The Coordinate (Dative) Bond — Explained with Examples
Most students learn two kinds of bond — ionic, where electrons are transferred, and covalent, where each atom contributes one electron to a shared pair. The coordinate bond is the third case that the syllabus quietly assumes you know: a shared pair in which both electrons came from the same atom. It is the bond that explains NH₄⁺, H₃O⁺, every coordination complex in Class 12, the Al₂Cl₆ dimer, and why BF₃ reacts eagerly with ammonia. This page sets out the rule, the formal-charge arithmetic examiners want to see, and the traps.
The definition and the two requirements
A coordinate bond (also called a dative bond or a dative covalent bond) is a covalent bond in which both shared electrons are supplied by one of the two atoms.
The arrow points from the donor to the acceptor.
Two things must be present, and an exam answer that omits either is incomplete:
- A donor atom or ion carrying at least one lone pair — nitrogen in NH₃, oxygen in H₂O, the halide in Cl⁻, carbon in CO.
- An acceptor with a vacant orbital of suitable energy — H⁺, boron in BF₃, aluminium in AlCl₃, or a transition metal ion with empty d, s and p orbitals.
In Lewis acid–base language the acceptor is the Lewis acid and the donor is the Lewis base. Coordinate bond formation is a Lewis acid–base reaction — the same event under two names, which is worth writing in an answer because it earns the link mark.
The electron configuration tells you who can do what
You can predict donor and acceptor behaviour straight from the ground-state configuration.
| Atom | Ground-state configuration | Role | Why |
|---|---|---|---|
| N (Z = 7) | 1s² 2s² 2p³ | Donor | After three N–H bonds, one lone pair is left over |
| O (Z = 8) | 1s² 2s² 2p⁴ | Donor | Two bonds leave two lone pairs |
| B (Z = 5) | 1s² 2s² 2p¹ | Acceptor | Only 3 valence electrons; in BF₃ the boron is 6-electron and has an empty p orbital |
| Al (Z = 13) | 1s² 2s² 2p⁶ 3s² 3p¹ | Acceptor | Same 6-electron shortfall in AlCl₃ |
| H⁺ | no electrons at all | Acceptor | Empty 1s orbital — the simplest acceptor there is |
Worked example 1 — the ammonium ion, NH₄⁺
Ammonia has three N–H covalent bonds and one lone pair on nitrogen. A proton H⁺ has an empty 1s orbital. The nitrogen lone pair is donated into it:
NH₃ + H⁺ → NH₄⁺
Formal charge check. Formal charge = (valence electrons) − (non-bonding electrons) − ½(bonding electrons).
Nitrogen in NH₄⁺: valence = 5, non-bonding = 0, bonding electrons = 8 (four bonds).
FC = 5 − 0 − ½(8) = 5 − 0 − 4 = +1
Each hydrogen: valence = 1, non-bonding = 0, bonding = 2.
FC = 1 − 0 − ½(2) = 1 − 0 − 1 = 0
Sum of formal charges = (+1) + 4(0) = +1, which matches the charge on the ion — the arithmetic check that your structure is drawn correctly.
The key experimental point: once formed, all four N–H bonds in NH₄⁺ are identical in length and strength. There is no way to tell which one was the "dative" one. A coordinate bond differs only in origin, never in the finished bond.
Worked example 2 — the ammonia–boron trifluoride adduct
In BF₃ the boron has only 6 electrons around it and an empty 2p orbital. Ammonia donates its lone pair:
H₃N: + BF₃ → H₃N → BF₃
Formal charges in the adduct:
Nitrogen: valence = 5, non-bonding = 0, bonding = 8 → FC = 5 − 0 − 4 = +1
Boron: valence = 3, non-bonding = 0, bonding = 8 → FC = 3 − 0 − 4 = −1
Overall = (+1) + (−1) = 0, correct for a neutral molecule.
This separation of formal charge — plus on the donor, minus on the acceptor — is the signature of a coordinate bond and is exactly what a marking scheme looks for.
Shape change worth mentioning: BF₃ on its own is trigonal planar (sp², bond angle 120°). After accepting the pair, boron has four electron domains and the fragment becomes tetrahedral (sp³, close to 109.5°). Ammonia's geometry changes little, but the boron end visibly pyramidalises.
Worked example 3 — counting in a complex ion
In coordination chemistry every metal–ligand bond is a coordinate bond: the ligand donates, the metal ion accepts. The Effective Atomic Number (EAN) rule counts the electrons the metal ends up surrounded by.
For [Fe(CN)₆]⁴⁻:
Iron is Z = 26. The oxidation state of Fe here is +2 (six CN⁻ contribute −6; overall charge is −4, so x + (−6) = −4, giving x = +2).
Electrons on Fe²⁺ = 26 − 2 = 24
Electrons donated by ligands = 6 ligands × 2 electrons = 12
EAN = 24 + 12 = 36, the atomic number of krypton — a noble gas configuration,
which is why this complex is unusually stable.
For [Ni(CO)₄]: nickel is Z = 28 and in the zero oxidation state here, so 28 electrons; four CO ligands donate 4 × 2 = 8. EAN = 28 + 8 = 36 again.
Note honestly: the EAN rule works well for many carbonyls and cyanides but is not universal — [Cu(NH₃)₄]²⁺, for instance, does not reach 36. Use it as a stability guide, not a law.
Where else coordinate bonds appear in your syllabus
| Species | Donor | Acceptor | Why it matters |
|---|---|---|---|
| H₃O⁺ | Lone pair on O of H₂O | H⁺ | The real form of "H⁺" in water; used in every pH calculation |
| Al₂Cl₆ | Lone pair on a bridging Cl | Al of the other AlCl₃ | Explains why anhydrous AlCl₃ exists as a dimer in the vapour and in non-polar solvents |
| [Cu(NH₃)₄]²⁺ | N lone pair of each NH₃ | Cu²⁺ | The deep blue colour test for Cu²⁺ |
| [Ag(NH₃)₂]⁺ | N lone pair | Ag⁺ | Tollens' reagent |
| Fe in haemoglobin | N of the porphyrin ring and of histidine; O₂ | Fe²⁺ | Oxygen transport in blood |
| Chelates such as EDTA complexes | Several donor atoms in one ligand | Metal ion | Hardness titrations, sequestering metal ions |
Mistakes that cost marks
- Treating a coordinate bond as weaker or different from a covalent bond. It is not. All four N–H bonds in NH₄⁺ are equivalent. Only the origin of the pair differs.
- Drawing the arrow the wrong way. The arrow runs donor → acceptor. An arrow from BF₃ to NH₃ is simply wrong and loses the mark.
- Forgetting the vacant orbital. A lone pair alone is not enough. CH₄ has no lone pair and no vacant low-energy orbital, so it neither donates nor accepts.
- Assigning formal charge to the wrong atom. The donor becomes formally positive, the acceptor formally negative — the opposite of what intuition suggests, because the donor gave away a share of its own pair.
- Confusing formal charge with oxidation state. In NH₄⁺ nitrogen has a formal charge of +1 but an oxidation state of −3. They are calculated by different rules and answer different questions.
- Assuming NH₃ can donate three pairs because it has three hydrogens. It has exactly one lone pair, so it is a one-site (monodentate) donor.
Where this is asked in exams
| Exam | Typical question |
|---|---|
| CBSE / ICSE Class 11 | Define a coordinate bond; draw NH₄⁺ and H₃O⁺ showing the dative bond |
| CBSE / ICSE Class 12 | Coordination compounds — ligands, coordination number, Werner's theory, EAN |
| JEE / NEET | Identify species containing a coordinate bond; Lewis acid–base pairs; AlCl₃ dimer |
| IIT-JAM / CUET-PG | Bonding theories for complexes, ligand donor atoms, chelate effect |
| GATE / CSIR-NET | Organometallics, the 18-electron rule, sigma donor and pi acceptor ligands |
Settle donor and acceptor from the configuration. Whether an atom has a spare lone pair or an empty orbital is written into its electron configuration. Type in any element and the Electron Configuration calculator fills the orbitals for you, so you can check N, O, B and Al for yourself instead of memorising a list.
Open the Electron Configuration Calculator →Bonding and coordination compounds are where Class 11 and Class 12 chemistry connect. ABC Chemistry teaches both together — Class 11–12 coaching at the Gurugram centre plus online classes across India, at abcchemistry.in. Students in Delhi, Noida or Gurgaon who prefer one-to-one teaching at home can arrange it through delhihometutor.com.