Group 13 (Boron family) — anomaly of B, diagonal B-Si; BX₃ Lewis acidity; diborane B₂H₆ (3c-2e bond), borazine: Visual walkthrough

Step 1 — Count what each atom brings

WHAT. Before drawing a single bond, we count how many valence electrons the whole molecule owns. A valence electron is an electron in the outermost shell — the only ones that make bonds. Think of each one as a hand an atom can hold with.

WHY. A bond is a shared pair of electrons. If we know the total number of electrons, we know exactly how many bonds we are allowed to draw — no more, no less. We must do this first, because everything strange about diborane comes from this number being too small.

PICTURE. Look at the figure: two boron atoms (cyan), six hydrogen atoms (white), each drawn with its hands (dots).

$two B atoms 2 \times each B has 3 valence e^{-} 3 + six H atoms 6 \times each H has 1 valence e^{-} 1 = total electrons available 12$

The $2$ counts the two borons.
The $3$ is boron's valence-electron count (group 13: $2 s^{2} 2 p^{1}$ → three outer electrons).
The $6$ counts the six hydrogens; each $\times 1$ because hydrogen has exactly one electron.
The grand total, 12 electrons, is our entire budget. Remember this number.

Step 2 — Try the "obvious" molecule and watch it fail

WHAT. We attempt the naïve structure everyone draws first: copy ethane $C_{2} H_{6}$ . Put the two borons in the middle, hang three hydrogens off each, and place a bond directly between the two borons. Now count the bonds themselves — not "bonds per atom", because a bond shared between two atoms must be counted once, not twice.

WHY. We test the simplest idea so its failure teaches us. A "normal" bond is a 2-centre 2-electron (2c–2e) bond: two atoms, one shared pair. To fill boron's octet the ethane way, each boron would need four bonds around it — but the B–B bond is shared, so we must not add it twice.

PICTURE. The figure draws the ethane-analogue and lists every distinct line: six B–H plus one B–B.

$six B-H lines 6 + one B-B line 1 = distinct lines 7 bonds$

$counted once each 7 bonds \times each 2c-2e bond 2 e^{-} / bond = electrons demanded 14 e^{-}$

$demanded 14 - available (Step 1) 12 = electrons SHORT 2$

The $7$ is the number of distinct two-electron bonds in the ethane-analogue (six B–H, one B–B), each line counted once.
Multiplying by $2$ turns bonds into electrons: this structure demands 14 electrons.
Subtracting our budget of $12$ leaves a deficit of 2 electrons.

We are two electrons short — one boron ends up with only 6 electrons around it, not the 8 it wants. The naïve ethane molecule is not allowed to exist as drawn. This shortfall is exactly what "electron-deficient" means: even the most economical normal structure cannot be paid for.

Step 3 — Spend the electrons you do have on the safest bonds first

WHAT. Since we cannot afford the ethane structure, we rebuild from scratch and buy the cheapest, most reliable bonds first: four ordinary terminal B–H bonds — two on each boron, pointing outward.

WHY. These four hydrogens have no reason to be special; a normal 2c–2e B–H bond is the natural, strong choice. We commit two hydrogens per boron this way and see what electrons remain in the wallet.

PICTURE. Four white terminal bonds drawn; the electron counter ticks down.

