Covalent Lewis structures

Take two chlorine atoms, each one electron short of a full outer shell. Neither can steal an electron from the other — they are identical, with identical pull. So instead they do something cleverer: they share a pair of electrons, and that shared pair counts toward both octets at once. This is covalent bonding, the second great bonding strategy in chemistry (after the electron-transfer trick of electrovalent bonds), and it is how almost every non-metal-to-non-metal molecule you know — H₂, O₂, N₂, HCl, H₂O, CH₄ — is held together. A Lewis structure (electron-dot diagram) is simply a bookkeeping picture of this sharing: every valence electron drawn as a dot, shared pairs sitting between two symbols, unshared "lone pairs" sitting on one symbol alone. This article gives you a reliable electron-counting method to construct any simple Lewis structure without guessing, applies it to a new molecule (O₂) to confirm the double bond, and sharpens the single/double/triple and polar/non-polar distinctions you met in Pass 1.

The core insight

A covalent bond is a shared pair of electrons that simultaneously completes the octet (or duplet, for hydrogen) of both bonded atoms, and the number of pairs shared between the same two atoms fixes whether the bond is single, double, or triple.

The core idea

Sharing, not transferring. Electrovalent (ionic) bonding works by one atom fully giving up electrons to another. Covalent bonding works differently: two non-metal atoms, each short of a full outer shell, place one or more electron pairs between their nuclei so that pair is claimed by both atoms at once. Neither atom loses its electron; both gain the benefit of counting it.

Counting how many pairs are shared. You do not need to guess how many pairs two atoms share — you can calculate it. Every atom in the structure "needs" a fixed number of electrons around it: 8 for atoms following the octet rule (C, N, O, Cl, and most non-metals), 2 for hydrogen (the duplet rule, since hydrogen's only shell holds a maximum of two electrons). Compare that total need against the electrons actually available (the sum of each atom's valence electrons), and the shortfall tells you how many electrons must be shared.

Classifying the bond by pair count. Once you know a molecule's total bonding electrons, dividing by two gives bonding pairs, and comparing that to the number of bonds needed tells you whether each bond is single (one shared pair), double (two shared pairs between the same two atoms), or triple (three shared pairs).

N_{\text{bond}} = \frac{N_{\text{needed}} - N_{\text{available}}}{2}

where: $N_{\text{needed}}$ is the sum of each atom's target electron count (8 per octet atom, 2 per hydrogen), $N_{\text{available}}$ is the sum of the valence electrons every atom actually brings to the molecule, and $N_{\text{bond}}$ is the total number of bonding (shared) pairs in the whole structure. Once $N_{\text{bond}}$ is found, the leftover electrons — $N_{\text{available}}$ minus the electrons used for bonding — are distributed as lone pairs so that every atom's final electron count (shared plus lone) reaches its target.

Polarity is a separate question from bond order. Whether a bond is single, double, or triple tells you how many pairs are shared. Whether it is polar or non-polar tells you how evenly that sharing happens — equally between identical atoms (non-polar, e.g. the N≡N of N₂ or the Cl–Cl of Cl₂), or unequally toward the stronger electron-attractor when the two atoms differ (polar, e.g. H–Cl or O–H). A bond of any order — single, double, or triple — can in principle be polar or non-polar; the two classifications are independent.

Figure

Electron dot diagram of oxygen O2 showing two shared pairs forming a double bond and two lone pairs on each oxygen atom

Derivation ACounting shared pairs from valence electrons

The goal is to derive, from nothing but each atom's valence electron count and its octet or duplet target, the exact number of bonding (shared) electron pairs a small covalent molecule must have — turning Lewis-structure drawing from guesswork into a two-line calculation that also predicts whether a bond is single, double, or triple.

Step 1. List what each atom needs and what it brings. For every atom in the molecule, write its target electron count ( $8$ for octet-following non-metals, $2$ for hydrogen) and its actual valence electron count (read off the group number, or the outer-shell electrons from its electron arrangement). Sum the targets across all atoms to get $N_{\text{needed}}$ , and sum the valence electrons to get $N_{\text{available}}$ .

Step 2. Find the electron shortfall. Every atom short of its target must gain electrons by sharing, and because a shared pair counts fully toward both atoms at once, the total shortfall $N_{\text{needed}} - N_{\text{available}}$ is exactly twice the number of bonding pairs needed (each pair fills one "slot" of need on each of its two atoms simultaneously).

Step 3. Divide by two to get bonding pairs. $N_{\text{bond}} = (N_{\text{needed}} - N_{\text{available}})/2$ . If the molecule has only one bond position (two atoms joined directly, as in a diatomic molecule), $N_{\text{bond}}$ itself tells you the bond order: $1$ means single, $2$ means double, $3$ means triple.

