Covalent bonding

The oxygen you breathe and the water you drink are both made of non-metal atoms, yet no electron is ever fully handed from one atom to another to build them. Instead the atoms share electrons — this sharing is covalent bonding, the second major bonding type in ICSE Class 9 Chemistry Chapter 4, "Atomic Structure and Chemical Bonding," studied right after electrovalent bonding. You are expected to describe, using atomic structure and the duplet (2-electron) and octet (8-electron) rules, how covalent bonds form in molecules such as $H_2$ , $Cl_2$ , $O_2$ , $N_2$ , $HCl$ , $H_2O$ , $NH_3$ , $CCl_4$ and $CH_4$ , to tell single, double and triple bonds apart, and to explain what a radical is. Coordinate bonding, VSEPR theory and hybridisation are beyond this syllabus and are not needed here — everything below stays inside the electron-counting picture you already have from electrovalent bonding.

The core insight

A covalent bond forms when two non-metal atoms each contribute one electron to a shared pair, so that pair counts toward both atoms' outer-shell total at once, letting each atom reach a stable duplet (2) or octet (8) without either atom fully gaining or losing an electron.

The core idea

A covalent bond is a chemical bond formed when two atoms share one or more pairs of electrons, so that each atom attains a stable outer-shell electron arrangement — a duplet of 2 for atoms close to helium (only hydrogen, in this syllabus), or an octet of 8 for every other atom here — without either atom fully gaining or losing electrons. Covalent bonds form between two non-metal atoms because both atoms have a strong pull on electrons and a tendency to gain, not lose, them. Neither atom can play the role of the electron-donating metal from electrovalent bonding, so instead each contributes one electron to a shared pair, and that pair is counted by both atoms simultaneously toward their own stability. Depending on how many electron pairs are shared between the same two atoms, a covalent bond is classified as:

Single bond — one shared pair, e.g. $H$ – $H$ in $H_2$ , $H$ – $Cl$ in $HCl$ .
Double bond — two shared pairs, e.g. $O=O$ in $O_2$ .
Triple bond — three shared pairs, e.g. $N \equiv N$ in $N_2$ .

How many bonds a given atom needs to form is not guesswork — it follows directly from how many electrons that atom is short of a stable shell. If an atom's own outer-shell electron count is $v$ , and its target is 8 (octet) or 2 (duplet, hydrogen only), then each covalent bond it forms supplies exactly one extra electron toward that target (the electron contributed by its partner), so the number of bonds it must form is:

n_b = T - v

where: $n_b$ is the number of covalent bonds (shared pairs) the atom forms, $T$ is its target — $T = 8$ for every atom here except hydrogen, for which $T = 2$ — and $v$ is the number of electrons already in that atom's own outer shell before any sharing. This single small rule is what tells you, before drawing anything, whether an atom will form one bond, two bonds, or more.

When a covalent bond breaks (for example, during a reaction), the two fragments can each keep one electron from the shared pair. Such a fragment — an atom or group of atoms carrying an unpaired electron, or more generally a charged group like $SO_4^{2-}$ or $OH^-$ formed on bond-breaking — is called a radical.

Derivation APredicting how many bonds an atom forms

The goal is to justify the rule $n_b = T - v$ stated above from nothing but the octet/duplet idea already used for $H_2$ and $Cl_2$ , and then check it correctly reproduces the number of bonds each atom forms in $H_2O$ , $NH_3$ , $CH_4$ , $O_2$ and $N_2$ — a check that matters because getting the bond count wrong is the single most common ICSE slip on this topic.

Step 1. Start from the target. Every atom in this chapter "wants" a fixed number of electrons in its outer shell: 2 for hydrogen (a duplet, since its only shell is the K shell) and 8 for every other atom here (an octet). Call this target $T$ .

Step 2. Count what the atom already has. Before any bond forms, the atom brings $v$ electrons of its own to its outer shell — for nitrogen ( $2,5$ ) that is $v=5$ ; for oxygen ( $2,6$ ), $v=6$ ; for carbon ( $2,4$ ), $v=4$ ; for chlorine ( $2,8,7$ ), $v=7$ ; for hydrogen, $v=1$ .

Step 3. Count what one bond supplies. In a single covalent bond the atom contributes one of its own electrons to the shared pair, and its partner contributes the other — the atom's own electron was already inside $v$ , so the new electron it gains access to from one bond is exactly 1 (its partner's contribution). Forming $n_b$ bonds therefore raises the atom's effective outer-shell count from $v$ to $v + n_b$ .

Step 4. Solve for $n_b$ . Setting the final count equal to the target, $v + n_b = T$ , gives $n_b = T - v$ . Checking every molecule in the syllabus: hydrogen, $n_b = 2-1=1$ (matches $H$ – $H$ and $H$ – $Cl$ ); chlorine, $n_b = 8-7=1$ (matches $Cl$ – $Cl$ ); nitrogen, $n_b = 8-5=3$ (matches the three N–H bonds in $NH_3$ ); oxygen, $n_b = 8-6=2$ (matches the two O–H bonds in $H_2O$ ); carbon, $n_b = 8-4=4$ (matches the four C–H bonds in $CH_4$ ). Every count matches the structures already established, so the rule is confirmed, not just plausible.

