Worked examples — Formal charge calculation — best resonance structure

Before anything, recall the three letters — each is a counting task you do by looking at the drawn structure:

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The scenario matrix

Here is the full space of cases this topic throws. Each worked example below is tagged with the cell(s) it covers.

#	Case class	What makes it tricky	Example
C1	All-neutral (charge 0, all FC = 0)	The "ideal" — confirms the tool	Ex 1 (CH₄)
C2	Positive FC present (cation)	Sum must be $+1$, place $+$ on least electronegative	Ex 2 (H₃O⁺)
C3	Negative FC present (anion)	Sum must be $-$charge, place $-$ on most electronegative	Ex 3 (OH⁻)
C4	Competing resonance forms tie	Use electronegativity tie-breaker	Ex 4 (OCN⁻)
C5	Magnitude ±2 vs ±1	Smaller magnitude wins	Ex 5 (ozone O₃)
C6	Expanded octet / degenerate "more bonds"	More bonds can lower FC below period-3 atoms	Ex 6 (SO₄²⁻)
C7	Zero / degenerate input ($L=0$ or $B=0$)	A lone ion, a bare atom	Ex 7 (Na⁺, Cl⁻)
C8	Real-world word problem	Translate a sentence into a structure	Ex 8 (nitrogen dioxide smell)
C9	Exam twist / isoelectronic	Same electrons, "which is real?"	Ex 9 (N₂ vs CO⁻ trap)

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Ex 1 — CH₄ (Cell C1: all-neutral)

1. Carbon. $V=4$ (group 14). Lone pairs? None on C, so $L=0$. Four single bonds → $B=4\times 2=8$. ${\text{FC}}_C=4-0-\frac{1}{2}(8)=4-0-4=0$ Why this step? C keeps half of its 8 bonding electrons (= 4), exactly matching its valence → neutral.
2. Each Hydrogen. $V=1$ (group 1). $L=0$. One single bond → $B=2$. ${\text{FC}}_H=1-0-\frac{1}{2}(2)=1-0-1=0$ Why this step? H brought 1 electron, keeps half of its one bond (= 1) → neutral.

Verify: checksum $\sum =0+4(0)=0$, matches CH₄'s neutral charge. ✔

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Ex 2 — H₃O⁺ (Cell C2: positive FC, cation)

1. Oxygen. $V=6$ (group 16). One LP → $L=2$. Three single bonds → $B=6$. ${\text{FC}}_O=6-2-\frac{1}{2}(6)=6-2-3=+1$ Why this step? O normally makes 2 bonds with 2 lone pairs. Here it made a third bond, giving away a lone-pair electron's worth of ownership → it comes up one electron short → $+1$.
2. Each Hydrogen. $V=1$, $L=0$, one bond → $B=2$. ${\text{FC}}_H=1-0-\frac{1}{2}(2)=0$ Why this step? H keeps half of its single bond (= 1), matching its valence → neutral.

Verify: checksum $=(+1)+3(0)=+1$, matches the ion's charge. ✔ On the placement rule: the arithmetic forces the $+1$ onto O. Rule 4 ("put $+$ on the least electronegative atom") is a tie-breaker between structures, not a licence to move a charge arbitrarily — here only one structure is sensible, so arithmetic wins.

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Ex 3 — OH⁻ (Cell C3: negative FC, anion)

1. Oxygen. $V=6$. Three LP → $L=6$. One single bond → $B=2$. ${\text{FC}}_O=6-6-\frac{1}{2}(2)=6-6-1=-1$ Why this step? O now owns 6 lone-pair electrons plus 1 shared = 7, one more than its valence 6 → excess electron → $-1$.
2. Hydrogen. $V=1$, $L=0$, $B=2$ → ${\text{FC}}_H=1-0-1=0$. Why this step? H keeps half of its one bond (= 1), matching its valence → neutral.

Verify: checksum $=(-1)+0=-1$. ✔ And by Rule 3 the negative sits on O, the more electronegative atom — consistent with reality.

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Ex 4 — OCN⁻ cyanate (Cell C4: resonance tie, electronegativity decides)

Valence: O=6, C=4, N=5.

1. Form A.
   • O ($L=4$, double bond → $B=4$): $6-4-\frac{1}{2}(4)=0$ Why this step? O owns 4 lone-pair + 2 shared = 6, exactly its valence → neutral.
   • C ($L=0$, two double bonds → $B=8$): $4-0-\frac{1}{2}(8)=0$ Why this step? C keeps half of 8 shared electrons (= 4) = its valence → neutral.
   • N ($L=4$, double bond → $B=4$): $5-4-\frac{1}{2}(4)=-1$ Why this step? N owns 4 lone-pair + 2 shared = 6, one more than its valence 5 → $-1$.
2. Form B.
   • O ($L=2$, triple bond → $B=6$): $6-2-\frac{1}{2}(6)=+1$ Why this step? O made a third bond and shed a lone pair, so it owns only 5 → one short → $+1$.
   • C ($L=0$, four bonds → $B=8$): $4-0-\frac{1}{2}(8)=0$ Why this step? Same as Form A — C keeps 4 = its valence → neutral.
   • N ($L=6$, single bond → $B=2$): $5-6-\frac{1}{2}(2)=-2$ Why this step? N owns 6 lone-pair + 1 shared = 7, two more than valence 5 → $-2$.
3. Compare. Form A has one nonzero FC of magnitude 1; Form B has magnitudes 1 and 2. Rule 1 (fewer charges) and Rule 2 (smaller magnitudes) already kill Form B before we even reach electronegativity.

Verify: checksum Form A $=0+0-1=-1$ ✔; Form B $=+1+0-2=-1$ ✔. Both are valid Lewis pictures but Form A is far superior. (Interesting twist: the best cyanate form actually puts the $-1$ on N, not O — because putting it on O would force the awkward Form B geometry. So Rule 3 bows to Rule 2 first: the ranking order matters!)

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Ex 5 — Ozone O₃ (Cell C5: ±1 present, geometry figure)

The figure below shows the bent molecule: the central oxygen (drawn in plum) sits at the top vertex, the double-bonded terminal oxygen (orange, joined by a two-line double bond) is at the lower left, and the single-bonded terminal oxygen (teal, joined by one line) is at the lower right. Each atom carries a labelled call-out giving its bond/lone-pair count and its resulting formal charge, and the box underneath restates the checksum.

1. Central O (1 LP → $L=2$; one double + one single bond → $B=4+2=6$). $\text{FC}=6-2-\frac{1}{2}(6)=6-2-3=+1$ Why this step? The central atom makes three bonds total — one more than oxygen "wants" — so it is a bond richer and an electron poorer → $+1$. This is the plum atom at the top of the figure.
2. Double-bonded terminal O (2 LP → $L=4$; double bond → $B=4$). $\text{FC}=6-4-\frac{1}{2}(4)=6-4-2=0$ Why this step? It owns 4 lone-pair + 2 shared = 6 = its valence → neutral (the orange atom, lower left).
3. Single-bonded terminal O (3 LP → $L=6$; single bond → $B=2$). $\text{FC}=6-6-\frac{1}{2}(2)=6-6-1=-1$ Why this step? Only one bond and three lone pairs → owns 7 electrons, one too many → $-1$ (the teal atom, lower right).

Verify: checksum $=(+1)+0+(-1)=0$, matches neutral O₃. ✔ Neither terminal ever reaches ±2, so no better all-zero structure exists — the charges are forced by the double bond having to sit on one side. Both resonance forms (double bond left vs right) are equally good — that's why ozone resonates.

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Ex 6 — Sulfate SO₄²⁻ (Cell C6: expanded octet lowers FC)

Valence: S=6, O=6.

1. Form X — Sulfur (four single bonds → $B=8$, $L=0$): ${\text{FC}}_S=6-0-\frac{1}{2}(8)=6-4=+2$ Why this step? S with only single bonds keeps just 4 shared electrons, two fewer than its valence 6 → $+2$. Each O (single bond → $B=2$, 3 LP → $L=6$): $6-6-1=-1$. Why this step? Each O owns 6 lone-pair + 1 shared = 7, one more than valence 6 → $-1$.
2. Form Y — Sulfur (two double + two single → $B=2(4)+2(2)=12$, $L=0$): ${\text{FC}}_S=6-0-\frac{1}{2}(12)=6-6=0$ Why this step? Two extra bonds let S "own" more shared electrons (6), dropping its FC from $+2$ all the way to $0$. This is allowed because S is period 3 and can hold more than 8 electrons (expanded octet). Double-bonded O ($B=4$, $L=4$): $6-4-2=0$. Why this step? Owns 4 lone-pair + 2 shared = 6 = valence → neutral. Single-bonded O ($B=2$, $L=6$): $6-6-1=-1$. Why this step? Owns 6 lone-pair + 1 shared = 7, one more than valence → $-1$.
3. Compare. Form X: charges $+2$ and four $-1$ (five charged atoms). Form Y: two atoms at $-1$, everything else zero. Fewer charges (Rule 1) and smaller magnitudes (Rule 2) → Form Y wins.

