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

The scenario matrix

Before working anything, let us list every class of case this topic can produce. Each worked example below is tagged with the cell it fills.

Cell	Case class	What makes it tricky	Filled by
A	Electron counting — is a normal structure possible?	Sign of the "electron deficit" (short / exact / surplus)	Ex 1, Ex 2
B	Lewis acidity ranking	Trend runs opposite to electronegativity	Ex 3
C	Back-bonding degenerate case	What if the "halogen" has no lone pair to donate?	Ex 4
D	Diagonal B–Si prediction	Predict an unknown reaction by analogy	Ex 5
E	Stoichiometry / limiting reagent (word problem)	Real numbers, grams, moles, a limiting reactant	Ex 6
F	Energy / thermochemistry limiting value	Sign of $Δ H$ , why $B^{3 +}$ never forms	Ex 7
G	Exam twist — borazine vs benzene reactivity	Isoelectronic yet behaves differently	Ex 8

We will hit A through G, every sign and every degenerate input.

Prerequisites you may want open: Lewis Acids and Bases, Back-bonding (pπ–pπ), Boron Trihalides, Ionisation Enthalpy Trends, Group 14 — Silicon, Benzene Aromaticity.

Cell A — Electron counting

Worked example Ex 1 — Is a plain 2c–2e structure possible for

$B_{2} H_{6}$ ? (Cell A, "short" sign) Forecast: Guess first — do we have more electrons than we need, exactly enough, or too few?

Count what we HAVE. Why this step? Every bonding argument starts from the total valence electron count — you cannot decide a structure without it. Each B brings 3, each H brings 1: $have = 2 (3) + 6 (1) = 12 electrons .$
Count what a "normal" all-2c–2e picture NEEDS. Why this step? A "normal" bond uses 2 electrons and joins exactly 2 atoms. With 2 B and 6 H, ethane-style bonding needs 8 B–H bonds (each B holding 3 terminal H + 1 shared). $need = 8 bonds \times 2 = 16 electrons .$
Take the difference (mind the sign). Why this step? The sign of (have − need) is the whole answer. Negative = deficient. $12 - 16 = - 4 \Rightarrow short by 4 electrons .$
Interpret. Four electrons short = two electron pairs missing. So two ordinary bonds must be replaced by two 3c–2e bridge bonds, each doing the work of a bond with only its own 2 electrons.

Verify: $4 terminal (2 c- 2 e) \times 2 + 2 bridge (3 c- 2 e) \times 2 = 8 + 4 = 12$ . ✔ Matches "have". The deficit sign was negative, confirming electron deficiency.

Cell B — Lewis acidity ranking

Worked example Ex 3 — Rank

$B F_{3}, B C l_{3}, B B r_{3}$ as Lewis acids (Cell B) Forecast: Fluorine is most electronegative — so does $B F_{3}$ come out strongest or weakest?

Identify what makes B a Lewis acid. Why this step? B has an empty $2 p_{z}$ orbital (only 6 electrons around it). A Lewis acid is an electron-pair acceptor; the emptier that orbital, the hungrier B is.
Ask what can fill that empty orbital already. Why this step? Each halogen has lone pairs. A lone pair can donate back into B's empty $2 p_{z}$ — this is $p\pi$–$p\pi$ back-bonding. The more it fills, the less acidic B becomes.
Compare overlap quality (this is the sign-flip step). Why this step? Overlap is best when the two p-orbitals are the same size. B is period 2 ( $2 p$ ).
- F is also period 2 ( $2 p$ ) → excellent overlap → strongest back-donation → B most satisfied.
- Cl ( $3 p$ ), Br ( $4 p$ ) are larger/more diffuse → poorer overlap → weaker back-donation → B stays hungry.
Rank. Weakest back-donation ⇒ strongest acid: $B F_{3} < B C l_{3} < B B r_{3} .$

Verify: The order runs opposite to electronegativity ( $F > C l > B r$ ), confirming the parent note's warning — back-bonding dominates over induction. Sanity check the endpoints: $B F_{3}$ has best overlap ⇒ weakest acid; $B B r_{3}$ has worst overlap ⇒ strongest of the three. ✔

Cell C — Degenerate back-bonding input

Worked example Ex 4 — Which is the stronger Lewis acid,

$B H_{3}$ or $B F_{3}$ ? (Cell C, "no lone pair" degenerate case) Forecast: F is electron-rich; H has just one electron and no lone pair. Guess which B is hungrier.

Check each substituent for a lone pair. Why this step? Back-bonding needs a filled p-orbital on the substituent. Test the degenerate input: does H have one?
- F: three lone pairs → can back-donate.
- H: only one electron, no lone pair → back-donation impossible (zero).
Consequence for $B H_{3}$ . With no back-bonding, B's empty $2 p_{z}$ stays fully empty → B is maximally electron-deficient.
Consequence for $B F_{3}$ . F partially fills the orbital → B less deficient.
Rank: $B F_{3} < B H_{3}$ in acidity. (Indeed $B H_{3}$ is so acidic/unstable alone that it instantly dimerises to $B_{2} H_{6}$ — Ex 1.)

Verify: The degenerate "no lone pair" case gives the hungriest boron, consistent with $B H_{3}$ not existing as a free monomer. This also closes the loop with Cell A: the same electron deficiency that forbids $B H_{3}$ forces the 3c–2e bridges of $B_{2} H_{6}$ . ✔

Cell D — Diagonal B–Si prediction

Worked example Ex 5 — Predict the hydrolysis of

$S i C l_{4}$ using the B–Cl analogy (Cell D) Forecast: We know $B C l_{3} + 3 H_{2} O \to H_{3} B O_{3} + 3 H C l$ . What will $S i C l_{4}$ + water give?

State the diagonal principle. Why this step? B and Si sit on a diagonal; their electronegativities (≈2.0 vs ≈1.8) and charge/size ratios nearly match, so their halides behave alike: volatile and readily hydrolysed.
Write the B template. $B C l_{3} + 3 H_{2} O ⟶ H_{3} B O_{3} + 3 H C l .$
Transfer to Si, balancing for its +4 covalency. Why this step? Si forms 4 bonds (one more than B's 3). Full hydrolysis of $S i C l_{4}$ ultimately gives the acidic oxide $S i O_{2}$ (as the parent note lists): $S i C l_{4} + 2 H_{2} O ⟶ S i O_{2} + 4 H C l .$
Interpret. Both halides fume in moist air, both give HCl + an acidic oxide/oxo-acid — direct evidence of the diagonal relationship.

Verify (atom balance): Left $= 1 S i, 4 C l, 4 H, 2 O$ ; right $= 1 S i + 2 O$ (in $S i O_{2}$ ) $+ 4 H + 4 C l$ (in $4 H C l$ ). Si:1=1, Cl:4=4, H:4=4, O:2=2. ✔

Cell E — Stoichiometry word problem (limiting reagent)

Worked example Ex 6 — Rocket-fuel burn (Cell E, real-world + limiting reactant)

Statement: A lab burns $2.00 mol$ of diborane in $5.00 mol$ of $O_{2}$ : $B_{2} H_{6} + 3 O_{2} ⟶ B_{2} O_{3} + 3 H_{2} O, Δ H = - 2165 kJ per mol B_{2} H_{6} .$ How much $B_{2} O_{3}$ forms, and how much heat is released? Forecast: Which runs out first — the diborane or the oxygen?

Find the limiting reagent. Why this step? You can never use more product-per-reactant than the scarcest reactant allows. The equation needs $3 O_{2}$ per $B_{2} H_{6}$ .
- $2.00 mol B_{2} H_{6}$ would need $2.00 \times 3 = 6.00 mol O_{2}$ .
- We only have $5.00 mol O_{2}$ → $O_{2}$ is limiting.
Convert limiting $O_{2}$ to product $B_{2} O_{3}$ . Why this step? Ratio $3 O_{2} : 1 B_{2} O_{3}$ . $n (B_{2} O_{3}) = \frac{5.00}{3} = 1.667 mol .$
Heat released. Why this step? $Δ H$ is quoted per mol $B_{2} H_{6}$ consumed, and $B_{2} H_{6} : O_{2} = 1 : 3$ , so mol of $B_{2} H_{6}$ actually burnt $= 5.00/3 = 1.667$ . $q = 1.667 \times (- 2165) = - 3608 kJ (about 3.61 \times 1 0^{3} kJ released) .$
Leftover diborane: $2.00 - 1.667 = 0.333 mol B_{2} H_{6}$ unburnt.

Verify (units + sign): $mol \times \frac{kJ}{mol} = kJ$ ✔. $Δ H < 0$ ⇒ exothermic, matches "very exothermic" fuel. Cross-check product: $1.667 mol B_{2} O_{3}$ needs $1.667 \times 3 = 5.00 mol O_{2}$ — exactly the oxygen we had. ✔

Cell F — Thermochemistry limiting value

Worked example Ex 7 — Why does no free

$B^{3 +}$ exist? Put a number on it (Cell F, sign of energy) Statement: The first three ionisation enthalpies of boron sum to about $6887 kJ/mol$ . A generous lattice/hydration energy might recover only $\sim 4000 kJ/mol$ . Does $B^{3 +}$ form? Forecast: Positive or negative net energy? Which way is favourable?

Cost of making $B^{3 +}$ . Why this step? Forming a +3 ion means paying all three ionisation enthalpies: $Δ_{i} H_{1} + Δ_{i} H_{2} + Δ_{i} H_{3} \approx + 6887 kJ/mol (energy IN) .$
Payback from lattice/hydration. Why this step? Only the surroundings' attraction to the ion can return energy; take the optimistic $- 4000 kJ/mol$ .
Net. $Δ E_{net} \approx 6887 - 4000 = + 2887 kJ/mol (still strongly positive) .$
Interpret the sign. A large positive net means the process is hugely unfavourable → no free $B^{3 +}$ ; boron bonds covalently instead.

Verify: The sign stays positive even with a generous payback, so the conclusion is robust — this is the limiting-behaviour argument behind "B is a metalloid, not a metal." ✔

Cell G — Exam twist: borazine vs benzene

Worked example Ex 8 — Borazine and benzene are isoelectronic, yet borazine adds HCl. Why? (Cell G)

Forecast: Same electron count, same shape — so why does one add HCl and the other refuse?

Confirm they are isoelectronic. Why this step? B (3 valence $e^{-}$ ) + N (5) average to 4 — same as two carbons. So $B_{3} N_{3} H_{6}$ matches $C_{6} H_{6}$ in electron count and planar hexagonal shape.
Locate the difference: polarity. Why this step? In benzene every ring atom is identical C → the $π$ -cloud is uniform, non-polar. In borazine, N is more electronegative than B, so the ring is polar: $N^{δ -}$ , $B^{δ +}$ .
Predict the addition. Why this step? A polar ring offers reactive $δ +$ / $δ -$ sites. $H^{δ +}$ of HCl attacks $N^{δ -}$ , $C l^{δ -}$ attacks $B^{δ +}$ : $B_{3} N_{3} H_{6} + 3 H C l ⟶ B_{3} N_{3} H_{6} (H C l)_{3} (an addition product) .$
Contrast with benzene. Benzene's non-polar, strongly delocalised ring resists addition (it substitutes instead). Borazine's polarity makes it more reactive.

Verify (H balance): LHS $= 6 H$ (borazine) $+ 3 H$ (HCl) $= 9 H$ ; RHS product $B_{3} N_{3} H_{6} \cdot 3 H C l$ carries $6 + 3 = 9 H$ and $3 C l = 3 C l$ from LHS. ✔

Recall One-line recap of every cell

Deficit sign (A) ::: $12 - 16 = - 4$ → electron-deficient → 3c–2e bridges. Ranking direction (B) ::: acidity rises F→Cl→Br as back-bonding overlap worsens. Degenerate input (C) ::: no lone pair on H ⇒ $B H_{3}$ hungrier than $B F_{3}$ . Diagonal transfer (D) ::: $S i C l_{4}$ hydrolyses like $B C l_{3}$ , giving $S i O_{2} + 4 H C l$ . Limiting reagent (E) ::: $O_{2}$ limits; $1.667 mol B_{2} O_{3}$ , $\sim 3608 kJ$ out. Energy sign (F) ::: net $\approx + 2887 kJ/mol$ ⇒ no free $B^{3 +}$ . Polarity twist (G) ::: borazine polar ⇒ adds $3 H C l$ ; benzene non-polar ⇒ resists.