Step 4. Place the remaining electrons as lone pairs. The electrons not used for bonding are $N_{\text{available}} - 2N_{\text{bond}}$ ; distribute these around the atoms, symmetrically where the atoms are equivalent, so every atom's final count (shared pairs it participates in, counted fully, plus its own lone pairs) reaches its target from Step 1. Check against Pass 1's own results: for N₂, $N_{\text{needed}} = 8+8=16$ , $N_{\text{available}} = 5+5=10$ , so $N_{\text{bond}} = (16-10)/2 = 3$ — a triple bond, exactly as found before. For HCl, $N_{\text{needed}} = 2+8=10$ , $N_{\text{available}} = 1+7=8$ , so $N_{\text{bond}} = (10-8)/2 = 1$ — a single bond, again matching Pass 1.

Worked example

Use the electron-counting method to construct the Lewis structure of oxygen, O₂, and confirm its bond type.

Step 1. Find $N_{\text{needed}}$ and $N_{\text{available}}$ . Oxygen (Z = 8) has electron arrangement 2, 6, so each oxygen atom brings 6 valence electrons and targets a full octet of 8. For two oxygen atoms: $N_{\text{needed}} = 8 + 8 = 16$ and $N_{\text{available}} = 6 + 6 = 12$ .

Step 2. Compute the bonding pairs. $N_{\text{bond}} = (16 - 12)/2 = 2$ . Two electron pairs must be shared between the two oxygen atoms.

Step 3. Classify the bond. Since O₂ has only one bond position (two atoms directly joined) and $N_{\text{bond}} = 2$ , this is a double covalent bond, written O=O — matching what Pass 1 stated by inspection, now confirmed by calculation.

Step 4. Place the lone pairs. Bonding uses $2 \times 2 = 4$ electrons out of the 12 available, leaving $12 - 4 = 8$ electrons as lone pairs, split evenly as 2 lone pairs on each oxygen (4 electrons each). Check each atom's total: 2 shared pairs (counted fully, 4 electrons) + 2 lone pairs (4 electrons) = 8 electrons — octet satisfied on both oxygens.

Step 5. Classify polarity. Both atoms are identical oxygen, so the two shared pairs are held equally by both nuclei: O=O is non-polar covalent, just like the N≡N of Pass 1's first example, and unlike the polar H–Cl of its second.

Competitive stretch

Apply the same method to methane, CH₄. Carbon (Z = 6, valence 4) targets an octet of 8; each of the 4 hydrogens targets a duplet of 2. So $N_{\text{needed}} = 8 + 4(2) = 16$ and $N_{\text{available}} = 4 + 4(1) = 8$ , giving $N_{\text{bond}} = (16-8)/2 = 4$ . With four separate C–H bond positions, those 4 pairs split one each into four single covalent bonds, with no lone pairs left over ( $8 - 2(4) = 0$ ) — consistent with CH₄'s familiar structure and its non-polar C–H bonds.

Common traps

Applying $N_{\text{bond}} = (N_{\text{needed}} - N_{\text{available}})/2$ to a multi-bond molecule (more than one bond position, like CH₄) and assuming the whole result is the order of a single bond — check how many separate bonds must share that total first.
Forgetting hydrogen's target is 2, not 8, and using 8 for every atom in the sum — this overstates $N_{\text{needed}}$ and gives the wrong bonding-pair count.
Leaving lone-pair electrons undrawn once the shared pairs are placed — every available electron must appear somewhere, either shared or as a lone pair, or the octet check will not balance.
Treating bond order (single/double/triple) and bond polarity (polar/non-polar) as the same property — a double bond between identical atoms (O=O) is still non-polar, while a single bond between different atoms (H–Cl) can be polar.

Ask the tutor: Want to try the electron-counting method on a molecule not covered here, like H₂O or Cl₂, and have your Lewis structure checked step by step?

Why does hydrogen only need 2 electrons while every other atom in these examples needs 8

Hydrogen has only one electron shell, and that shell (like helium's) is full with just 2 electrons — there is no second shell to fill up to 8. This is why hydrogen follows a duplet rule rather than the octet rule that governs carbon, nitrogen, oxygen, chlorine, and other non-metals with an outer shell that holds up to 8 electrons. Always check which rule an atom follows before adding its target to $N_{\text{needed}}$ .

Does a higher bond order like a triple bond mean the bond is automatically stronger or shorter than a single bond

Within the same pair of elements, yes — more shared pairs pull the two nuclei closer and hold them more tightly, so a triple bond is generally shorter and stronger than a double bond, which is in turn shorter and stronger than a single bond between the same two atoms. This ICSE syllabus stops at identifying and drawing the bond types, though, so you are not expected to calculate or compare exact bond energies or bond lengths — just to recognise which type a Lewis structure shows.

If I already know a molecule's structure by memory, like H2O having two O-H bonds, do I still need to use the counting method

For familiar molecules, recognising the structure by memory is fine and faster in an exam. The counting method is most useful as a check when you are unsure, or for a molecule you have not memorised — it guarantees you get the correct number of bonding pairs and lone pairs instead of guessing, and it is exactly the kind of reasoning examiners reward when they ask you to "justify" a Lewis structure rather than just draw one.

Covalent Lewis structures

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads