Coordinate bond

A coordinate bond needs exactly two ingredients on two different atoms, and both must be present simultaneously:

• The donor must have a lone pair of electrons it is willing to share — a filled, non-bonding orbital.
• The acceptor must have a vacant orbital of accessible energy to receive that pair — an electron-deficient centre such as $\text{H}^+$ (no electrons at all) or boron in $\text{BF}_3$ (an incomplete octet).

The bond is written with an arrow from donor to acceptor, $\text{D}\rightarrow\text{A}$, distinguishing its origin from the plain line ($\text{D}-\text{A}$) of a normal shared-pair bond. That distinction is only notational bookkeeping about where the electrons came from; the resulting bond has the same length, the same strength, and obeys VSEPR exactly like any other bond around that atom. This is why all four N–H bonds of $\text{NH}_4^+$ are experimentally indistinguishable even though they were not all made the same way. Formal charge is the tool that keeps track of the donation without needing to relabel every atom "special." For any atom in a Lewis structure,

where: $V$ is the number of valence electrons the free atom normally has, $N$ is the number of non-bonding (lone-pair) electrons it owns in the structure, and $B$ is the total number of electrons in the bonds around it (so $B/2$ is the number of bonds, each bond credited half to each atom). A donor atom, having given up ownership of a lone pair without gaining a new bond partner "for itself," typically ends up with a positive formal charge; an acceptor gains a bond it did not have before and often ends up negative, or the two charges cancel exactly when the adduct is neutral overall.

The reason this donation succeeds — and produces a bond indistinguishable from an ordinary one — is that a covalent bond is defined purely by the presence of a shared electron pair in a bonding orbital between two nuclei, not by how the two electrons were assembled there. Quantum mechanically, once the pair occupies the shared orbital, both electrons are indistinguishable and belong equally to both atoms; the orbital does not "remember" that both electrons started on nitrogen. This is why VSEPR, which counts electron domains around a central atom to predict shape, treats a coordinate bond exactly like any other bonding pair: $\text{NH}_4^+$ has four bonding pairs and no lone pair on nitrogen, so it is tetrahedral, full stop — the fact that one pair arrived by donation changes nothing about the geometry. Formal charge is what lets us keep the electron count honest through the donation without inventing a different bonding rule. Before donation, the donor atom already "owned" both electrons of its lone pair (each lone-pair electron counts fully toward that atom in the $N$ term of Equation 1). After donation, those same two electrons sit in a bond, so the donor now owns only one of them (the $B/2$ term credits half a bond to each atom) — the donor's electron count on paper drops by one, which is exactly a $+1$ shift in formal charge. The acceptor's count rises by exactly one bond's worth, a $-1$ shift (or, from an ion with zero electrons like $\text{H}^+$, a shift from $+1$ actual charge to $0$ formal charge on that atom, with the $+1$ now sitting on the donor). Charge is conserved overall — it has only moved from being a "real" ionic charge on the acceptor to a "formal" bookkeeping charge on the donor atom.

Confirming that $\text{H}_3\text{N}\rightarrow\text{BF}_3$ is a neutral adduct.

Step 1. Draw the two pre-donation fragments. In $\text{NH}_3$, nitrogen ($V = 5$) has 3 bonds and 1 lone pair, so by Equation 1, $\text{FC}_{\text{N}} = 5 - 2 - \tfrac{6}{2} = 0$. In $\text{BF}_3$, boron ($V = 3$) has 3 bonds and no lone pair (an incomplete octet — this is exactly the vacant-orbital acceptor condition), so $\text{FC}_{\text{B}} = 3 - 0 - \tfrac{6}{2} = 0$. Both fragments are neutral before they meet.

Step 2. Identify donor and acceptor. Nitrogen has a lone pair to give; boron has an empty $p$ orbital to receive it. The lone pair on nitrogen is donated into that empty orbital, forming a new N–B bond: $\text{H}_3\text{N} + \text{BF}_3 \rightarrow \text{H}_3\text{N}\rightarrow\text{BF}_3$.

Step 3. Recompute formal charge on nitrogen after donation. Nitrogen now has 4 bonds and 0 lone pairs: $\text{FC}_{\text{N}} = 5 - 0 - \tfrac{8}{2} = 5 - 4 = +1$.

Step 4. Recompute formal charge on boron after donation. Boron now has 4 bonds and 0 lone pairs: $\text{FC}_{\text{B}} = 3 - 0 - \tfrac{8}{2} = 3 - 4 = -1$.

Step 5. Add the two shifted charges: $(+1) + (-1) = 0$. The adduct $\text{H}_3\text{N}\rightarrow\text{BF}_3$ is overall neutral, exactly as expected for a molecule assembled from two neutral fragments — the coordinate bond redistributes formal charge onto the donor and acceptor atoms without changing the total.

• Treating a coordinate bond as weaker or "different" in the final molecule. It is not — once formed, it is a normal covalent bond in every measurable respect; only its electron-counting origin is special.
• Forgetting that both conditions are needed together. A lone pair with no nearby empty orbital, or an empty orbital with no nearby lone pair, cannot form a coordinate bond — you need a donor and an acceptor present at once.
• Drawing the arrow backwards. The arrowhead always points from the donor (lone-pair source) toward the acceptor (empty-orbital owner), never the reverse.
• Skipping the formal-charge recount after donation. Applying $\text{FC} = V - N - B/2$ only to the pre-donation fragments, and forgetting to redo it for the product, is the most common reason students get the wrong overall charge on an adduct or ion.

Ask the tutor: Want to walk through the formal-charge calculation for a coordinate bond in a species not covered here, or see how this connects to VSEPR shape predictions?