Formal Charge, Limitations of Octet Rule & Exceptions

Formal Charge: Bookkeeping for Electrons

When a molecule or ion can be drawn in more than one way, we need a fair method to assign the 'leftover' charge to each atom. That method is formal charge.

Definition: The formal charge on an atom in a Lewis structure is the charge it would have if all bonding electrons were shared equally. $$\text{Formal charge} = (\text{valence electrons in free atom}) - (\text{non-bonding electrons}) - \tfrac{1}{2}(\text{bonding electrons})$$

A shorthand version: $$FC = V - L - \tfrac{B}{2}$$ where $V$ = valence electrons, $L$ = lone-pair (non-bonding) electrons, $B$ = bonding electrons (twice the number of bonds).

[JEE Tip] The sum of all formal charges in a molecule equals the overall charge (0 for a neutral molecule, the ion's charge for an ion). Use this as a quick check.

Choosing the Best Lewis Structure

Formal charge helps us pick the most plausible of several possible structures. The rules, in order of importance:

1. Smallest formal charges win. The structure in which atoms carry formal charges closest to zero is the most stable.
2. Negative formal charge should sit on the most electronegative atom.
3. Avoid like charges on adjacent atoms.

Example — carbon dioxide

The structure $O = C = O$ gives every atom a formal charge of 0, whereas $O \equiv C - O$ would place $+1$ and $-1$ charges. So the symmetric double-bonded structure is preferred.

Key Point: Formal charge is a bookkeeping device, not a real charge. The actual electron distribution is governed by electronegativity, but formal charge is an excellent tie-breaker between candidate structures.

[NEET Important] In $CO_2$, the preferred Lewis structure has zero formal charge on all atoms — a very common exam question.

Limitation 1 & 2: Incomplete Octet and Odd-Electron Molecules

The octet rule is a brilliant guide, but nature breaks it in three well-defined ways.

Incomplete octet (electron-deficient molecules)

Some central atoms are stable with fewer than 8 electrons. These are typically elements of Groups 1, 2 and 13:

• $BeCl_2$ — Be has only 4 electrons.
• $BCl_3$, $BF_3$ — B has only 6 electrons.
• $AlCl_3$ — Al has only 6 electrons.

Such molecules are electron-deficient and act as Lewis acids (they accept lone pairs).

Odd-electron molecules

Molecules with an odd total number of valence electrons cannot possibly pair every electron, so an octet is impossible for at least one atom:

• Nitric oxide $NO$ (11 valence electrons)
• Nitrogen dioxide $NO_2$ (17 valence electrons)
• Chlorine dioxide $ClO_2$

These are called free radicals and are usually very reactive.

[JEE Tip] If the total valence-electron count is odd, the octet rule must be violated — recognise this instantly in exams.

Limitation 3: Expanded Octet & Other Shortcomings

Expanded octet (more than 8 electrons)

Elements of period 3 and beyond can form molecules in which the central atom has more than eight electrons. Examples include:

• $PCl_5$ — phosphorus has 10 electrons.
• $SF_6$ — sulphur has 12 electrons.
• $IF_7$ — iodine has 14 electrons.
• $XeF_4$, $H_2SO_4$, $HClO_4$ also feature expanded octets.

Key Point: Second-period elements (C, N, O, F) never exceed an octet.

Other shortcomings of the octet rule

• It does not explain the shape of molecules (we need VSEPR for that).
• It says nothing about the relative stability or energy of molecules.
• It cannot explain why $O_2$ is paramagnetic — only Molecular Orbital Theory does (Section 11).
• It treats all bonds as localised, ignoring resonance (Section 7).

[NEET Important] 'Octet rule cannot explain paramagnetism of $O_2$' is a frequently asked one-liner.

Solved Examples

Example 1: Formal charge on each oxygen in ozone

For the structure $O = O - O$ (with the usual lone pairs), find the formal charge on the central and terminal oxygens.

Solution:

1. Formula: $FC = V - L - B/2$, with $V = 6$ for oxygen.
2. Central O (double bond + single bond, 1 lone pair): $L = 2$, bonds $= 3$ so $B = 6$. $FC = 6 - 2 - 3 = +1$.
3. Doubly bonded terminal O (2 lone pairs): $L = 4$, $B = 4$. $FC = 6 - 4 - 2 = 0$.
4. Singly bonded terminal O (3 lone pairs): $L = 6$, $B = 2$. $FC = 6 - 6 - 1 = -1$.
5. Check: $(+1) + 0 + (-1) = 0$. ✓ (neutral molecule)

Takeaway: The central O carries $+1$ and one terminal O carries $-1$ — the origin of ozone's overall dipole.

Example 2: Formal charge in the cyanide ion

Find the formal charges on C and N in $CN^-$ ($:C \equiv N:$).

Solution:

1. Carbon: $V=4$, lone pair $L=2$, triple bond $B=6$. $FC = 4 - 2 - 3 = -1$.
2. Nitrogen: $V=5$, lone pair $L=2$, $B=6$. $FC = 5 - 2 - 3 = 0$.
3. Check: $(-1) + 0 = -1$ = ion charge. ✓

Takeaway: The negative charge in cyanide sits on carbon, not nitrogen — a classic counter-intuitive result confirmed by formal charge.

Example 3: Choosing the better structure for CO$_2$

Between $O=C=O$ and $O \equiv C-O$, which is preferred and why?

Solution:

1. $O=C=O$: every atom has formal charge 0.
2. $O \equiv C-O$: gives formal charges of $+1$ (triple-bonded O) and $-1$ (single-bonded O), with 0 on C.
3. Rule: the structure with the smallest formal charges is preferred.
4. Conclusion: $O=C=O$ is the favoured Lewis structure.

Takeaway: Minimise formal charges to find the most stable Lewis structure.

Example 4: Identifying an incomplete octet

How many electrons surround boron in $BF_3$, and what does this make boron?

Solution:

1. Bonds: three B–F single bonds → 3 shared pairs → 6 electrons around B.
2. Octet? No — only 6, an incomplete octet.
3. Consequence: electron-deficient boron is a Lewis acid; it accepts a lone pair (e.g. from $NH_3$) to form $F_3B \leftarrow NH_3$.

Takeaway: Group-13 trihalides are classic electron-deficient Lewis acids.

Example 5: Spotting an odd-electron molecule

Show that $NO$ must violate the octet rule.

Solution:

1. Valence electrons: $N(5) + O(6) = 11$ — an odd number.
2. Pairing: 11 electrons cannot all be paired; at least one electron is unpaired.
3. Conclusion: $NO$ is a free radical and cannot satisfy the octet on both atoms.

Takeaway: Odd total valence electrons ⇒ guaranteed octet violation (free radical).

Example 6: Expanded octet in SF$_6$

How many electrons surround sulphur in $SF_6$, and why is this allowed?

Solution:

1. Bonds: six S–F single bonds → 6 shared pairs → 12 electrons around S.
2. Allowed because: sulphur is a period-3 element and can accommodate more than eight electrons.
3. Conclusion: $SF_6$ has an expanded octet (12 electrons), giving an octahedral shape.

Takeaway: Only period-3+ central atoms can expand the octet; second-period atoms never can.

Example 7: Why can't nitrogen form NF$_5$?

Explain why $NF_5$ does not exist while $PF_5$ does.

Solution:

1. Nitrogen is a second-period element and cannot expand its octet beyond 8 electrons.
2. Phosphorus is in period 3 and can form 5 bonds (10 electrons).
3. Conclusion: $PF_5$ exists (expanded octet) but $NF_5$ cannot.

Takeaway: Second-period elements are limited to a maximum covalence of 4.

Example 8: Formal charge to locate charge in NH$_4^+$

Find the formal charge on nitrogen in the ammonium ion.

Solution:

1. Nitrogen: four N–H bonds, no lone pair. $V=5$, $L=0$, $B=8$.
2. Compute: $FC = 5 - 0 - 4 = +1$.
3. Check: all H atoms have $FC=0$, so total $= +1$ = ion charge. ✓

Takeaway: The positive charge of $NH_4^+$ formally resides on nitrogen.

Example 9: Counting electrons around P in PCl$_5$

State the number of electrons around phosphorus in $PCl_5$ and classify the octet status.

Solution:

1. Bonds: five P–Cl single bonds → 5 shared pairs → 10 electrons.
2. Status: expanded octet (more than 8).
3. Reason: phosphorus can accommodate more than an octet, giving a trigonal bipyramidal shape.

Takeaway: $PCl_5$ = 10 electrons around P = expanded octet.

Example 10: Which limitation explains O$_2$ paramagnetism?

The octet rule predicts $O_2$ to be diamagnetic, yet it is paramagnetic. Which limitation is this?

Solution:

1. Lewis structure $O=O$ shows all electrons paired → predicts diamagnetic.
2. Experiment: liquid $O_2$ is attracted by a magnet → paramagnetic (2 unpaired electrons).
3. Limitation: the octet/Lewis approach cannot account for unpaired electrons or magnetic behaviour — only Molecular Orbital Theory can.

Takeaway: Paramagnetism of $O_2$ is the textbook failure of the octet rule, resolved by MOT (Section 11).

Example 11: Formal charge in the sulphate ion

In the all-single-bond Lewis structure of $SO_4^{2-}$, find the formal charge on sulphur and on each oxygen.

Solution:

1. Sulphur (4 single bonds, no lone pair): $V=6$, $L=0$, $B=8$. $FC = 6 - 0 - 4 = +2$.
2. Each oxygen (1 single bond, 3 lone pairs): $V=6$, $L=6$, $B=2$. $FC = 6 - 6 - 1 = -1$.
3. Check: $(+2) + 4(-1) = -2$ = ion charge. ✓ (Drawing two S=O double bonds lowers S's formal charge to 0 — the preferred expanded-octet structure.)

Takeaway: Expanded-octet structures with double bonds reduce formal charges and are favoured.

Example 12: Lewis acid behaviour from incomplete octet

Predict the product when electron-deficient $BF_3$ reacts with $NH_3$.

Solution:

1. $BF_3$: electron-deficient (6 electrons on B) → Lewis acid.
2. $NH_3$: has a lone pair on N → Lewis base.
3. Reaction: nitrogen donates its lone pair into boron's empty orbital, forming a coordinate bond: $H_3N \rightarrow BF_3$ (an adduct).
4. Result: boron now has a complete octet.

Takeaway: Incomplete-octet molecules complete their octet by accepting a lone pair — the basis of Lewis acid-base chemistry.