positive charges.

**Example 1: Carbon Dioxide (CO₂) **

• Carbon (C): Group 14 (6), so 4 valence electrons
• Oxygen (O): Group 16 (6), so 6 valence electrons
• Total valence electrons: (1 C atom * 4 valence electrons/C atom) + (2 O atoms * 6 valence electrons/O atom) = 4 + 12 = 16 valence electrons

Example 2: Sulfate Ion (SO₄²⁻)

• Sulfur (S): Group 16 (6), so 6 valence electrons
• Oxygen (O): Group 16 (6), so 6 valence electrons
• Charge: -2, so add 2 electrons
• Total valence electrons: (1 S atom * 6 valence electrons/S atom) + (4 O atoms * 6 valence electrons/O atom) + 2 (from the 2- charge) = 6 + 24 + 2 = 32 valence electrons

Step 2: Draw the Skeletal Structure

This step involves arranging the atoms in a way that shows which atoms are connected to each other.

• Identify the central atom: The central atom is usually the least electronegative element (excluding hydrogen). Carbon is almost always a central atom. If you have multiple atoms of the same element, the one present in the smallest number is usually the central atom.
• Connect the atoms: Draw single bonds (lines representing shared electron pairs) between the central atom and the surrounding atoms.

Example 1: Carbon Dioxide (CO₂)

• Central atom: Carbon (C)
• Skeletal structure: O - C - O

Example 2: Sulfate Ion (SO₄²⁻)

• Central atom: Sulfur (S)

• Skeletal structure:

  O | O - S - O | O

Step 3: Distribute Electrons to Outer Atoms

Now it's time to start placing the valence electrons.

• Complete the octets (or duets for hydrogen) of the outer atoms: Add lone pairs (pairs of dots) around each outer atom until it has eight electrons (or two for hydrogen). Remember, each single bond already represents two electrons.
• Subtract the used electrons: Subtract the number of electrons used in bonding and lone pairs from the total number of valence electrons you calculated in Step 1.

Example 1: Carbon Dioxide (CO₂)

• Start with O - C - O
• Add lone pairs to oxygen atoms: :O - C - O: becomes :Ö - C - Ö: (Each oxygen now has 8 electrons: 2 from the bond and 6 from the lone pairs)
• Electrons used: 4 bonds * 2 electrons/bond + 12 lone pair electrons = 16 electrons
• Remaining electrons: 16 (total) - 16 (used) = 0

Example 2: Sulfate Ion (SO₄²⁻)

• Start with:

  O | O - S - O | O

• Add lone pairs to oxygen atoms:

  :Ö: | :Ö - S - Ö: | :Ö:

• Electrons used: 4 bonds * 2 electrons/bond + 24 lone pair electrons = 32 electrons

• Remaining electrons: 32 (total) - 32 (used) = 0

Step 4: Place Remaining Electrons on the Central Atom

If you still have electrons left after completing the octets of the outer atoms, place them as lone pairs on the central atom.

Example: Let's modify CO₂ to make it illustrative.

Imagine we incorrectly calculated CO₂ to have 18 valence electrons. After step 3, we'd have 2 electrons remaining. We would then place those two electrons as a lone pair on the carbon atom:

:Ö - C: - Ö: with a lone pair on Carbon. (This is INCORRECT for CO₂, but illustrates the point)

Step 5: Form Multiple Bonds If Necessary

If, after placing all the electrons, the central atom does not have an octet, you'll need to form multiple bonds (double or triple bonds).

• Move lone pairs: Move one or more lone pairs from an outer atom to form a double or triple bond with the central atom.
• Prioritize achieving octets: Focus on getting an octet around the central atom.

Example 1: Carbon Dioxide (CO₂) - Continuing from Step 3

• We had :Ö - C - Ö: with 0 remaining electrons.
• Carbon only has 4 electrons (2 from each single bond). It needs 4 more to achieve an octet.
• Move one lone pair from each oxygen to form double bonds: Ö=C=Ö
• Now, each oxygen has 2 lone pairs and 2 bonds (8 electrons), and carbon has 4 bonds (8 electrons).

