Formal Charge — Reading Electrons Off a Structure

Formal charge trips up a lot of JEE aspirants. It shouldn't. By the end of this post you'll see it's just a head-count of electrons — and once you can count, you're done.

Why bother? Because formal charge shows you the relative electron density on an atom. A plus means that atom is short of electrons. A minus means it's carrying extra. You'll need this every time you draw an organic structure, and you'll lean on it hard when you reach resonance.

First, one idea everything rests on.

Valence electrons and bonds

Every atom keeps its electrons in shells. The ones in the outermost shell do the bonding, and we call them valence electrons.

Carbon has 4. Nitrogen has 5. Quick one — how many does oxygen have? Six.

Valence electrons also tell you how many bonds an atom forms:

• 4 or fewer valence electrons → bonds = valence electrons.
• More than 4 → bonds = $8 -$ valence electrons.

So carbon makes 4 bonds, nitrogen makes 3, oxygen makes 2. Hold that thought.

So where does charge come in?

Carbon is supposed to have 4 valence electrons. But the carbon sitting in front of you in some structure might own a little less, or a little more.

That gap — the deficiency or the excess — is the formal charge. Fewer electrons than it should have? Positive. More? Negative. That's the whole idea.

The formula

Two steps. No magic.

1. Find how many valence electrons the atom should have.
2. Look at the structure and count how many it actually has.

Subtract:

$FC = V - N - \tfrac{B}{2}$

where $V$ is valence electrons, $N$ is non-bonding (lone-pair) electrons, and $B$ is bonding electrons.

If counting whole bonds is easier in your head, use this version — it's the same thing:

$FC = V - (2 \times \text{lone pairs} + \text{bonds})$

Let's put it to work.

Worked example 1 — an oxygen

Picture an oxygen with three lone pairs and one single bond to it (think of the O in a hydroxide ion):

Draw it without the charge and it looks wrong, doesn't it? Something's missing. Let's find it.

• Oxygen should have 6 valence electrons.
• It actually has: $2 \times 3$ lone-pair electrons $+ 1$ bond $= 6 + 1 = 7$.

So:

$FC = 6 - 7 = -1$

One extra electron, one minus charge. Park it on the oxygen and the structure is finally honest.

Worked example 2 — a nitrogen

Now a nitrogen with no lone pairs and four bonds (the N in an ammonium ion):

Pen and paper. Work it with me:

• Nitrogen should have 5 valence electrons.
• It actually has: $0$ lone-pair electrons $+ 4$ bonds $= 4$.

So:

$FC = 5 - 4 = +1$

One electron short, one plus charge. The nitrogen carries a $+$.

Your turn. Take the carbon in a carbanion $:CH_3^{-}$ — it has one lone pair and three bonds. What's its formal charge?

Check: $V = 4$, electrons present $= 2 + 3 = 5$, so $FC = 4 - 5 = -1$. The carbon is negative.

The takeaway

Formal charge is just bookkeeping for electrons. Does this atom have more than it's owed, or less? Count, subtract, label.

$FC = V - N - \tfrac{B}{2}$

It looks like a small thing, and people skip it. Don't. Get it right and your structures stop being wrong in the one way examiners always check — and resonance, when it arrives, won't rattle you.