Worked example

Use the bond-count rule to explain why oxygen forms a double bond in $O_2$ but only single bonds in $H_2O$ , and why nitrogen forms a triple bond in $N_2$ .

Step 1. Find oxygen's required bond count. Oxygen has arrangement $2,6$ , so $v=6$ and $T=8$ , giving $n_b = 8-6=2$ : every oxygen atom must form exactly 2 bonds, however those 2 are distributed.

Step 2. Apply it to water. In $H_2O$ , oxygen spends its 2 required bonds on two separate single bonds, one to each of two different hydrogen atoms: $O(2,6) + 2\,H(1) \longrightarrow H_2O$ . Oxygen ends with $6+2=8$ electrons (a full octet), and each hydrogen ends with $1+1=2$ (a full duplet).

Step 3. Apply it to $O_2$ . Here oxygen's only partner is another oxygen atom, which also needs 2 bonds, so both of oxygen's required bonds go to that same atom, giving one double bond, $O=O$ , instead of two single bonds to two different atoms. Each oxygen atom still ends with $6+2=8$ electrons; the only difference from water is how the 2 required bonds are distributed.

Step 4. Apply the same logic to nitrogen. Nitrogen has $v=5$ , so $n_b = 8-5=3$ . With three separate hydrogen partners, nitrogen forms three single N–H bonds, as in $NH_3$ . With only one nitrogen partner, as in $N_2$ , all 3 required bonds go to that single atom, giving a triple bond, $N \equiv N$ . Answer: oxygen and nitrogen always form exactly 2 and 3 bonds respectively; whether those bonds appear as several single bonds to several atoms, or as one multiple bond to a single identical partner, depends only on how many partner atoms are available, not on the total bond count required.

Competitive stretch

The same rule tells you carbon in $CH_4$ needs $n_b = 8-4=4$ bonds and, having four separate hydrogen partners, forms four single C–H bonds rather than any multiple bond — carbon has no single partner left over to double up with, unlike oxygen and nitrogen bonding to themselves.

Common traps

Confusing covalent bonding with electrovalent bonding — in covalent bonding electrons are shared and stay associated with both atoms; in electrovalent bonding electrons are fully transferred from one atom to the other.
Forgetting that a shared electron pair counts toward the outer-shell total of both bonded atoms at once, not just one — this is why "sharing 1 electron each" gives each atom 2 electrons' worth of stability, not 1.
Applying the octet rule to hydrogen — hydrogen only ever needs a duplet (2 electrons, $T=2$ ), since its single K shell is full at 2; every other atom in this syllabus follows the octet rule ( $T=8$ ).
Assuming a single shared pair is always enough — when $n_b = T-v$ works out to more than 1 for a single remaining partner (as with two oxygen atoms or two nitrogen atoms), the atoms must form a double or triple bond, not just a single one, to fully satisfy both octets.

Ask the tutor: Want a step-by-step electron-dot walkthrough for $H_2O$ , $CCl_4$ , or any other molecule from this chapter — just name the molecule and ask.

Why does hydrogen only need a duplet while every other atom here needs an octet?

It comes down to which shell is being filled. Hydrogen has only one electron shell, the K shell, and the K shell is completely full at 2 electrons — there is no room for more, so 2 is its natural stopping point, matching helium's stable arrangement. Every other atom here (chlorine, oxygen, nitrogen, carbon) is filling its outermost shell beyond the K shell, and those outer shells are stable at 8 electrons, the octet. The target is simply "however many electrons make that shell full."

If oxygen needs 2 bonds and nitrogen needs 3, why doesn't oxygen form a triple bond or nitrogen a quadruple one?

Because the bond count is fixed by how many electrons the atom is short of its octet — forming more would overshoot a stable octet, and forming fewer would leave it incomplete. What can vary is only how those required bonds are distributed: over several different single-bonded partners (as in $H_2O$ or $NH_3$ ), or concentrated onto one identical partner as a multiple bond (as in $O_2$ or $N_2$ ). The total is fixed by $n_b = T - v$ for that atom.

Is a radical the same thing as an ion?

No, though both can carry an unpaired electron or a charge. An ion, from electrovalent bonding, forms when an atom fully gains or loses electrons, leaving a whole-number charge. A radical forms differently — when an existing covalent bond breaks and each fragment keeps just one electron from the shared pair, leaving an unpaired electron on a fragment that may be a single atom or a group of atoms. Some radicals are neutral and some are charged groups like $OH^-$ or $SO_4^{2-}$ formed the same way; the defining feature is that they come from a covalent bond breaking apart, not from a full electron transfer.

Covalent bonding

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads