Lewis structures: the counting method that always works, with formal charge and resonance worked on the nitrate ion

To draw a Lewis structure, count before you draw: add up the valence electrons (plus one for each negative charge, minus one for each positive), connect the atoms with single bonds and subtract two electrons per bond, give the outer atoms their octets, and put whatever is left on the central atom. If the central atom is short of eight, move a lone pair from an outer atom into a double bond. Below, the nitrate ion is drawn this way line by line, its formal charges computed, its three resonance structures explained, and a sibling ion left for you.

Count first: the five steps

Most students who get stuck on how to draw Lewis structures are drawing before they count. They put dots around atoms until it looks right, run out of electrons or end with too many, and start over. The method below never starts over, because every step is arithmetic and the arithmetic tells you when you are done.

1. Count the valence electrons. Add the valence electrons of every atom (the group number for the main group: 1 for H, 4 for C, 5 for N, 6 for O, 7 for the halogens). Add one for each unit of negative charge; subtract one for each unit of positive charge. This number does not change for the rest of the problem.
2. Draw the skeleton with single bonds. The central atom is the least electronegative one that is not hydrogen, usually the one that appears once in the formula. Hydrogen is always an outer atom, and a halogen usually is (the exception is a halogen bonded to oxygen, as in the chlorate ion, ClO₃⁻). Each single bond uses 2 electrons; subtract them from the count.
3. Complete the outer atoms' octets. Each outer atom already has 2 electrons from its bond and needs 6 more as three lone pairs (hydrogen needs none). Subtract them.
4. Put what is left on the central atom as lone pairs.
5. If the central atom has fewer than eight, make a double bond. Move one lone pair from an outer atom into a second bond to the centre. The count is unchanged; the electrons have moved. Repeat if the centre is still short.

Then recount: bonds × 2 plus lone pairs × 2 must equal the number from step 1. If it does, the structure is complete.

In Ontario this is SCH4U's Ionic and covalent bonding; Lewis structures; electronegativity, in strand C; the SCH4U course map shows where it sits, and every chemistry idea worked, in course order is the section this page belongs to.

Worked example: the nitrate ion, NO₃⁻

Every line carries its count.

1. Count. N contributes 5, each O 6: 5 + 3 × 6 = 23. Add 1 for the single negative charge: 24 valence electrons.
2. Skeleton. Nitrogen is central (it appears once, and oxygen is more electronegative). Three N–O single bonds use 3 × 2 = 6 electrons. 24 − 6 = 18 left.
3. Outer octets. Each oxygen has 2 electrons from its bond and takes 6 more as three lone pairs: 3 × 6 = 18. 18 − 18 = 0 left.
4. The central atom. Nitrogen has three bonds, 6 electrons, short of eight, and there is nothing left to give it.
5. The fix. Move one lone pair from one oxygen into a second N–O bond. Nitrogen now has four bonds (one double, two single): 4 × 2 = 8. That oxygen has a double bond (4 electrons) and two lone pairs (4): 8. The other two oxygens are unchanged at a single bond and three lone pairs each.

Final count: the double bond 4, the two single bonds 4, the two lone pairs on the double-bonded oxygen 4, the six lone pairs on the two single-bonded oxygens 12. Total 4 + 4 + 4 + 12 = 24. Complete.

In words, since this page draws nothing: nitrogen in the centre; one oxygen joined by a double bond carrying two lone pairs; two oxygens each joined by a single bond carrying three lone pairs; no lone pair on the nitrogen. The ion carries a 1− charge overall.

The wrong lines. The first is counting 23, the charge forgotten. An odd count leaves one electron unpaired and no way to finish, and the student concludes the ion "cannot be drawn". The second is stopping at step 3 and handing in nitrogen with six electrons. Step 4 exists so that you look at the central atom before you stop; a central nitrogen, carbon or oxygen with fewer than eight is unfinished.

Formal charge, and why it tells you the structure is right

Formal charge is bookkeeping: it asks how many electrons each atom "owns" in the drawing compared with how many it brought.

Formal charge = valence electrons − nonbonding electrons − ½ × bonding electrons.

For the nitrate structure above:

• Nitrogen: 5 − 0 − 8 ÷ 2 = 5 − 0 − 4 = +1
• The double-bonded oxygen: 6 − 4 − 4 ÷ 2 = 6 − 4 − 2 = 0
• Each single-bonded oxygen: 6 − 6 − 2 ÷ 2 = 6 − 6 − 1 = −1, and there are two of them

Sum: +1 + 0 + (−1) + (−1) = −1, which is the ion's charge. That equality is the check, and it must hold for every structure you draw. When more than one structure satisfies the count, prefer the one with the smallest formal charges and with any negative charges on the more electronegative atoms; here the negative charges sit on oxygen, where they belong.

The wrong line. Halving every electron around the atom, lone pairs included, instead of only the bonding electrons. For the double-bonded oxygen that gives 6 − (4 + 4) ÷ 2 = +2 instead of 0, and for each single-bonded oxygen 6 − (6 + 2) ÷ 2 = +2 instead of −1. The sum comes to +1 + 2 + 2 + 2 = +7 for an ion whose charge is −1, so the check catches it at once. Lone-pair electrons belong wholly to their atom and are subtracted whole; only the shared ones are halved.

Resonance: three drawings, one ion

The double bond in nitrate can be drawn to any of the three oxygens, and the three drawings are equally good: same count, same formal charges, same shape. They are resonance structures.