Verify: Form X checksum $=(+2)+4(-1)=-2$ ✔. Form Y checksum $=0+2(0)+2(-1)=-2$ ✔. Both equal the ion's $-2$; Form Y is the major contributor.

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Ex 7 — Na⁺ and Cl⁻ (Cell C7: degenerate inputs, $B=0$)

1. Na⁺. Free Na has $V=1$. As Na⁺ it has lost its valence electron, so it has no lone pairs and no bonds: $L=0$, $B=0$. $\text{FC}=1-0-0=+1$ Why this step? This is the degenerate case: no bonds means FC = valence electrons brought minus lone electrons kept. Na kept nothing of its one electron → $+1$.
2. Cl⁻. Free Cl has $V=7$. Chloride gained one electron → 8 electrons, all as four lone pairs (4 LP): $L=8$, $B=0$. $\text{FC}=7-8-0=-1$ Why this step? It owns 8, brought 7, so it has one extra → $-1$.

Verify: FC of a bare ion equals its actual ionic charge — the formula degrades gracefully to the obvious answer. $+1$ ✔, $-1$ ✔.

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Ex 8 — NO₂ word problem (Cell C8: real-world)

1. Nitrogen. $V=5$. It has one unpaired electron → $L=1$ (a single electron still counts as one non-bonding electron). One double + one single bond → $B=4+2=6$. ${\text{FC}}_N=5-1-\frac{1}{2}(6)=5-1-3=+1$ Why this step? The lone single electron contributes only 1 to $L$, not 2 — this is the trap. N still ends up electron-poor → $+1$.
2. Double-bonded O ($L=4$, $B=4$): $6-4-2=0$. Why this step? Owns 4 lone-pair + 2 shared = 6 = valence → neutral.
3. Single-bonded O ($L=6$, $B=2$): $6-6-1=-1$. Why this step? Owns 6 lone-pair + 1 shared = 7, one more than valence → $-1$.

Verify: checksum $=(+1)+0+(-1)=0$, matching neutral NO₂. ✔ By Rules 3–4 the $+1$ sits on the less-electronegative N and $-1$ on the more-electronegative O — exactly what the placement rules prefer, so this structure is real. That unpaired electron makes NO₂ a reactive radical.

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Ex 9 — Isoelectronic trap: CO vs the "N₂" reflex (Cell C9: exam twist)

1. N₂ check. Each N: $V=5$, one LP → $L=2$, triple bond → $B=6$: $5-2-3=0$. Symmetric → both zero. Why this step? Identical atoms split identically → no charge separation.
2. CO — Carbon ($V=4$, $L=2$, $B=6$): $4-2-3=-1$. Why this step? C brought only 4 valence electrons but owns 2 lone-pair + 3 shared = 5, one extra → $-1$.
3. CO — Oxygen ($V=6$, $L=2$, $B=6$): $6-2-3=+1$. Why this step? Same bonding pattern as N₂, but O brought 6 and owns only 5 → one short → $+1$. Electron count matches N₂; electron ownership vs valence does not.

Verify: N₂ checksum $=0+0=0$ ✔. CO checksum $=(-1)+(+1)=0$ ✔, matching neutral CO. So the student is false: isoelectronic species share electron totals, not formal charges. (The odd $-1$ on the less-electronegative carbon is exactly why CO is such a reactive, unusual gas — consistent with Worked Example 2 in the parent.)

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Recall

Recall Which ranking rule beats which?

Order: (1) fewest nonzero FCs → (2) smallest magnitudes → (3) negative on most electronegative → (4) positive on least electronegative → (5) checksum must hold. Rules 1–2 come before electronegativity — that's why Ex 4 and Ex 6 were decided by magnitude, not by which atom is more electronegative.

Recall The degenerate case

For a bare ion with no bonds ($B=0$): FC $=V-L$, which just equals the actual ionic charge. The formula never breaks.

Recall The odd-electron trap

A single unpaired electron counts as $L=1$, not $2$. (See NO₂, Ex 8.)

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Connections

• Parent topic — this page drills every case of it.
• Lewis Structures — every example starts from a drawn Lewis diagram.
• Resonance Structures — Ex 4, 5, 6 are all "which resonance form wins?" cases.
• Electronegativity — the tie-breaker used in Ex 4 and placement in Ex 8, 9.
• Oxidation Number — contrast: FC splits bonds evenly; oxidation number gives them all to the electronegative atom.
• Octet Rule — Ex 6 shows S breaking the octet to lower its FC (expanded octet).
• VSEPR Theory — the winning structure (bent O₃, bent NO₂) then feeds shape prediction.