In the molecule O(A)=C=O(B), the formal charge on O(A), C and O(B) are respectively. (a) -1, 0, +1 (b) +1, 0,

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Q. In the molecule O(A)=C=O(B), the formal charge on O(A), C and O(B) are respectively.

(a) -1, 0, +1

(b) +1, 0, -1

(d) 0, 0, 0

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Concept understanding — Formal Charge

The Intuition: Who "Owns" the Electrons?

Imagine a molecule as a small company. Each atom is an employee, and the electrons are the company's resources. Some resources are kept privately (lone pairs), and some are shared between employees (bonding pairs). But here's the catch: in a real molecule, the bossier atoms (more electronegative ones) pull the shared resources closer to themselves. That's the real charge distribution.

Formal charge ignores that bossiness. It asks a simpler, hypothetical question: If every bond were a perfect 50-50 split, and every atom started with its neutral number of valence electrons, what charge would each atom end up with?

That's the core idea. It's a bookkeeping tool, not a measurement of reality. It helps us draw the most plausible Lewis structures by penalizing structures where atoms end up with unreasonably large hypothetical charges.

The Precise Statement

Formal charge ( $FC$ ) on an atom in a Lewis structure is:

FC = V - N_{nb} - \frac{1}{2} B

Where:

$V$ = number of valence electrons on the free, neutral atom (from the periodic table: 1 for H, 4 for C, 5 for N, 6 for O, 7 for F, etc.)
$N_{nb}$ = number of non-bonding electrons on that atom in the structure (the lone pairs, counted as full electrons)
$B$ = number of bonding electrons on that atom in the structure (each single bond contributes 2, a double bond 4, a triple bond 6)

FC = V - N_{nb} - \frac{1}{2}B

How It Works: A Worked Example

Take the cyanide ion, $CN^-$ . Carbon has 4 valence electrons ( $V=4$ ), nitrogen has 5 ( $V=5$ ). The ion has a total of $4+5+1=10$ valence electrons.

A plausible Lewis structure is $:C \equiv N:$

Let's calculate formal charges.

For Carbon:

$V = 4$
$N_{nb} = 2$ (one lone pair)
$B = 6$ (triple bond)
$FC = 4 - 2 - \frac{1}{2}(6) = 4 - 2 - 3 = -1$

For Nitrogen:

$V = 5$
$N_{nb} = 2$ (one lone pair)
$B = 6$ (triple bond)
$FC = 5 - 2 - \frac{1}{2}(6) = 5 - 2 - 3 = 0$

The formal charges are $-1$ on carbon and $0$ on nitrogen. Notice that the sum of formal charges ( $-1 + 0 = -1$ ) equals the overall charge on the ion. This is always true.

Why This Matters: Choosing the Best Structure

Formal charge is a tiebreaker. When multiple Lewis structures are possible (resonance), the preferred one usually follows these rules:

Smallest formal charges are best.
Negative formal charge on the more electronegative atom is better.
Like charges should not be on adjacent atoms.

Consider the cyanate ion, $OCN^-$ . Two major resonance structures exist:

Structure	Formal Charges
$: \ddot{O} - C \equiv N:$	O: $6 - 6 - 1 = -1$ ; C: $4 - 0 - 4 = 0$ ; N: $5 - 2 - 3 = 0$	Preferred (negative on O, the most electronegative)
$: O = C = \ddot{N}:$	O: $6 - 4 - 2 = 0$ ; C: $4 - 0 - 4 = 0$ ; N: $5 - 4 - 2 = -1$	Less preferred (negative on N, less electronegative than O)

Watch out

Formal charge is not the same as oxidation state. Oxidation state assigns all bonding electrons to the more electronegative atom. Formal charge splits them equally. They are different bookkeeping systems for different purposes.

The Bottom Line

Formal charge is a hypothetical number that assumes perfect democracy in electron sharing. It's a tool for choosing between possible Lewis structures, not a description of real charge. The best structure is the one that keeps these hypothetical charges as small and as sensibly placed as possible.

Searches like 'formal charge formula', 'how to calculate formal charge class 11 chemistry', and 'formal charge important questions' point to this being a frequently revised topic within the NCERT/CBSE Class 11 Chemistry Chemical Bonding chapter. Using formal charge to pick the best Lewis structure is a recurring question style in JEE Main and NEET inorganic and bonding sections.

Step 1. Formal charge $= N_v - N_l - N_b/2$ . CO₂'s stable structure is O(A)=C=O(B), two C=O double bonds, each terminal oxygen carrying two lone pairs.

Step 2. Carbon: $N_v = 4$ , $N_l = 0$ , $N_b = 8$ (two double bonds = 4 bond pairs = 8 electrons) → FC $= 4 - 0 - 8/2 = 0$ .

Step 3. Each oxygen (O(A) and O(B)): $N_v = 6$ , $N_l = 4$ (two lone pairs), $N_b = 4$ (one double bond = 2 bond pairs) → FC $= 6 - 4 - 4/2 = 0$ .

Step 4. So the formal charges on O(A), C, O(B) are 0, 0, 0 respectively.

✓Final answer

(d) 0, 0, 0

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Q.In the molecule O(A)=C=O(B), the formal charge on O(A), C and O(B) are respectively.

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