Lewis Acid–Base Adducts

Introduction

When a Lewis base donates a pair to an electron-deficient molecule, the combined product is an adduct. BF₃·NH₃ is the classic example: nitrogen's lone pair binds to boron, changing boron's coordination from three to four. The example is valuable because it connects Lewis counting, geometry, formal charge and equilibrium. It also shows why “coordinate bond” describes how a bond formed, not a permanent marker of an unusual electron pair.

Core explanation

Free BF₃ has three B–F bonds and a trigonal-planar boron centre. A simple valence count gives only six electrons around B, with an available accepting orbital. NH₃ has three N–H bonds and one nitrogen lone pair. In BF₃ + NH₃ ⇌ F₃B←NH₃, the electron pair moves from N toward B. The arrow notation identifies donor and acceptor. After binding, boron has four sigma-bond directions and is approximately tetrahedral in the adduct, though actual bond angles depend on ligand sizes and bonding.

Formal charge bookkeeping can appear surprising. In a Lewis depiction with four bonds around B, boron is assigned −1 formal charge; nitrogen with four bonds is assigned +1. The total adduct remains neutral. This assignment does not imply the B–N bond is a pair of separated ions. Formal charge is a way to distribute bonding electrons in a drawing; measured electron density depends on polarised covalent interactions among all atoms.

The BF₃–NH₃ bond is often called dative because NH₃ supplied both electrons of the new shared pair at the moment of formation. Once the bond exists, those electrons occupy a bonding interaction and cannot be tagged as permanently “owned” by nitrogen. A conventional covalent bond and a coordinate covalent bond can have similar physical descriptions after formation. The arrow is especially useful for mechanism and donor–acceptor identity, not for claiming a fundamentally separate bond species.

Adduct formation is an equilibrium. Its position depends on temperature, solvent and competing bases. Water, ethers or halide donors can bind BF₃, and a donor already solvated may be less available. Steric crowding can weaken binding even when electron deficiency suggests strong Lewis acidity. For a general equilibrium A + B ⇌ AB, a formation constant can be written K = a(AB)/(a(A)a(B)) in activity terms. A favourable K does not by itself prove rapid formation; kinetics is separate.

BF₃ is a useful caution against simple acidity rankings. Strongly electronegative fluorine withdraws electron density, suggesting a Lewis-acidic B centre, but B–F bonding and ligand donation complicate comparisons with BCl₃ or BBr₃. One should not predict exact adduct strength solely from the halogen electronegativity. The same logic applies to AlCl₃ dimers and metal complexes: electron-pair acceptance is general, but the actual adduct structure depends on all energy terms.

Step-by-step reasoning

1. Draw BF₃ and NH₃ separately with boron's vacancy and nitrogen's lone pair. 2. Draw the electron-flow arrow from N to B, not the reverse. 3. Recount bonds and formal charges in the neutral adduct. 4. Predict boron geometry before and after binding. 5. Discuss formation as a solvent-dependent equilibrium rather than an automatic irreversible step.

Visual explanation

Draw planar BF₃ on the left with an open site above B and pyramidal NH₃ with a lone pair on N. A curved arrow from the lone pair to B leads to a four-coordinate F₃B←NH₃ sketch. Label B trigonal planar before and approximately tetrahedral after.

Real-world analogy

One person brings a shared tool to a collaboration; after both begin using it, the tool is part of their joint work rather than physically stamped with its original owner. The nitrogen pair begins on NH₃ but becomes part of the B–N bonding interaction in the adduct.

Real-world example

BF₃ can be supplied in donor-complexed forms such as ether adducts for synthetic chemistry. Binding a donor changes handling and reactivity relative to free BF₃. Chemists choose an adduct partly for controlled delivery of Lewis acidity, while accounting for the donor's competition with a target substrate.

Why?

Why can boron's geometry change upon NH₃ binding? Free BF₃ has three bonding domains; donation creates a fourth B–N bond direction. Four domains favour an approximately tetrahedral arrangement, though the three B–F bonds remain attached.

Common misconception

“The B–N bond is weaker or not real because it is dative” is unwarranted. Dative describes the electron-pair origin in the formation step. The resulting bond is a genuine shared-electron interaction whose strength must be measured or calculated.

Worked example

Count boron's formal electrons in BF₃·NH₃. Boron contributes three valence electrons and is assigned half of eight bonding electrons in the four bonds, or four; its formal charge is 3 − 4 = −1. Nitrogen contributes five, owns no lone pair after donation and is assigned half of eight bonding electrons, or four, so its formal charge is +1. The charges cancel for a neutral adduct, consistent with BF₃ + NH₃ → BF₃·NH₃.

Quick check

1. Which direction should the Lewis electron-pair arrow point in BF₃·NH₃ formation? Answer: From the nitrogen lone pair of NH₃ toward the electron-deficient boron of BF₃.

Exam focus

Show donor, acceptor, electron-flow arrow, formal charges and geometry change. Distinguish the formation history from the nature of the formed bond. If comparing Lewis acidity, mention solvent and competing donor effects rather than one-factor electronegativity claims.

Advanced insight

Molecular orbital analysis views adduct formation as interaction of an occupied donor orbital with an empty or low-energy acceptor orbital. Electrostatic attraction, orbital overlap, reorganisation energy and solvent stabilisation all contribute to binding free energy. This explains why a simple octet argument predicts possibility but not a precise equilibrium constant.

Summary

BF₃·NH₃ forms when nitrogen donates a lone pair to boron. Boron changes from three- to four-coordinate, and formal-charge bookkeeping yields B−/N+ while the adduct remains neutral. Dative notation identifies pair origin; equilibrium and bond strength depend on conditions.

Practice questions

1. What is boron's geometry in free BF₃ versus the adduct? Answer: Trigonal planar in free BF₃ and approximately tetrahedral in BF₃·NH₃. 2. Why is BF₃ a Lewis acid? Answer: Its boron centre has an accessible electron-pair-accepting site in the simple six-electron structure. 3. Does a favourable formation constant prove adduct formation is fast? Answer: No. The constant describes equilibrium favourability; rate depends on the activation barrier and conditions.