Intuition What this page isThe parent note gave you the algorithm . This page is the drill ground : we walk through every kind of molecule the octet rule can throw at you, so that when an exam gives you a formula you have never seen, you already know which "cell" it belongs to.
We use exactly one toolkit the whole way — four counting steps and one formal-charge formula — but we push it against every awkward case: neutral, charged, too few electrons, too many, an odd number, and a real-world twist.
Before any example, let us restate the machinery in plain words so no symbol is used before it is defined .
Definition The five quantities we count
N total total number of valence electrons the atoms bring to the table. "Valence electrons" = the electrons in the outermost shell of an atom, the only ones that do bonding. We read this off the periodic group number.N need if each sat alone happy (8 for most atoms, 2 for hydrogen). It is a wish-list, not reality.N shared = N need − N total shared (put into bonds). Why subtract? A shared pair is counted by both atoms, so sharing "creates" the electrons the wish-list is short by.bonds = N shared /2 pair = 2 electrons.Leftover / lone-pair electrons N lone = N total − N shared not used up in bonds. Why this subtraction? Every electron either goes into a shared bond or stays home as a lone pair, so whatever survives after the bonds are drawn must sit as lone pairs. We place these leftovers as pairs (2 e⁻ each), filling the octets of the outer atoms first, then piling any remainder on the central atom. The count L one particular atom .F C formal charge on one atom = V − L − 2 1 B V L B
Mnemonic Read the group, get the valenceGroup 1 → 1 valence e⁻, Group 2 → 2, Group 13 → 3, Group 14 → 4, Group 15 → 5, Group 16 → 6, Group 17 → 7, Group 18 → 8. (For the old "1–8" column numbering: the column number IS the valence count for main-group atoms.)
Every Lewis-structure problem lives in one of these cells. Our worked examples below are tagged with the cell they cover, and together they touch all of them .
Cell
Case class
What makes it tricky
Example that hits it
C1 Neutral, obeys octet
baseline — sanity check the method
H₂O (Ex 1)
C2 Neutral, multiple bonds needed
N shared N₂ (Ex 2)
C3 Anion (charge correction, add e⁻)must add electrons for negative charge
OH⁻ (Ex 3)
C4 Cation (charge correction, subtract e⁻)must remove electrons for positive charge
NH₄⁺ (Ex 4)
C5 Resonance / formal-charge tie-break
several valid sketches; pick best
NO₃⁻ (Ex 5)
C6 Incomplete octet (too few e⁻)arithmetic "wants" a bond that FC forbids
BeCl₂ (Ex 6)
C7 Expanded octet (too many e⁻)central atom holds >8
XeF₄ (Ex 7)
C8 Odd-electron (degenerate/impossible octet)odd total → unpaired e⁻
NO₂ (Ex 8)
C9 Real-world word problem
translate a story into a structure
ozone air-quality (Ex 9)
C10 Exam twist (which is wrong ?)
spot the illegal expanded octet
N vs P (Ex 10)
Worked example Draw the Lewis structure ofH 2 O Forecast: guess first — how many bonds and how many lone pairs on oxygen?
N total O ( 6 ) + 2 × H ( 1 ) = 8 Why this step? Every dot we draw must be a real electron; H is group 1 (1 e⁻), O is group 16 (6 e⁻).N need 8 + 2 × 2 = 12 Why? Hydrogen chases helium's duet, not an octet.N shared = 12 − 8 = 4 = 4/2 = 2 Why? Two O–H single bonds. That matches the two hydrogens exactly.Lone pairs: N lone = N total − N shared = 8 − 4 = 4 2 lone pairs , both on oxygen.
Why? Every electron not spent in a bond must stay home as a lone pair; hydrogen is already full at 2, so the 4 leftovers pile onto the central oxygen.
This is the picture we are building — look at figure s01: the two mint H atoms bonded to the coral O, with the two lavender lone pairs drawn as dot-pairs on top of oxygen. Trace each O–H line to one shared pair, then count the four leftover dots as the two lone pairs.
Verify: O has 2 bonds (4 + 4 = 8 F C = 6 − 4 − 2 1 ( 4 ) = 0
N 2 Forecast: single, double, or triple bond between the two nitrogens?
N total 2 × N ( 5 ) = 10 Why? N is group 15 → 5 valence each.N need 8 × 2 = 16 Why? Two non-H atoms, both want octets.N shared = 16 − 10 = 6 = 3 Why? All 3 bonds sit on the single N–N link → a triple bond .Lone pairs: N lone = N total − N shared = 10 − 6 = 4 2 lone pairs , one on each N.
Why? The 6 bonding electrons are locked in the triple bond; the 4 that remain cannot vanish, so they settle as one lone pair per nitrogen — the only atoms left to hold them.
Look at figure s02: the two lavender N atoms joined by the triple bond (three parallel slate lines), with one mint lone pair pointing away from each nitrogen. The three lines are the 6 shared electrons; the two outward pairs are the 4 leftovers.
Verify: each N: triple bond (6 + 2 = 8 F C = 5 − 2 − 2 1 ( 6 ) = 0
Worked example Draw the hydroxide ionO H − Forecast: does the extra electron change the number of bonds, or just the lone pairs?
N total O ( 6 ) + H ( 1 ) + 1 (charge) = 8 Why the + 1 A charge of − 1 one extra electron the ion carries. Anion → add, cation → subtract.N need 8 + 2 = 10 N shared = 10 − 8 = 2 = 1 Lone pairs: N lone = 8 − 2 = 6 3 lone pairs on oxygen.
Why? Only 2 electrons went into the single bond; the remaining 6 must lodge as lone pairs, and hydrogen is full, so all three pairs sit on oxygen.
Verify: O: 1 bond (2 + 6 = 8 F C = 6 − 6 − 2 1 ( 2 ) = − 1 − 1 Electronegativity ).
Worked example Draw the ammonium ionN H 4 + Forecast: will nitrogen have a lone pair left over?
N total N ( 5 ) + 4 × H ( 1 ) − 1 (charge) = 8 Why the − 1 Charge + 1 missing one electron . Cation → subtract.N need 8 + 4 × 2 = 16 N shared = 16 − 8 = 8 = 4 Lone pairs: N lone = 8 − 8 = 0 no lone pair on nitrogen.
Why? All eight electrons were consumed by the four bonds, so nothing remains to place as lone pairs.
Verify: N: 4 bonds (8 = 8 F C = 5 − 0 − 2 1 ( 8 ) = + 1 + 1 F C = 1 − 0 − 2 1 ( 2 ) = 0
N O 3 − Forecast: three O's, but how many double bonds? And where?
N total N ( 5 ) + 3 × O ( 6 ) + 1 (charge) = 24 N need 8 × 4 = 32 N shared = 32 − 24 = 8 = 4 Why? Three N–O links must share 4 bonds → one N=O double, two N–O single .Lone pairs: N lone = 24 − 8 = 16 Why? The 8 bonding electrons are fixed; the 16 survivors distribute as lone pairs on the oxygens, giving the double-bonded O two pairs and each single-bonded O three pairs.Formal charges: double-bonded O: 6 − 4 − 2 1 ( 4 ) = 0 6 − 6 − 2 1 ( 2 ) = − 1 5 − 0 − 2 1 ( 8 ) = + 1 = 0 + ( − 1 ) + ( − 1 ) + ( + 1 ) = − 1 Resonance: the double bond can sit on any of the three oxygens equally. The true ion is the average of the three sketches — see Formal charge & resonance .
Look at figure s03: three copies of the ion side by side, the double bond (twin slate lines) parked on a different oxygen in each, linked by mint double-headed arrows. Those arrows are the resonance flip — no atom moves, only the double bond slides — so all three N–O bonds are really the same averaged bond.
Verify: total FC = − 1 = 4 bonds /3 links = 4/3 ≈ 1.33
B e C l 2 Forecast: does beryllium reach an octet, or fall short?
N total B e ( 2 ) + 2 × C l ( 7 ) = 16 Why? Be is group 2 → 2 valence; Cl is group 17 → 7.N need 8 × 3 = 24 N shared = 24 − 16 = 8 = 4 suggests 4 bonds — two Be=Cl doubles.Steel-man the trap, then reject it: a Be=Cl double bond would give F C ( Be ) = 2 − 0 − 2 1 ( 8 ) = − 2 F C ( Cl ) = 7 − 4 − 2 1 ( 4 ) = + 1 Positive charge on chlorine (very electronegative) is absurd. Fix: keep only 2 single Be–Cl bonds . Beryllium ends with just 4 electrons — an incomplete octet .
Verify: with singles, F C ( Be ) = 2 − 0 − 2 1 ( 4 ) = 0 F C ( Cl ) = 7 − 6 − 2 1 ( 2 ) = 0 Lewis acid .
X e F 4 Forecast: how many electrons crowd around xenon — 8, 10, or 12?
N total X e ( 8 ) + 4 × F ( 7 ) = 36 Why? Xe is group 18 → 8 valence; F is group 17 → 7.N need 8 × 5 = 40 N shared = 40 − 36 = 4 = 2 4 Xe–F bonds (four fluorines to attach). The naive count under-shoots because xenon will hold more than 8 .Build directly: make 4 Xe–F single bonds (8 24 N lone,Xe = 36 − 8 − 24 = 4 2 lone pairs on Xe .Count around Xe: 4 bonds (8 + 4 = 12 expanded octet , allowed because Xe is in period 5 with accessible vacant d Hypervalency & d-orbital participation , Periodic trends ).
Look at figure s04: the butter-coloured Xe in the middle, four mint F atoms bonded around it in a cross, each F wearing its three lone pairs, and the two lavender lone pairs sitting on Xe itself. Count around the centre: four bonds plus those two extra pairs give the twelve-electron crowd.
Verify: F C ( Xe ) = 8 − 4 − 2 1 ( 8 ) = 0 F C ( F ) = 7 − 6 − 2 1 ( 2 ) = 0 12 > 8
N O 2 Forecast: can every atom get a full even octet here?
N total N ( 5 ) + 2 × O ( 6 ) = 17 An odd number!
Why does odd matter? Electrons pair up two-at-a-time to make bonds and lone pairs; an odd total means one electron cannot find a partner .N need 8 × 3 = 24 Why? Three non-H atoms, all wishing for octets.N shared = 24 − 17 = 7 = 7/2 = 3.5 Why the half? The odd electron count breaks the neat "shared pairs" arithmetic — you cannot have half a bond. Round down to 3 whole bonds (one N=O double, one N–O single) and carry the surviving single electron separately.Lone pairs + the lone electron: bonds use 6 N lone = 17 − 6 = 11 1 unpaired electron is left over on nitrogen.
Why on N, not O? Placing the odd electron on N gives F C ( N ) = 0 Electronegativity ).Formal-charge check: with N=O double, N–O single, and the lone e⁻ on N — F C ( N ) = 5 − 1 − 2 1 ( 6 ) = 1 − 2 1 ( 6 ) … L = 1 B = 6 F C ( N ) = 5 − 1 − 3 = + 1 F C ( N ) = 5 − 3 − 2 1 ( 4 ) = 0 Resonance: just like nitrate, the N=O double flips between the two oxygens → NO₂ has two equivalent resonance forms , so both N–O bonds are identical (bond order 1.5
Verify: 17 17 mod 2 = 1 paramagnetic radical , which is why it is coloured and reactive. Bond order = 3/2 = 1.5
Worked example Air-quality sensors report ground-levelozone, O 3 , a lung irritant. An engineer needs its bonding to model how it absorbs UV. Draw its Lewis structure and give the average bond order.
Forecast: are the two O–O bonds identical or different?
Translate the story → a formula: ozone = O 3 N total 3 × O ( 6 ) = 18 N need 8 × 3 = 24 N shared = 24 − 18 = 6 = 3 one double, one single .Lone pairs: N lone = 18 − 6 = 12 FC: central O: F C = 6 − 2 − 2 1 ( 6 ) = + 1 6 − 4 − 2 1 ( 4 ) = 0 6 − 6 − 2 1 ( 2 ) = − 1 = + 1 + 0 − 1 = 0 Resonance: the double bond flips between the two ends → both O–O bonds are really identical, bond order = 3/2 = 1.5 . This "in-between" bond is exactly what lets ozone absorb UV — the delocalised electrons are easy to excite.
Verify: total FC = 0 = 3/2 = 1.5
Worked example A student drawsP C l 5 and N C l 5 Forecast: which central atom is forbidden from 5 bonds?
N total P N 5 + 5 × 7 = 40 N need 8 × 6 = 48 N shared = 48 − 40 = 8 = 4 demands 5 bonds → the central atom must expand past 8 . The arithmetic itself flags an expanded octet.Apply the FC formula to the 5-bond structure: central atom with 5 single bonds → F C = 5 − 0 − 2 1 ( 10 ) = 0 = 7 − 6 − 2 1 ( 2 ) = 0 perfect — so FC alone would happily allow both molecules. The tie-breaker is not charge, it is orbital availability .P C l 5 period 3 ; it has accessible vacant 3 d can physically hold the 10 > 8 expanded octet . Real molecule.N C l 5 period 2 . Its valence shell has only 2 s 2 p at most 8 electrons. There is no low-energy 2 d it can never hold 10 . N C l 5
Verify: N total = 40 10 F C ( centre ) = 5 − 0 − 2 1 ( 10 ) = 0 = 3 ≥ 3 = 2 < 3 parent note 's "N and O never expand" rule in action.
Recall Which cell is each formula in? (test yourself)N H 4 + B e C l 2 N O 2 X e F 4 O 3 N C l 5 d
Recall The one-line procedureCount N total N need − N total = N shared N total − N shared F C
Formal charge & resonance — the tie-breaker used in Ex 5 and Ex 9.Electronegativity — why negative FC sits on O/Cl (Ex 3, Ex 6).Lewis acids and bases — the incomplete-octet Be/B species (Ex 6).Hypervalency & d-orbital participation — expanded octets (Ex 7, Ex 10).Free radicals — odd-electron NO₂ (Ex 8).Periodic trends — the period-≥3 expansion rule (Ex 10).VSEPR theory — the next step: these structures → 3-D shapes. 