The real ion is not any one of them. It is the average of the three. Each N–O bond is a double bond in one drawing and a single bond in the other two, so its bond order is (2 + 1 + 1) ÷ 3 = 4/3, a little more than a single bond. All three bonds are the same length, and that length is shorter than a typical N–O single bond and longer than a typical N=O double bond, which is the measured evidence that the average, not any one drawing, is the ion.

What resonance is not: the ion does not flip between the drawings. The drawings are three incomplete pictures of one thing, the way three photographs from different sides are three pictures of one building. The shape every structure on this page takes comes from the count of electron domains, and resonance does not change it, because each drawing has the same number of domains around the nitrogen: three, trigonal planar.

The wrong line. Drawing one structure and marking the other two "wrong", or drawing all three and writing that the ion "resonates" between them. The first loses the resonance marks; the second describes a motion that does not happen.

When the octet rule bends

The octet rule is a rule for carbon, nitrogen, oxygen and fluorine. Four cases bend it, and a Grade 12 test asks about the first three.

Hydrogen takes two, not eight. It has one orbital to fill.

Boron is content with six. BF₃: 3 + 3 × 7 = 24 electrons; three B–F bonds use 6; the three fluorines take 18 as nine lone pairs; nothing is left and boron sits at six. Do not make a double bond here; the structure with boron at six and every atom at formal charge 0 is the accepted one.

Period-3 and heavier central atoms can hold more than eight, because they have room. SF₆: 6 + 6 × 7 = 48 electrons, six S–F bonds (12) and eighteen lone pairs on the fluorines (36), sulfur with twelve electrons around it. PCl₅: 5 + 5 × 7 = 40, five bonds (10) and fifteen lone pairs (30), phosphorus with ten.

Odd-electron molecules cannot give everyone an octet. Nitrogen monoxide, NO, has 5 + 6 = 11 electrons, and one atom is left with seven however you draw it. This fourth case is a note, not a test question, at this level.

Lewis structures on one screen

• Count first: valence electrons, +1 per negative charge, −1 per positive. The number never changes.
• Skeleton: the least electronegative non-hydrogen atom in the centre; single bonds; 2 electrons each.
• Outer octets, then the centre, then double bonds if the centre is short.
• Recount: bonds × 2 + lone pairs × 2 = the count from step 1.
• Formal charge = valence − nonbonding − ½ bonding. The sum equals the charge, every time.
• Resonance: several drawings, one average; bond order = total bonds ÷ positions.
• Bends: H takes 2; B is content with 6; period 3 and below can exceed 8.

Why one ion is worked in full before any is set: Sweller and Cooper (1985) found that students who studied worked examples and then solved matching problems learned the procedure faster than students who only solved problems. The matching problem is next, and chapter 3 of the free playbook is the habit of redoing it cold.

Try one cold: the carbonate ion, CO₃²⁻

Cover everything above. Draw the carbonate ion by the five steps, in words if you like: the count, which atom carries the double bond, the lone pairs on each oxygen. Compute the formal charge on every atom and check the sum. Say how many resonance structures there are and the bond order. Then a shorter one: sulfur dioxide, SO₂, count and structure only.

Check

• Count: C 4 + 3 × O 6 = 22, plus 2 for the 2− charge: 24 electrons. Same skeleton and same fix as nitrate: one C=O double bond, two C–O single bonds, two lone pairs on the double-bonded oxygen, three on each single-bonded one, none on carbon.
• Formal charges: C 4 − 0 − 8 ÷ 2 = 0; the double-bonded O 6 − 4 − 2 = 0; each single-bonded O 6 − 6 − 1 = −1. Sum 0 + 0 − 1 − 1 = −2, the ion's charge.
• Three resonance structures; bond order (2 + 1 + 1) ÷ 3 = 4/3.
• SO₂: 6 + 2 × 6 = 18 electrons. Sulfur central with one S=O double bond, one S–O single bond and one lone pair on the sulfur; two lone pairs on the double-bonded oxygen, three on the single-bonded one. Count: 4 + 2 + 2 + 4 + 6 = 18. Two resonance structures. Three domains on sulfur with one of them a lone pair, so the molecule is bent.

If your carbonate count was 22, the charge was forgotten; if carbon came out +1, a lone pair was halved.

Where this sits in the course

In the SCH4U pack this is idea 8 of 24, Ionic and covalent bonding; Lewis structures; electronegativity, in strand C. It follows Electron configurations and periodic trends, which is where the valence count comes from: the electrons in the outermost shell of each atom's configuration are the ones you count in step 1. It is followed by Molecular shape (VSEPR) and polarity, where the Lewis structure is the input to the shape, and by Intermolecular forces and physical properties, where the chain from polarity to boiling point starts from the structure and the shape together. The organic strand's structures are Lewis structures with the lone pairs left off; the functional groups page draws them that way.

How to practise it

Draw ten structures from the formula alone: neutral molecules, ions and at least one that needs a double bond, mixed so that you cannot tell from the order which kind is next. Check each one by the electron count and the formal-charge sum rather than against an answer key. That matters for two reasons. The count and the sum are checks you will have in the exam room, where the key is not. And checking by rule is a test of the method, which Roediger and Karpicke (2006) found builds longer-lasting recall than rereading a finished drawing does. How to test yourself honestly has the marking rules, and how to study chemistry as procedures has the routine for a whole unit of this kind.