Octet rule — Lewis structures, exceptions (incomplete octet, expanded octet, odd-electron species): Exercises

Definition The three numbers we count (rebuilt from zero)

Everything on this page uses three counts. Let's define each in plain words, tied to a picture.

Valence electrons of an atom = the electrons in its outermost shell. For a main-group atom this equals its group number's ones-digit (group 14 → 4, group 16 → 6, group 17 → 7). These are the only electrons that bond.
$N_{total}$ = the grand total of valence electrons the whole molecule brought to the table. Every dot and every line you draw must be paid for by one of these.
$N_{need}$ = the electrons needed if every atom stood alone and demanded a full shell: $8$ per non-H atom, $2$ per H atom.

Then the magic subtraction: $N_{shared} = N_{need} - N_{total}, number of bonds = \frac{N _{shared}}{2} .$ Why this works: a shared pair is counted twice (once for each atom) in $N_{need}$ but paid for once in $N_{total}$ . The gap between "need" and "have" is exactly the double-counting caused by sharing — so it equals the shared electrons.

Figure — Octet rule — Lewis structures, exceptions (incomplete octet, expanded octet, odd-electron species)

Level 1 — Recognition

Recall Solution L1.1

What we do: add each atom's group-valence, then correct for charge (add for negative, subtract for positive). Why: every electron we draw must be a real electron.

(a) $H_{2} O$ : $2 (1) + 6 = 8$ .
(b) $N H_{3}$ : $5 + 3 (1) = 8$ .
(c) $S O_{4}^{2 -}$ : $6 + 4 (6) + 2 = 32$ . The " $+ 2$ " is because the $2 -$ charge means two extra electrons.

Recall Solution L1.2

$B e C l_{2}$ : Be brings only 4 valence electrons → central Be ends with 4 → incomplete octet.
$P C l_{5}$ : 5 P–Cl bonds → 10 electrons on P → expanded octet (P is period 3, has vacant $d$ ).
$N O_{2}$ : $5 + 2 (6) = 17$ , an odd total → odd-electron radical.
$C H_{4}$ : $4 + 4 (1) = 8$ , four C–H bonds give C exactly 8 → obeys octet.

Level 2 — Application

Recall Solution L2.1

$N_{total} = 4 + 2 (6) = 16$ .
$N_{need} = 8 \times 3 = 24$ (three non-H atoms).
$N_{shared} = 24 - 16 = 8 \Rightarrow bonds = 8/2 = 4$ .
What 4 bonds mean: only two C–O connections exist, so $4$ bonds split as $2 + 2$ → two double bonds, $O = C = O$ .
Octet check: each O has 1 double bond ( $4$ shared e⁻) $+$ 2 lone pairs ( $4$ e⁻) $= 8$ ✓. C has two double bonds $= 8$ ✓.

Recall Solution L2.2

$N_{total} = 5 + 4 (1) - 1 = 8$ . The " $- 1$ " is because the $+ 1$ charge means one electron removed.
$N_{need} = 8 (N) + 2 \times 4 (H) = 8 + 8 = 16$ .
$N_{shared} = 16 - 8 = 8 \Rightarrow bonds = 4$ .
Four N–H single bonds. N: $4$ bonds $= 8$ e⁻ ✓ (octet). Each H: 1 bond $= 2$ e⁻ ✓ (duet). No lone pairs left ( $8 - 8 = 0$ ).

Recall Solution L2.3

$N_{total} = 3 + 3 (7) = 24$ ; $N_{need} = 8 \times 4 = 32$ ; bonds $= (32 - 24) /2 = 4$ .
The trap: 4 bonds over 3 links suggests one B=F double bond, which would give B a full octet.
Formal-charge test of that double-bond structure: on B, $F C = 3 - 0 - \frac{1}{2} (8) = 3 - 4 = - 1$ ; on the double-bonded F, $F C = 7 - 4 - \frac{1}{2} (4) = 7 - 4 - 2 = + 1$ .
A $+ 1$ on fluorine (the most electronegative atom of all) is energetically absurd. Correct structure: three plain B–F single bonds; B keeps only $6$ electrons — incomplete octet. B is electron-deficient, a Lewis acid.

Level 3 — Analysis

Recall Solution L3.1

$N_{total} = 6 + 4 + 5 + 1 = 16$ ; $N_{need} = 24$ ; bonds $= (24 - 16) /2 = 4$ . Four bonds over two links (O–C, C–N).

Structure (I): $O = C = N$ (double–double). O has 2 lone pairs, N has 2 lone pairs.

O: $F C = 6 - 4 - \frac{1}{2} (4) = 0$ .
C: $F C = 4 - 0 - \frac{1}{2} (8) = 0$ .
N: $F C = 5 - 4 - \frac{1}{2} (4) = - 1$ .
Sum $= - 1$ ✓ matches ion charge.

Structure (II): $O \equiv C - N$ (triple–single). O has 1 lone pair, N has 3 lone pairs.

O: $F C = 6 - 2 - \frac{1}{2} (6) = + 1$ .
C: $F C = 4 - 0 - \frac{1}{2} (8) = 0$ .
N: $F C = 5 - 6 - \frac{1}{2} (2) = - 2$ .
Sum $= - 1$ ✓.

Verdict: Structure (II) has $∣ F C ∣$ values of ${1, 0, 2}$ — bigger and it puts $+ 1$ on oxygen (more electronegative than N). Structure (I) has ${0, 0, 1}$ with the $- 1$ on nitrogen. Since O is more electronegative than N, we'd prefer the negative charge on O — but (II) is disqualified by its large magnitudes and the wrong-signed charge on O. Structure (I) dominates. (Full analysis lives in Formal charge & resonance.)

Recall Solution L3.2

$S F_{6}$ : six S–F bonds → $12$ e⁻ around S. Sulfur is in period 3, so it has accessible vacant $3 d$ orbitals (and is large enough to hold 6 neighbours) → expanded octet is allowed. See Hypervalency & d-orbital participation.
$O F_{6}$ : oxygen is in period 2. Its valence shell is only $2 s 2 p$ — no low-energy $d$ orbitals — and O is far too small to physically fit 6 fluorines. So O is capped at $8$ electrons and $O F_{6}$ cannot form. General rule: only period $\geq 3$ elements expand (Periodic trends).

Level 4 — Synthesis

Recall Solution L4.1

$N_{total} = 4 + 3 (6) + 2 = 24$ ; $N_{need} = 8 \times 4 = 32$ ; bonds $= (32 - 24) /2 = 4$ .
Four bonds over three C–O links → one C=O, two C–O.
Formal charges: double-bonded O: $6 - 4 - \frac{1}{2} (4) = 0$ . Each single-bonded O: $6 - 6 - \frac{1}{2} (2) = - 1$ . C: $4 - 0 - \frac{1}{2} (8) = 0$ . Sum $= 0 + (- 1) + (- 1) + 0 = - 2$ ✓.
Resonance: the double bond can sit on any of the 3 oxygens equally → 3 equivalent structures.
Bond order: total bonds spread over 3 links $= 4/3 \approx 1.33$ . Every C–O is identical in the real ion — the average of one double and two singles.

Recall Solution L4.2

$N_{total} = 1 + 5 + 3 (6) = 24$ . Connectivity: H–O–N, with N also bonded to two other O's.
The tempting (wrong) structure: N with two N=O and one N–O → that's $2 + 2 + 1 = 5$ bonds $= 10$ electrons on N. Nitrogen is period 2 — no $d$ orbitals — so it can never exceed $8$ electrons (4 bonds). This structure is impossible regardless of how nice the formal charges look.
Correct structure: N has one N=O double bond and two N–O single bonds (one going to the OH). Then N: $F C = 5 - 0 - \frac{1}{2} (8) = + 1$ . The singly-bonded terminal O (not the OH one) carries $F C = 6 - 6 - \frac{1}{2} (2) = - 1$ .
So $H N O_{3}$ genuinely has a $+ 1$ on N and $- 1$ on one O — these formal charges are physically correct, not to be "fixed" by breaking the period-2 limit. The $+ 1/ - 1$ pair delocalizes over the two terminal oxygens by resonance.

