Electron-Deficient Compounds

Introduction

Not every stable molecule satisfies the simple octet rule in a localized Lewis picture. BF₃ has only six electrons around boron in its straightforward three-bond structure, while diborane B₂H₆ has too few electrons to make every B–H and B–B link an ordinary two-centre two-electron bond. Their structures explain why BF₃ accepts electron pairs and why diborane uses bridging hydrogens.

Core explanation

BF₃ contains 3 + 3(7) = 24 valence electrons. Three B–F single bonds use six, and each fluorine receives three lone pairs, using the remaining eighteen. Boron has three bonding domains and no lone pair, so the molecule is trigonal planar with approximate 120° F–B–F angles. Boron has only six electrons in the formal Lewis valence count. An empty acceptor orbital lets BF₃ bind a Lewis base such as NH₃: BF₃ + NH₃ → F₃B←NH₃. The donation raises the number of bonds at boron and can alter geometry toward tetrahedral.

The B–F bonds are strongly polar, and fluorine lone-pair donation can contribute to bonding descriptions, so BF₃ is not a bare isolated B³⁺ centre. Nevertheless, the electron-deficient Lewis structure predicts its pair-accepting behaviour. Comparing Lewis acidity among BF₃, BCl₃ and related compounds requires more than electronegativity alone; back-donation, bond strengths and solvent effects matter. State BF₃'s Lewis-acid function without assuming a simple one-factor ranking.

Diborane has 2(3) + 6(1) = 12 valence electrons. If six ordinary B–H bonds alone used all twelve, no electrons would remain for a B–B bond, and the observed bridged structure would not be explained by two-centre lines. Instead, four hydrogens are terminal, forming ordinary B–H bonds, and two hydrogens bridge the boron atoms. Each bridge is a B–H–B three-centre two-electron interaction: one electron pair is spread over three nuclei. Two such bridges, one above and one below the B–B axis in a common drawing, hold the electron-deficient framework together.

Do not describe each bridging hydrogen as making two independent conventional B–H bonds with four electrons. The bridge contains only two electrons in total in the simple bonding model. Also do not treat the sketch as a standard single B–B bond plus six normal B–H bonds; that would overcount available electrons. The central concept is electron sharing over more than two centres.

Electron deficiency is not the same as universal instability. BF₃ and B₂H₆ are isolable under suitable conditions, yet they are reactive toward electron-pair donors or other reagents. Their actual chemistry depends on partner and environment. College-level boron-hydride teaching at https://chem.libretexts.org/Bookshelves/Inorganic Chemistry/Chemistry of the Main Group Elements %28Barron%29/06%3A Group 13/6.04%3A Boron Hydrides shows the bridging 3c–2e description, while Calvin University's inorganic notes at https://chem.libretexts.org/Courses/Calvin University/Chem 230%3A Essential Inorganic Chemistry/02%3A Bonding Theory Overview/2.01%3A Covalent- Lewis Octet/2.1.04%3A Lewis- Electron-Deficient Octets of B and Be discuss BF₃ and diborane as octet exceptions.

Step-by-step reasoning

1. Count available valence electrons before drawing any bond network. 2. For BF₃, assign three bonds and complete fluorine octets to reveal boron's six-electron count. 3. Infer electron-pair acceptance and test it with a donor adduct equation. 4. For B₂H₆, allocate four terminal B–H bonds and two B–H–B bridges. 5. Check that each bridge uses one pair over three centres, preserving the total electron count.

Visual explanation

Draw planar BF₃ with three F atoms around B and an empty orbital direction above the plane. Draw B₂H₆ with four terminal H atoms and two shared bridge H atoms between the borons, one above and one below. Label each bridge “3 centres, 2 electrons.”

Real-world analogy

Two people can hold one shared handle between them rather than each owning a separate handle. The diborane bridge similarly shares one electron pair across two borons and one hydrogen. The analogy illustrates sharing, but the real stabilization comes from quantum electron density.

Real-world example

BF₃ is used as a Lewis-acid catalyst or reagent in synthetic chemistry because it binds electron-pair donors and can activate substrates. Diborane chemistry illustrates a different kind of electron economy that appears throughout borane clusters. Neither practical use should obscure that both materials require controlled professional handling.

Why?

Why can NH₃ bind BF₃? Nitrogen has a lone pair available to donate, while boron in BF₃ has an electron-deficient acceptor site. The resulting N→B coordinate bond uses the donor pair and makes a four-coordinate boron adduct.

Common misconception

“Every stable compound must have a conventional octet at every atom” fails for BF₃ and diborane. BF₃ is a Lewis acid with six formal boron electrons, and diborane uses multicentre bonds that a simple two-centre octet scheme misses.

Worked example

Count electrons in B₂H₆: two borons contribute six and six hydrogens six, total twelve. Four terminal B–H bonds use eight electrons. Two electrons occupy each of the two B–H–B bridges, using the remaining four. This assignment accounts for all twelve electrons and explains the bridged skeleton without inventing an ordinary B–B bond.

Quick check

1. How many electrons occupy one B–H–B bridge in the simple diborane model? Answer: Two electrons shared across three atomic centres.

Exam focus

State BF₃'s trigonal-planar shape and six-electron boron count, then identify it as a Lewis acid. For diborane, draw two bridging H atoms and label each 3c–2e bond. Show an electron count rather than only saying “octet exception.”

Advanced insight

Three-centre bonding can be analysed with molecular orbitals: combinations of three atomic contributions create a bonding orbital occupied by the two shared electrons. The simple bridge picture predicts topology, but a full calculation provides electron density and bond-order distribution without assigning wholly separate pairs to each B–H segment.

Summary

BF₃ is electron deficient at boron and accepts donor pairs. B₂H₆ shares electrons across two B–H–B bridges, each a three-centre two-electron bond. Both structures show where localized octet bookkeeping must be extended to describe real main-group compounds.

Practice questions

1. What molecular geometry does BF₃ have before binding a donor? Answer: Trigonal planar around boron, with three bonding domains. 2. Which atom donates the pair in BF₃ + NH₃? Answer: Nitrogen in NH₃ donates its lone pair to the electron-deficient boron centre. 3. Why is a drawing of six normal B–H bonds plus one normal B–B bond impossible for B₂H₆? Answer: Seven ordinary two-electron bonds would require fourteen electrons, but B₂H₆ has only twelve valence electrons.