Coordinate bonding

In an ordinary covalent bond, each atom contributes exactly one electron, and the pair is shared roughly equally. A coordinate bond is chemically the same kind of bond — a shared pair sitting between two nuclei — but it is built differently.

• The donor is an atom that already has a full octet and at least one lone pair — electrons not already used in any bond. Nitrogen in NH₃ and oxygen in H₂O both qualify.
• The acceptor is a species with a vacant position that can hold two more electrons. The syllabus's standard acceptor is H⁺, a bare proton with no electrons of its own at all, so it cannot form a normal covalent bond by contributing an electron — it has none to contribute.
• The donor's lone pair moves in wholesale to fill the acceptor's vacancy, and a new bond appears where none of the acceptor's own electrons are involved.

where: the donor is the lone-pair-bearing atom (N in NH₃, O in H₂O), and the acceptor is the electron-deficient species (H⁺ in both syllabus examples). No new electrons are created — the pair that ends up shared already belonged, in full, to the donor before the reaction. Two conditions must both hold before a coordinate bond can form: the donor must have a spare lone pair, and the acceptor must have room to receive it. Miss either condition and no bond forms. Once formed, a coordinate bond is completely indistinguishable from an ordinary covalent bond in strength, length and behaviour. The label "coordinate" describes only the bond's origin — where its two electrons came from — not anything about the finished bond itself. This is why, in the product ion, chemists cannot point to "the" coordinate bond; all the equivalent bonds have merged into one identical set.

The syllabus's two examples both follow the same logic, and understanding why removes the need to memorise them separately. Nitrogen has five valence electrons. Three go into ordinary N–H covalent bonds in NH₃ (one electron from N, one from each H), which uses up three electrons and leaves nitrogen with exactly one pair left over — a lone pair, already a complete pair, not needing to borrow from anywhere. That lone pair sits ready and available. A hydrogen ion, H⁺, is what remains after a hydrogen atom loses its single electron. It is a bare proton: no electron shell, nothing to contribute, but an empty 1s orbital ready to accept two electrons. When NH₃ and H⁺ meet, nitrogen's lone pair is pulled entirely into that empty orbital, shared between N and the incoming H exactly like any other covalent pair, only both electrons started out on nitrogen. The same reasoning explains H₂O + H⁺ → H₃O⁺. Oxygen has six valence electrons; two form ordinary covalent bonds to the two hydrogens in H₂O, leaving two lone pairs (not one). Either lone pair can donate to an incoming H⁺, giving oxygen three O–H bonds (two ordinary, one coordinate) and one lone pair remaining in H₃O⁺. This is why chemists say acids release hydronium ions, H₃O⁺, in water rather than "naked" protons — a bare H⁺ is captured by the nearest available lone pair the instant it is released, which in water means a lone pair on a neighbouring H₂O molecule.

Track the electron count through NH₃ + H⁺ → NH₄⁺ and confirm both the donor–acceptor identification and the final charge on the product ion, step by step.

Step 1. Start from nitrogen's valence electron count: 5. In NH₃, three of these pair up with the single electrons of three separate hydrogen atoms, forming three ordinary N–H covalent bonds. That accounts for 3 of nitrogen's 5 electrons; the remaining 2 sit together as one lone pair.

Step 2. Identify the acceptor. H⁺ is a hydrogen atom that has lost its one electron, so it brings 0 electrons and an empty orbital to the reaction — it cannot form a bond the ordinary way because it has nothing to contribute.

Step 3. Apply the two conditions for coordination: nitrogen has a spare lone pair (condition 1, satisfied) and H⁺ has a vacancy (condition 2, satisfied). The lone pair on nitrogen is donated wholesale into that vacancy, forming the fourth N–H bond using two electrons that both originated on nitrogen.

Step 4. Count bonds and charge in the product. Nitrogen now has four N–H bonds (an octet), so NH₄⁺ is correctly saturated. Charge bookkeeping: the neutral NH₃ molecule combined with a species carrying a +1 charge (H⁺ has one more proton than electron), and no electrons were created or destroyed in the process — so the whole ion NH₄⁺ carries that same net +1 charge.

Step 5. Confirm indistinguishability. All four N–H bonds in NH₄⁺ are the same bond length, the same strength, and chemically identical — there is no experiment that identifies "the" coordinate bond after formation, since the label only ever described how one particular bond originated, not what it currently is.

The ammonium ion NH₄⁺ carries a +1 charge, with four identical N–H bonds, one of which formed by nitrogen donating both electrons of its lone pair to H⁺.

Competitive stretch

Repeat the electron count for H₂O + H⁺ → H₃O⁺. Oxygen starts with 6 valence electrons; 2 form ordinary covalent bonds to the two hydrogens in H₂O, leaving 2 lone pairs (4 electrons). One lone pair donates to H⁺, using 2 of those 4 electrons and leaving oxygen with exactly one lone pair in H₃O⁺, alongside three O–H bonds (two ordinary, one coordinate). H₃O⁺ carries a +1 charge and retains one lone pair on oxygen — the reason a dot-and-cross diagram of hydronium always shows exactly one unshared pair on O, never two.

• Calling a coordinate bond "weaker" or fundamentally different once formed — it is not; the special description applies only to how the bond's electrons originated, and the finished bond behaves exactly like an ordinary covalent bond.
• Forgetting to check that the donor genuinely has a spare lone pair before assigning it as donor — a coordinate bond simply cannot form without one.
• Swapping donor and acceptor — the donor is the electron-rich atom with a complete octet and lone pair(s) (N in NH₃, O in H₂O); the acceptor is the electron-deficient species with the vacancy (H⁺ in both examples).
• Believing the four N–H bonds in NH₄⁺ (or the three O–H bonds in H₃O⁺) stay distinguishable in the product — once formed, they merge into an identical set, and the coordinate origin only ever mattered during formation.

Ask the tutor: Want to work through the electron-dot diagram for NH₄⁺ or H₃O⁺ step by step, or check whether you have the donor and acceptor right in a new example — ask the AI tutor to walk it through with you.