Formal charge calculation — best resonance structure: Question bank

First: what do the three letters even mean?

Before any trap, pin down the symbols. Look at the figure — it draws one oxygen atom with its electrons sorted into two piles.

Figure — Formal charge calculation — best resonance structure

We build the formula, one honest step at a time. The figure walks the same three panels.

Intuition The whole derivation is "started − kept"

Step 1 — what the atom started with. A neutral free atom brought $V$ valence electrons to the party. Why $V$ ? Because that is what makes the isolated atom neutral in the first place.

Step 2 — count what it keeps inside the molecule. Two kinds of electrons surround it:

Lone-pair electrons are its own — it keeps all of them: that's $L$ .
Bonding electrons are shared with a partner. Split perfectly evenly (the FC fiction), the atom owns exactly half of each bond: that's $\frac{1}{2} B$ .

So electrons kept $= L + \frac{1}{2} B$ .

Step 3 — charge is the shortfall. If an atom now owns fewer electrons than it started with, it effectively gave some away → positive; if it owns more, it took some → negative: $FC = started with V - now owns (L + \frac{1}{2} B) = V - L - \frac{1}{2} B$ No magic — just subtraction of "kept" from "started."

True or false — justify

Formal charge is the actual, physical charge sitting on an atom in the molecule.

False — FC pretends every bond is split perfectly evenly, which ignores Electronegativity; the real partial charge is different and leans toward the more electronegative atom.

If every atom in a structure has a formal charge of zero, the sum of formal charges is zero, so the molecule must be neutral.

True in the sense that the sum equals the overall charge — all zeros sum to zero, so this structure can only describe a neutral species, never an ion.

Two resonance structures of the same species can have different sums of formal charge.

False — the sum of FCs always equals the overall charge of the species, and that charge is fixed, so every valid resonance form of it must give the same total.

A structure with formal charges of

+ 1

and

- 1

is always worse than one with all zeros.

Generally true by Rule 1 (fewer charged atoms is better), but only when both are otherwise valid — sometimes an all-zero structure is impossible without breaking the Octet Rule or overall charge.

Adding a lone pair to an atom always makes its formal charge more negative.

True — in

FC = V - L - \frac{1}{2} B

, increasing

L

by 2 (one lone pair) lowers FC by 2, since owning more electrons alone pushes charge negative.

Formal charge and Oxidation Number are just two names for the same quantity.

False — FC splits each bond equally (½ each), while oxidation number gives both bonding electrons to the more electronegative atom; they answer different questions.

The best resonance structure is simply the one that obeys the octet rule.

False — octet compliance is important but not sufficient; among octet-satisfying forms we still rank by minimising formal charge and placing charges on the right atoms.

If a structure's formal charges don't sum to the ion's charge, one of the individual FCs must be wrong.

True — the checksum is exact; a mismatch means you miscounted lone-pair or bonding electrons somewhere, so recount before trusting any single FC.

An unpaired (radical) electron on an atom counts toward

L

in the formula.

True — a lone single electron is still non-bonding, so it adds 1 to

L

; the formula

V - L - \frac{1}{2} B

works unchanged for odd-electron species.

Spot the error

"For oxygen in water, I counted all 4 bonding electrons plus its 4 lone-pair electrons as owned, so FC

= 6 - 8 = - 2

The error is counting all bonding electrons; oxygen owns only half of them. Correct:

6 - 4 - \frac{1}{2} (4) = 0

"A C=O double bond has 4 bonding electrons, so I used

B = 2

because there are 2 bonds."

B

counts electrons, not bonds — a double bond is

B = 4

. Only after dividing,

\frac{1}{2} B = 2

, do you get "2 bonds' worth" owned.

"Nitrogen has 5 valence electrons, and it's in a molecule, so I used

V = 8

to give it a full octet."

V

is the valence-electron count of the free atom (its group number), always 5 for N — the octet describes electrons in the molecule, not

V

"The negative formal charge in

SCN^{-}

should go on carbon because carbon is in the middle."

Position in the skeleton is irrelevant; Rule 3 says negative charge belongs on the most electronegative atom, which is nitrogen, not the central carbon.

"CO has FC of

- 1

on carbon, which is impossible because carbon is less electronegative than oxygen."

It's unusual but not impossible — nature accepts the "wrong" placement to keep magnitudes at

\pm 1

instead of a larger

\pm 2

; this oddity is exactly why CO is so reactive.

"I found FC

= + 1

on an atom, meaning that atom physically lost a whole electron."

FC is a bookkeeping label from equal-sharing arithmetic, not a literal missing electron; the true charge is a fractional partial charge shaped by electronegativity.

"NO has 11 valence electrons total, an odd number, so formal charge can't be calculated for it."

FC works fine — the odd electron simply sits as a single unpaired electron (part of

L

) on one atom; you still compute

V - L - \frac{1}{2} B

atom by atom.

Why questions

Why do we take half the bonding electrons rather than all of them or none?

Because a bond is shared — under the equal-sharing fiction each of the two bonded atoms owns exactly one of the two electrons, so an atom's share is

\frac{1}{2} B

Why does formal charge deliberately ignore electronegativity, when we know electronegativity is real?

To make FC a fair, symmetric comparison across atoms; if it favoured the electronegative atom it would just reproduce oxidation number and couldn't independently rank resonance structures.

Why is a structure with charges

\pm 1

preferred over one with

\pm 2

Larger separated charges cost more energy to create and are less stable; Rule 2 encodes that smaller charge magnitudes describe a lower-energy, more realistic distribution.

Why does the sum of formal charges have to equal the overall charge of the species?

FC just redistributes the molecule's electrons among atoms without creating or destroying any, so the atom-by-atom bookkeeping must add back up to the true total charge.

Why can two resonance forms both be "correct" yet one still dominate?

Resonance Structures are all imperfect snapshots of one real molecule; the form with the lowest, best-placed formal charges resembles reality most, so it contributes more to the true blend.

Why does putting a negative charge on the more electronegative atom make a structure better?

An electronegative atom holds extra electron density comfortably (low energy), so that placement describes a more stable, more realistic charge arrangement.

Why does a single unpaired electron count as 1 (not 2) toward

L

Because

L

counts electrons, and a radical simply has one non-bonding electron in that spot; you add exactly what is there — one.

Edge cases

For a neutral free atom drawn all by itself (no bonds, all valence electrons as lone pairs), what is its formal charge?

Zero — with

L = V

and

B = 0

, we get

FC = V - V - 0 = 0

, confirming an isolated neutral atom carries no formal charge.

For

H^{+}

(a bare proton with no electrons and no bonds), what does the formula give?

V = 1

L = 0

B = 0

, so

FC = 1 - 0 - 0 = + 1

— matching the real

+ 1

charge, since it lost its only electron.

For the radical NO with structure

\dot{N} = O

(N: double bond + 1 lone pair + 1 unpaired e⁻; O: double bond + 2 lone pairs), what are the formal charges?

5 - 3 - \frac{1}{2} (4) = 0

and O:

6 - 4 - \frac{1}{2} (4) = 0

— the odd electron just adds 1 to N's

L

, and both atoms come out neutral.

For a chlorine radical atom on its own (

Cl \cdot

, 7 valence electrons, all non-bonding), what is its FC?

FC = 7 - 7 - 0 = 0

— a lone radical atom is neutral; the unpaired electron makes it reactive but does not change its formal charge.

If an atom has a lone pair and the same total bonds in two resonance forms, can its FC differ between them?

No — FC depends only on

V

L

, and

B

for that atom; if all three match, the FC is identical regardless of what neighbouring atoms do.

Can a single atom ever legitimately carry a formal charge larger than

\pm 2

Yes, arithmetically (e.g. highly charged or hypervalent situations), but such structures rank poorly and rarely describe reality — large magnitudes signal a bad or minor resonance form.

In a homonuclear molecule like

O_{2}

, does the electronegativity tie-breaker (Rules 3–4) ever apply?

No — both atoms are identical, so any formal charges must be assigned symmetrically or cancel; equal electronegativity leaves nothing to break the tie.

If a species is an anion, must at least one atom carry a negative formal charge?

Yes — the FCs must sum to the negative overall charge, so it is impossible for every atom to be zero; some atom must bear the deficit.

For an atom with zero lone pairs and four single bonds (like carbon in methane), what is its FC and why?

FC = 4 - 0 - \frac{1}{2} (8) = 0

— it owns half of eight bonding electrons (four), exactly its valence, so it is neutral.

Connections

Formal charge calculation — best resonance structure — the parent note with the full derivation and worked numbers.
Electronegativity — the tie-breaker behind Rules 3 and 4.
Oxidation Number — the unequal-sharing cousin to contrast against FC.
Resonance Structures — why ranking imperfect forms matters at all.
Octet Rule — the constraint that often competes with FC minimisation.
Lewis Structures — the diagrams FC is computed on top of.
VSEPR Theory — where the chosen structure goes next.