Level 5 — Mastery

Recall Solution L5.1

(a) $N O$ : $N_{total} = 5 + 6 = 11$ — an odd number. You cannot pair 11 electrons into an even $8 + 8$ or place them without one being unpaired. One atom (nitrogen) keeps a single unpaired electron → $N O$ is a paramagnetic free radical (see Free radicals).
(b) $N O^{+}$ : removing one electron gives $N_{total} = 11 - 1 = 10$ — now even. With $N \equiv O^{+}$ (a triple bond), N and O each reach an octet with no unpaired electron. $N_{need} = 16$ , $N_{shared} = 16 - 10 = 6 \Rightarrow 3$ bonds = a triple bond ✓. Removing the lone unpaired electron is exactly what cures the radical — that's why $N O$ is easily oxidized to $N O^{+}$ .

Recall Solution L5.2

Isolated molecule: $N_{total} = 2 + 2 (7) = 16$ . Two Be–Cl single bonds use $4$ electrons; Be ends with only 4 electrons (2 bonds) → deeply incomplete octet. $N_{need} = 24$ would suggest 4 bonds, but Be has only 2 valence electrons to offer, so it genuinely stays electron-deficient — a strong Lewis acid.
Reaching an octet: because Be is electron-hungry, in the solid it forms a chain where each Cl donates a lone pair into Be's empty orbital (a coordinate/dative bond). Now Be has 4 bonds = 8 electrons. The extra 2 bonds are gifts from neighbouring chlorines, not new electrons — the counting still balances. This is the same "Lewis acid accepts a lone pair" mechanism that lets $B F_{3}$ react with $N H_{3}$ .

Recall Solution L5.3

$N_{total} = 8 + 4 (7) = 36$ . Four Xe–F single bonds use $8$ electrons; the remaining $36 - 8 = 28$ electrons: $24$ go as $6$ lone pairs on the four F's ( $3$ each on... actually $18$ e⁻ = 3 lone pairs × ... let's count carefully): each F needs 3 lone pairs = 6 e⁻; four F's take $4 \times 6 = 24$ e⁻. Left: $28 - 24 = 4$ e⁻ = 2 lone pairs on Xe.
Xe now has $4$ bonds ( $8$ e⁻) $+$ $2$ lone pairs ( $4$ e⁻) $= 12$ electrons → expanded octet. Xe is period 5 (well past period 2), so expansion is allowed. The 2 lone pairs plus 4 bonds give the square-planar shape predicted by VSEPR theory.

Recall Feynman recap: the whole ladder in one breath

Count the marbles (valence electrons), correct for charge, subtract "want" from "have" to see how many are shared, draw the bonds, then let formal charge be the referee. If an atom is boron/beryllium it may stop short of 8; if it's period-3-or-below it may hold more than 8; if the total is odd, one marble stays lonely and you've got a radical.

Connections

Formal charge & resonance — the referee used in L2.3, L3.1, L4.1–L4.2.
Electronegativity — decides where negative $F C$ belongs (L3.1).
Lewis acids and bases — electron-deficient $B F_{3}$ , $B e C l_{2}$ (L2.3, L5.2).
Hypervalency & d-orbital participation — expanded octets $S F_{6}$ , $X e F_{4}$ (L3.2, L5.3).
Free radicals — odd-electron $N O$ , $N O_{2}$ (L1.2, L5.1).
Periodic trends — why period ≥3 expands (L3.2).
VSEPR theory — Lewis structure → shape (L5.3).