Example 2: Let's consider Carbon Monoxide (CO)

• Total valence electrons: 4 (C) + 6 (O) = 10
• Skeletal structure: C - O
• Add lone pairs to oxygen: C - Ö:
• Electrons used: 2 (bond) + 6 (lone pairs on O) = 8
• Remaining electrons: 10 - 8 = 2
• Place remaining electrons on carbon: :C - Ö:
• Now, carbon has 3 electrons and oxygen has 7.
• Form a triple bond: :C≡Ö:
• To fulfill octets, add a lone pair to each atom: :C≡Ö:
• This is the correct Lewis structure for carbon monoxide.

Step 6: Calculate Formal Charges (Optional but Recommended)

Formal charge helps to determine the most stable Lewis structure when multiple structures are possible.

For more on this topic, read our article on x 2 2 3x 6 or check out write the fraction that is represented by the x.

• Formula: Formal Charge = (Valence Electrons) - (Non-bonding Electrons) - (1/2 Bonding Electrons)
• Goal: Minimize formal charges. The most stable structure generally has formal charges as close to zero as possible. Negative formal charges should be on the most electronegative atoms.

Example 1: Carbon Dioxide (CO₂) - Ö=C=Ö

• Carbon: FC = 4 (valence) - 0 (non-bonding) - 1/2(8 bonding) = 4 - 0 - 4 = 0
• Oxygen: FC = 6 (valence) - 4 (non-bonding) - 1/2(4 bonding) = 6 - 4 - 2 = 0

All formal charges are zero, making this a very stable Lewis structure.

Example 2: Ozone (O₃)

• Total valence electrons: 3 * 6 = 18
• Possible Structure 1: Ö=Ö-Ö:
• Possible Structure 2: :Ö-Ö=Ö

Let's calculate formal charges for Structure 1:

• Oxygen (double bond): FC = 6 - 4 - 1/2(4) = 0
• Oxygen (single bond): FC = 6 - 6 - 1/2(2) = -1
• Oxygen (central): FC = 6 - 2 - 1/2(6) = +1

Now for Structure 2: It's just the reverse!

Since ozone is symmetrical, the actual structure is a resonance hybrid, meaning the true structure is an average of the two. Both oxygens have a partial double bond character.

Step 7: Consider Resonance Structures

Sometimes, more than one valid Lewis structure can be drawn for a molecule or ion. These are called resonance structures.

• Delocalization: Resonance structures represent the delocalization of electrons, meaning the electrons are spread out over a larger area.
• Representing Resonance: Resonance structures are connected by a double-headed arrow (↔).
• True Structure: The true structure of the molecule is a hybrid of all resonance structures.

Example: Ozone (O₃) - continued

As we saw above, Ö=Ö-Ö: and :Ö-Ö=Ö are both valid Lewis structures. The real ozone molecule is a hybrid of these two, where the bond length between all three oxygen atoms is actually the same.

Example: Benzene (C₆H₆)

Benzene has two major resonance structures: alternating single and double bonds within the ring. This electron delocalization gives benzene its unique stability.

Exceptions to the Octet Rule

While the octet rule is a useful guideline, there are exceptions:

• Incomplete Octet: Some atoms, like beryllium (Be) and boron (B), can be stable with fewer than eight valence electrons. Take this: in BF₃, boron only has six valence electrons.
• Expanded Octet: Atoms in the third row of the periodic table and beyond (e.g., S, P, Cl) can accommodate more than eight valence electrons. This is because they have access to d orbitals. Examples include SF₆ and PCl₅.
• Odd Number of Electrons: Molecules with an odd number of valence electrons (called free radicals) cannot satisfy the octet rule for all atoms. An example is nitrogen monoxide (NO).

