Visual walkthrough — Formal charge calculation — best resonance structure

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Step 1 — Count what a lonely atom brings to the party

WHAT: We start with a single atom, alone, no bonds. We just count its outer dots.

WHY: Charge is a comparison. To say an atom is "$+$" or "$-$" we need a baseline — the number of electrons it had before anything happened. That baseline is $V$. Every later step compares against this number.

PICTURE: Look at figure s01. Oxygen sits alone with 6 lavender dots around it — that is $V=6$. This is our "before" snapshot. Remember this pile; everything we do next is measured against it.

$V=\underset{\text{electrons O brought, before bonding}}{\underbrace{6}}$

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Step 2 — Inside a molecule, split every bond down the middle

WHAT: We draw a bond and cut it in half with a dashed line, giving one electron of the pair to each side.

WHY: We need a rule for "how many electrons does this atom own now?" The only choice that is symmetric and simple is: half of each shared pair. This is the choice that makes $\text{FC}$ ignore electronegativity — which is exactly what distinguishes it from Oxidation Number (which gives all the bond to the greedier atom).

PICTURE: In figure s02, a single bond of 2 coral dots is cut by a dashed line. The left atom keeps 1, the right atom keeps 1. A double bond (4 dots) cut the same way gives 2 each.

$\text{electrons an atom owns from bonds}=\frac{1}{2}\underset{\text{total bonding electrons touching it}}{\underbrace{B}}$

Here $B$ = every electron sitting in a bond that touches this atom. One single bond → $B=2$. One double bond → $B=4$. One triple bond → $B=6$.

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Step 3 — Lone pairs belong entirely to their atom

WHAT: We circle the dots that are not in any bond. They count fully.

WHY: These electrons are shared with no one, so the "split in half" rule of Step 2 does not apply. Ownership of a private pair is 100%, not 50%. We must add them in full, or we'd rob the atom of electrons it clearly keeps.

PICTURE: Figure s03 shows an oxygen with two private mint pairs (a full circle around each) and one bond going off to the right. The circled pairs are $L=4$; only the bond gets the dashed half-cut.

$\text{electrons owned as lone pairs}=\underset{\text{private dots, counted in full}}{\underbrace{L}}$

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Step 4 — Add up what the atom owns NOW

WHAT: We total the two ownership contributions into one number: the "after" pile.

WHY: To compare against the "before" pile ($V$ from Step 1), we need the whole "after" pile as a single number. Nothing else contributes — an electron is either private (Step 3) or shared (Step 2), never both.

PICTURE: Figure s04 stacks the two contributions into a bar: the mint block is $L$, the coral block is $\frac{1}{2}B$. Their combined height is the atom's current holdings.

$\text{owned now}=\underset{\text{all lone-pair }{e}^{-}}{\underbrace{L}}+\underset{\text{half of bonding }{e}^{-}}{\underbrace{\frac{1}{2}B}}$

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Step 5 — Charge = what you started with − what you kept

WHAT: We subtract the "after" pile from the "before" pile.

WHY: If an atom gave away electrons it now owns fewer than it brought → it's short of negative charge → positive FC. If it gained, it owns more → negative FC. So charge is literally $\text{(brought)}-\text{(kept)}$. That subtraction is the entire idea; the algebra just tidies the minus sign onto both terms.

PICTURE: Figure s05 places the "before" bar ($V$) next to the "after" bar ($L+\frac{1}{2}B$). The gap between the two bar-tops, drawn as a labelled arrow, is the formal charge. Gap pointing down (owns less) = $+$; gap pointing up (owns more) = $-$.

$\boxed{\text{FC}=V-L-\frac{1}{2}B}\text{equivalently}\text{FC}=V-L-n$

where $n=\frac{1}{2}B$ is just the number of bonds on the atom.

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Step 6 — Every case: positive, negative, and exactly zero

WHAT: We run the finished formula on three oxygens in different situations so no scenario surprises you later.

WHY: The parent note's ranking rules care about sign and size. You must be able to produce all three outcomes. We use oxygen ($V=6$) so only $L$ and $B$ change.

PICTURE: Figure s06 shows three oxygens side by side with their bar-gaps: one gap points down ($+1$), one is flat ($0$), one points up ($-1$).

Oxygen as...	$L$	$B$	$\text{FC}=6-L-\frac{1}{2}B$	Gap
3 bonds, 1 lone pair (e.g. in ${\text{H}}_3{\text{O}}^{+}$)	$2$	$6$	$6-2-3=+1$	owns less → $+$
2 bonds, 2 lone pairs (water)	$4$	$4$	$6-4-2=0$	balanced → $0$
1 bond, 3 lone pairs (e.g. ${\text{OH}}^{-}$)	$6$	$2$	$6-6-1=-1$	owns more → $-$

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Step 7 — The formula chooses the best resonance form (SCN⁻)

WHAT: We apply the finished tool to two rival Lewis pictures of thiocyanate $[\text{SCN}{]}^{-}$ and let it pick the winner — the exact job the parent promised.

WHY: This is the payoff. A single overall charge of $-1$ can be drawn on different atoms. FC + one tie-breaker tells us which drawing best matches reality. Valences: $\text{S}=6$, $\text{C}=4$, $\text{N}=5$.

PICTURE: Figure s07 draws both forms with each atom's gap-arrow. In Form A the $-1$ gap sits on N; in Form B it sits on S.

Form A $[\text{S}=\text{C}=\text{N}{]}^{-}$: $\text{S}:6-4-\frac{1}{2}(4)=0,\text{C}:4-0-\frac{1}{2}(8)=0,\text{N}:5-4-\frac{1}{2}(4)=-1$

Form B $[\text{S}-\text{C}\equiv \text{N}{]}^{-}$: $\text{S}:6-6-\frac{1}{2}(2)=-1,\text{C}:4-0-\frac{1}{2}(8)=0,\text{N}:5-2-\frac{1}{2}(6)=0$

Both pass the checksum ($\sum \text{FC}=-1$ = the ion's charge). Both have one atom at $-1$. So we go to the tie-breaker: the negative gap should sit on the more electronegative atom. From Electronegativity, $\text{N}(3.0)>\text{S}(2.6)$, so Form A wins.

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The one-picture summary

Figure s08 compresses the whole journey: the before pile $V$ on the left, the after pile $L+\frac{1}{2}B$ in the middle, and the gap arrow on the right that is $\text{FC}$ — with the three sign-outcomes labelled. If you can regenerate this one image in your head, you can derive formal charge for any atom in any structure.

Recall Feynman retelling — the whole walkthrough in plain words

Every atom shows up to the party carrying a certain number of toys — that's $V$ (Step 1). When two atoms hold hands over a shared pile of toys (a bond), they agree to each keep half of that pile — that's the $\frac{1}{2}B$ rule (Step 2). Any toys an atom is sitting on all by itself (lone pairs) stay 100% theirs — that's $L$ (Step 3). Add up "half of what I share" plus "all of what's private" and you get how many toys the atom holds now (Step 4). Finally, compare: toys-brought minus toys-now-held. If the atom gave some away it's grumpy and positive; if it grabbed extra it's happy and negative; if it's exactly even it's content and zero (Steps 5–6). To pick the best drawing of a molecule, draw the pictures where the fewest kids are grumpy or happy — mostly content zeros — and if two pictures tie, hand the leftover negative to the greediest kid, the most electronegative atom (Step 7). That's it: $\text{FC}=V-L-\frac{1}{2}B$, start minus keep.

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Connections

• Lewis Structures — every step here sits on top of a Lewis diagram.
• Resonance Structures — Step 7 is the resonance-ranking payoff.
• Electronegativity — the Step 7 tie-breaker.
• Oxidation Number — the "give the whole bond to the greedy atom" cousin of Step 2's half-split.
• Octet Rule — the SCN⁻ forms both obey octets; FC breaks the tie.
• VSEPR Theory — the winning structure feeds shape prediction.
• Hinglish version →