AP Chemistry: Lewis Structure Drawing

Mastering Lewis structures—diagrams that show the bonding between atoms and the lone pairs of electrons in a molecule—is a foundational skill in AP Chemistry. It provides the essential first step for predicting molecular shapes, bond polarities, reactivity, and hybridization. A systematic approach transforms what can seem like guesswork into a reliable, step-by-step procedure for mapping a molecule’s electron arrangement.

The Core Step-by-Step Method

The process for drawing a correct Lewis structure is a sequence of logical decisions. For a molecule like sulfur dioxide ($S{O}_2$), you would apply these steps methodically.

Step 1: Count All Valence Electrons This is the total pool of electrons you have to distribute. Valence electrons are the electrons in the outermost shell of an atom. For a neutral molecule, sum the valence electrons from each atom. For example, sulfur is in Group 16 and has 6 valence electrons. Oxygen is also in Group 16, so each oxygen contributes 6. The total for $S{O}_2$ is $6+(2\times 6)=18$ valence electrons. For polyatomic ions, you must adjust this count: add one electron for each negative charge, and subtract one for each positive charge.

Step 2: Connect Atoms with Single Bonds Identify the central atom, usually the least electronegative element (excluding hydrogen, which is always terminal). Place the other atoms around it, connecting each to the central atom with a single bond, which uses 2 electrons per bond. In $S{O}_2$, sulfur is central. We draw it with two oxygen atoms attached: S-O and S-O. These two single bonds use $2\times 2=4$ electrons, leaving $18-4=14$ electrons to place.

Step 3: Distribute Remaining Electrons to Complete Octets Place the leftover electrons as lone pairs on the outer (terminal) atoms first to satisfy the octet rule—the tendency of atoms to seek eight electrons in their valence shell. Each lone pair uses 2 electrons. Each oxygen in $S{O}_2$ needs 6 more electrons to reach an octet (they already have 2 from the single bond). Adding three lone pairs (6 electrons) to each oxygen uses $2\times 6=12$ electrons. We have placed $4+12=16$ electrons, leaving just 2 electrons remaining.

Step 4: Place Any Leftover Electrons on the Central Atom If you have electrons left after giving octets to the outer atoms, place them on the central atom as lone pairs. For $S{O}_2$, the final 2 electrons go on sulfur as a lone pair. At this stage, the central atom may not have an octet.

Step 5: Check and Convert Lone Pairs to Bonds to Satisfy Octets If the central atom lacks an octet, you must form multiple bonds. This is done by converting a lone pair from an outer atom into a bonding pair with the central atom, creating a double or triple bond. In our $S{O}_2$ sketch, sulfur only has 4 electrons (two single bonds and one lone pair). To solve this, we take a lone pair from one oxygen and convert the S-O single bond into a double bond. This uses the same number of electrons but gives sulfur two more. Even after creating one double bond, sulfur still only has 6 electrons. We repeat the process with the other oxygen, creating a second double bond. The final, stable Lewis structure shows sulfur double-bonded to both oxygens, with each oxygen having two lone pairs.

Evaluating Structures with Formal Charge

When multiple valid Lewis structures are possible, formal charge helps identify the most plausible one. It is a bookkeeping tool that compares an atom's number of valence electrons to the electrons assigned to it in the structure. $$FormalCharge=(Valence{e}^{-})-(Nonbonding{e}^{-}+\frac{1}{2}Bonding{e}^{-})$$ The most stable structure typically has:

1. Formal charges as close to zero as possible.
2. Any negative formal charge on the most electronegative atom.

For our $S{O}_2$ candidate with two double bonds, calculate formal charge:

• Sulfur: Valence = 6. Assigned electrons: 4 lone pair + $\frac{1}{2}(8bonding)=4+4=8$. Formal Charge = $6-8=-2$.
• Each Oxygen: Valence = 6. Assigned: 4 lone pair + $\frac{1}{2}(4bonding)=4+2=6$. Formal Charge = $6-6=0$.

This gives a total charge of -2, but $S{O}_2$ is a neutral molecule. The structure we actually observe is a hybrid of two resonance structures, where one bond is a double and the other is a single. This delocalization of electrons spreads out the formal charge, making the molecule more stable than any single drawing can depict.

Handling Exceptions to the Octet Rule

Reliably drawing structures requires recognizing when the standard octet does not apply. There are three primary categories of exceptions.

1. Expanded Octets Elements in Period 3 and below (e.g., phosphorus, sulfur, chlorine, xenon) can accommodate more than eight electrons in their valence shell because they have accessible d-orbitals. In sulfate ion ($S{O}_4^{2-}$), sulfur is surrounded by 12 electrons (four double bonds or a combination of bonds and lone pairs). The step-by-step method still works: the total valence electron count for $S{O}_4^{2-}$ is $6+(4\times 6)+2=32$. You'll find that satisfying the octets of the four oxygens forces sulfur to hold 12 electrons.

2. Electron-Deficient Species Some molecules, like boron trifluoride ($B{F}_3$), have a central atom with fewer than eight electrons. Boron has only 3 valence electrons. After forming three single bonds with fluorine atoms, boron is surrounded by just 6 electrons. It is stable this way because its small size and high charge density make it difficult to accommodate a lone pair.

3. Molecules with Odd Numbers of Electrons Free radicals, like nitrogen dioxide ($N{O}_2$), have an odd total electron count, making a perfect octet for all atoms impossible. In $N{O}_2$, with 17 valence electrons, one atom will inevitably have 7 electrons. In such cases, the Lewis structure will show one unpaired electron.

Common Pitfalls

Miscounting Valence Electrons for Ions: Forgetting to add or subtract electrons for the ion's charge is a frequent error. For the phosphate ion ($P{O}_4^{3-}$), you must calculate: P (5) + O (4 x 6) + 3 = 32 valence electrons. Drawing it with only 29 will make it impossible to complete the structure correctly.

Incorrect Central Atom Placement: Choosing hydrogen or the most electronegative atom (like fluorine or oxygen) as the central atom will lead to a dead end. The central atom is generally the one with the lowest electronegativity (other than H). In $Cl{O}_2^{-}$, chlorine is central, not oxygen.

Forcing an Octet on an Expanding Atom: Trying to cram sulfur into an 8-electron octet in $S{F}_6$ is impossible and ignores its ability to expand its valence. Conversely, forcing an octet onto an electron-deficient atom like beryllium (in $BeC{l}_2$) by adding fictional lone pairs is also incorrect. Let the electron count guide you.

Confusing Formal Charge with Oxidation State: These are different concepts. Formal charge assumes equal sharing in bonds, while oxidation state assumes complete electron transfer. In the nitrate ion ($N{O}_3^{-}$), the nitrogen has a formal charge of +1 in its resonance structures, but its oxidation state is +5. Use formal charge to evaluate the quality of a Lewis structure, not to discuss redox chemistry.

Summary

• The step-by-step method—count electrons, connect atoms with single bonds, complete outer atom octets, place leftovers on the central atom, and create multiple bonds as needed—provides a reliable framework for constructing any Lewis structure.
• Formal charge is a critical tool for choosing the most stable structure among possibilities, and resonance describes electron delocalization in molecules where multiple valid structures exist.
• Key exceptions to the octet rule include expanded octets for Period 3+ elements, electron-deficient molecules like $B{F}_3$, and odd-electron free radicals.
• Always double-check your work by verifying that the total electrons in the drawing match your initial count and that formal charges sum to the molecule's or ion's overall charge.