\qquad\Longrightarrow\qquad \underbrace{12}_{\text{budget}} - \underbrace{8}_{\text{spent}} = \underbrace{4 \text{ e}^-\text{ left}}$$ - The four terminal bonds swallow $8$ electrons — two-thirds of the whole budget. - We have **4 electrons left**, and still **two hydrogens and both borons** are unbonded. Here is the crisis: two hydrogens need to be attached, both borons still need more bonding, but we own only **two pairs** of electrons. Two pairs cannot make four separate normal bonds. --- ## Step 4 — The trick: let one pair hold *three* atoms **WHAT.** We stop insisting each pair sit between two atoms. Instead, we place a single hydrogen *between* the two borons and let **one electron pair bind all three** — the atom chain $B\!-\!H\!-\!B$. This is a **three-centre two-electron (3c–2e) bond**, nicknamed a **banana bond** because its shared cloud curves like a banana. **WHY.** We are short exactly the electrons two ordinary bonds would need, but we still must attach two hydrogens and knit the borons together. The only way to bind *more atoms with fewer electrons* is to make each pair do double duty — spread over three centres instead of two. With **4 electrons left**, we can afford **two** such 3c–2e bridges. **PICTURE.** Two amber bridge clouds, one above the molecular plane and one below, each cradling a hydrogen between the borons. $$\underbrace{2}_{\text{bridges}} \times \underbrace{2}_{\substack{\text{e}^- \text{ in each}\\ \text{3c–2e bond}}} = \underbrace{4 \text{ e}^-}_{\text{exactly what remained}}$$ - Each bridge is *one pair* ($2$ electrons) shared over the three atoms $B$, $H$, $B$. - Two bridges use the final $4$ electrons — **the budget closes to zero.** > [!definition] Three-centre two-electron (3c–2e) bond > ==One electron pair (2e) shared among three atoms (3c)==. In diborane each bridge is a $B\!-\!H\!-\!B$ unit holding a single pair. There is **no direct B–B bond** — the borons are stitched together *through* the bridging hydrogens. --- ## Step 5 — Check the books balance **WHAT.** Add up every electron we spent and confirm it equals the budget from Step 1. **WHY.** A structure is only valid if it uses *exactly* the electrons available — not one more, not one fewer. This is the accountant's final audit. **PICTURE.** The full molecule, colour-coded: white terminal bonds, amber bridges, with a running total. $$\underbrace{4 \times 2}_{\substack{\text{terminal}\\ \text{B–H} = 8}} \;+\; \underbrace{2 \times 2}_{\substack{\text{bridge}\\ \text{3c–2e} = 4}} \;=\; \underbrace{12}_{\text{matches Step 1 ✓}}$$ The books balance perfectly. Twelve electrons in, twelve electrons placed. The bridged structure is not a guess — it is the **only** arrangement the electron count permits. > [!formula] What the geometry looks like (consequence of the count) > - **4 terminal** H lie in one plane; **2 bridging** H sit above and below it. > - Terminal $B\!-\!H \approx 119$ pm (normal). Bridging $B\!-\!H \approx 133$ pm — **longer**, because a 3c–2e bond spreads its single pair thinner, so it grips each atom more weakly. > - Each boron uses **four $sp^3$ hybrid orbitals** (two toward terminal H, two toward the bridges), so its bonding is *based on* $sp^3$. But the angles are **distorted** away from the ideal $109.5^\circ$: the terminal $\angle H_t\text{–}B\text{–}H_t \approx 121^\circ$ (opened up) while the bridge $\angle H_b\text{–}B\text{–}H_b \approx 97^\circ$ (squeezed). **Why:** the strong terminal bonds repel each other more than the weak, electron-poor bridges do, so the terminal angle widens past tetrahedral and the bridge angle closes below it. "Roughly $sp^3$" means the hybridisation scheme is tetrahedral-type, not that every angle is exactly $109.5^\circ$. --- ## Step 6 — Where do the bridge orbitals actually come from? (the cleavage corner) **WHAT.** We zoom into one bridge and ask: *which* orbitals overlap to make the banana? Each boron offers an $sp^3$ hybrid; the bridging hydrogen offers its $1s$. Three orbitals, one pair of electrons. **WHY.** Someone will object: "if only one pair lives there, isn't the bond half-empty?" We must show the geometry is stable and cover the **extreme case** — the moment the two borons are pulled apart. This is [[Back-bonding (p\u03c0\u2013p\u03c0)]] thinking applied to a bridge: orbital overlap decides everything. **PICTURE.** The three overlapping lobes forming one banana cloud, with the electron pair drawn inside it. - Two boron $sp^3$ lobes reach *toward* the same hydrogen; the H $1s$ sits in the overlap. - The single pair occupies the **bonding combination** of all three — lowest energy, so the arrangement is stable. - **The pull-apart (cleavage) limit:** stretch the two borons apart and the three-way overlap dies; with only one pair you *cannot* form two independent B–H bonds, so the molecule would rather stay bridged than split. Symmetric cleavage only happens when an outside base (e.g. $NMe_3$) donates *fresh* electron pairs: $B_2H_6 + 2NMe_3 \to 2\,Me_3N\!\cdot\!BH_3$. The base supplies the electrons the bridge could not. > [!mistake] "The bridge is just two half-bonds" > **Why it feels right:** One pair, two B–H links — sounds like sharing one bond between two. > **The fix:** It is **one bond over three atoms**, not two weak bonds. The whole $B\!-\!H\!-\!B$ unit is a single quantum object. That is why breaking it needs *external* electrons, not just heat. --- ## The one-picture summary The whole derivation on one blueprint: budget **12 e⁻** → the cheapest normal structure (ethane-analogue, **7 distinct bonds**) demands **14** → **2 short** → so we rebuild: spend 8 on four terminal bonds → **4 left** → two 3c–2e bridges spend exactly those 4 → **books balance**. Electron-deficiency *forces* the banana bridges; nothing was assumed. > [!recall]- Feynman: the whole walk in plain words > Boron is a small atom that shows up to the party with only three hands each. Two borons and six hydrogens together have **twelve hands**. To build the ordinary molecule everybody expects — the ethane look-alike, with six B–H links and one B–B link, seven handshakes in all — you'd need **fourteen hands**. You're **two hands short**, so that molecule can't be built as drawn. So the atoms rebuild from scratch: first they make four normal, strong outward B–H handshakes, using up eight hands. Now only **four hands are left**, but there are still two hydrogens with nobody to hold and two borons drifting apart. The clever fix: put one hydrogen *between* two borons and let a single handshake grip **all three** at once — a "banana" grip. Do that twice (one above, one below), using your last four hands. Now everything is held, and you've spent exactly twelve hands — not one more. That "one grip, three atoms" trick is the famous **3c–2e bond**, and it exists purely because boron never brought enough electrons to bond the normal way. > [!recall]- > Total valence electrons in $B_2H_6$ and why ::: 12 = 2×3 (B) + 6×1 (H); it's short of what a normal structure needs. > How many distinct bonds in the ethane-analogue, and how many electrons? ::: 7 bonds (6 B–H + 1 B–B), counted once each = 14 electrons — 2 more than available. > How are the 12 electrons distributed in the real molecule? ::: 8 in four terminal 2c–2e B–H bonds + 4 in two bridging 3c–2e B–H–B bonds. > Why is the bridging B–H longer than terminal? ::: The 3c–2e bond spreads one pair over three atoms, so it binds each more weakly → ~133 pm vs ~119 pm. > Why can't heat alone split diborane symmetrically? ::: A base must donate fresh electron pairs (e.g. NMe₃) because the bridges themselves are electron-poor.