Tips for Drawing Accurate Lewis Structures

• Be Neat and Organized: Use a pencil and eraser. Draw clear dots and lines.
• Double-Check Your Work: Recount the number of valence electrons and make sure all atoms have the correct number of electrons around them.
• Practice, Practice, Practice: The more you practice, the easier it will become to draw Lewis structures.
• Use the Periodic Table: Keep a periodic table handy to quickly find the number of valence electrons for each element.
• Memorize Common Exceptions: Be aware of common exceptions to the octet rule.
• Consider Formal Charge: Use formal charge to distinguish between possible Lewis structures.
• Don't be Afraid to Erase: It's common to make mistakes. Erase and try again!

Common Mistakes to Avoid

• Incorrectly Counting Valence Electrons: This is the most common mistake. Double-check your work!
• Forgetting Lone Pairs: Make sure to include all lone pairs around the atoms.
• Giving Hydrogen More Than Two Electrons: Hydrogen can only hold two electrons.
• Violating the Octet Rule Unnecessarily: Only violate the octet rule if it is necessary (e.g., with elements in the third row or beyond).
• Ignoring Formal Charges: Not considering formal charges can lead to incorrect structures.

Examples: Putting it All Together

Let's work through a few more examples to solidify your understanding.

Example 1: Hydrogen Cyanide (HCN)

1. Valence Electrons: 1 (H) + 4 (C) + 5 (N) = 10
2. Skeletal Structure: H - C - N
3. Distribute Electrons: H - C - :N: (Nitrogen gets 6 lone pair electrons)
4. Remaining Electrons: 10 - 2 (bond) - 6 (N lone pairs) = 2. Place on carbon: H-:C-:N:
5. Octet Rule: Hydrogen has 2, Carbon has 4, Nitrogen has 6. Carbon needs 4 more electrons and Nitrogen needs 2 more.
6. Multiple Bonds: Form a triple bond between C and N. H-C≡N:
7. Final Structure: H-C≡N: (Carbon and Nitrogen both have octets; Hydrogen has its duet.)

Example 2: Phosphorus Pentachloride (PCl₅)

1. Valence Electrons: 5 (P) + 5 * 7 (Cl) = 40

2. Skeletal Structure:

   Cl | Cl - P - Cl | Cl | Cl

3. Distribute Electrons: Add lone pairs to each chlorine atom. Each Cl gets 3 lone pairs for a total of 6 electrons (plus the 2 from the bond = 8)

4. Octet Rule: All chlorine atoms have an octet. Phosphorus has 10 electrons around it (expanded octet).

5. Formal Charge: P has a formal charge of +1 if you assume it must obey the octet rule, but it's more stable with the expanded octet.

Example 3: Nitrate Ion (NO₃⁻)

1. Valence Electrons: 5 (N) + 3 * 6 (O) + 1 (charge) = 24

2. Skeletal Structure:

   O | O - N - O

3. Distribute Electrons: Add lone pairs to each oxygen atom.

   :Ö: | :Ö - N - Ö:

4. Octet Rule/Remaining Electrons: All Oxygens have octets. Nitrogen only has 6. Took long enough.

5. Multiple Bonds: Form a double bond with one of the oxygen atoms.

Ö=N-Ö | Ö

6. Resonance: The double bond can be in any of the three positions:

Ö=N-Ö Ö-N=Ö Ö-N-Ö | | | Ö Ö Ö

The actual nitrate ion is a resonance hybrid where each N-O bond has a bond order of 1 1/3.

Conclusion

Drawing Lewis dot structures is a fundamental skill in chemistry. Remember to follow the steps carefully, pay attention to the octet rule (and its exceptions), and use formal charges to help you determine the most stable structure. While it may seem challenging at first, with practice and a clear understanding of the steps involved, you can master this essential tool. With these tools, you'll gain a deeper understanding of chemical bonding and molecular structure.