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

Level 1 — Recognition

Q1 (L1) — The odd one out

Which of these is the correct maximum covalency of boron, and why can it not go higher? (a) 2 (b) 3 (c) 4 (d) 6

Recall Solution

Answer: (c) 4. Boron's valence shell is only $2 s 2 p$ — it has no d-orbitals, so it cannot expand its octet beyond 8 electrons (= 4 shared pairs = 4 bonds). That is why $B F_{4}^{-}$ exists but $B F_{6}^{3 -}$ does not. Contrast: aluminium (period 3) has empty $3 d$ orbitals, so $[A l F_{6}]^{3 -}$ is fine. This is the essence of the first-element anomaly.

Q2 (L1) — Name the bond

In $B_{2} H_{6}$ , the two bridging hydrogens each sit in a bond described by "one electron pair shared among three atoms." Name this bond type and state how many electrons it holds.

Recall Solution

It is a three-centre two-electron (3c–2e) bond, often drawn as a banana bond. It holds exactly 2 electrons shared over the three atoms $B - H - B$ .

Level 2 — Application

Q3 (L2) — Rank the acids

Arrange in order of increasing Lewis acidity: $B B r_{3}$ , $B F_{3}$ , $B C l_{3}$ , $B I_{3}$ . Justify in one sentence.

Recall Solution

$B F_{3} < B C l_{3} < B B r_{3} < B I_{3}$ Acidity increases down the halogens because back-bonding into boron's empty $2 p_{z}$ orbital weakens as the halogen orbitals grow larger and more diffuse (best overlap for F's $2 p$ , worst for I's $5 p$ ). Weaker back-donation → hungrier boron → stronger Lewis acid. See Figure below for the overlap picture.

Q4 (L2) — Count the electrons

Verify by counting that $B_{2} H_{6}$ cannot be built from eight ordinary 2c–2e B–H bonds. How many valence electrons does it actually have, and how many would eight normal bonds require?

Recall Solution

Valence electrons available: $2 \times B 3 + 6 \times H 1 = 6 + 6 = 12.$ Eight normal 2c–2e bonds would need $8 \times 2 = 16$ electrons. We are $16 - 12 = 4$ electrons short → the molecule is electron-deficient → it must use bridge bonds. The actual split: 4 terminal (2c–2e) = 8 e⁻, plus 2 bridges (3c–2e) = 4 e⁻, total $8 + 4 = 12$ ✔.

Q5 (L2) — Balance a preparation

Balance the industrial-style preparation of diborane: $L i A l H_{4} + B F_{3} \to B_{2} H_{6} + L i F + A l F_{3}$

Recall Solution

Balance boron first (need even number for $B_{2} H_{6}$ ), then fluorine, then Li/Al/H: $3 L i A l H_{4} + 4 B F_{3} \to 2 B_{2} H_{6} + 3 L i F + 3 A l F_{3}$ Check each element:

B: left $4$ , right $2 \times 2 = 4$ ✔
F: left $4 \times 3 = 12$ , right $3 + 3 \times 3 = 12$ ✔
Li: $3 = 3$ ✔ Al: $3 = 3$ ✔
H: left $3 \times 4 = 12$ , right $2 \times 6 = 12$ ✔

Level 3 — Analysis

Q6 (L3) — Why no free B³⁺?

Aluminium happily forms $A l^{3 +}$ in salts, but boron never gives a free $B^{3 +}$ ion. Explain quantitatively using the idea of an energy balance. (Given: $Δ_{i} H_{1} + Δ_{i} H_{2} + Δ_{i} H_{3} \approx 6887$ kJ/mol for B.)

Recall Solution

Forming an ion requires paying the ionisation cost and recovering it via lattice or hydration energy.

Boron is tiny (covalent radius ≈ 85 pm), so its three ionisation enthalpies sum to a huge ≈ 6887 kJ/mol.
The energy returned when a small $B^{3 +}$ forms ionic bonds cannot match this enormous outlay (lattice/hydration energies of realistic salts are far smaller than 6887 kJ/mol).
Net: the books never balance for ionic bonding, so boron always chooses covalent bonds instead. Aluminium is larger with lower IE sums, so its $A l^{3 +}$ energy budget does balance.

Q7 (L3) — The diagonal disguise

Boron sits above aluminium, yet chemically it "acts like" silicon. Give the underlying reason and two experimental pieces of evidence.

Recall Solution

Reason: Moving right across a period raises electronegativity; moving down a group lowers it. Along the B→Si diagonal these two trends cancel, giving similar electronegativity (B ≈ 2.0, Si ≈ 1.8) and similar charge/size ratio (polarising power). Evidence (any two):

Both oxides are acidic, covalent, glassy: $B_{2} O_{3}$ and $S i O_{2}$ .
Both halides are volatile and hydrolyse readily: $B C l_{3} + 3 H_{2} O \to H_{3} B O_{3} + 3 H C l$ ; $S i C l_{4} + 2 H_{2} O \to S i O_{2} + 4 H C l$ .
Both form covalent flammable hydrides (boranes / silanes) and extended borate/silicate networks. See Group 14 — Silicon for the silicon side of the analogy.

Q8 (L3) — Reading the geometry

In $B_{2} H_{6}$ , the bridging B–H bond is ≈ 133 pm while the terminal B–H is ≈ 119 pm. Explain why the bridge bond is longer, and what this tells you about its strength.

Recall Solution

The terminal bond is an ordinary 2c–2e bond: 2 electrons glue 2 atoms, a full-strength single bond → short (119 pm). The bridge is a 3c–2e bond: the same 2 electrons must now hold 3 atoms together, so the "electron glue per link" is diluted. Weaker bonding → atoms sit farther apart → longer (133 pm). Longer = weaker is the general rule. See figure.

Level 4 — Synthesis

Q9 (L4) — Build the borazine story

Borazine $B_{3} N_{3} H_{6}$ is called "inorganic benzene." (a) Write its balanced synthesis from diborane and ammonia. (b) Explain, using electron counting and back-bonding, why it is isoelectronic with benzene. (c) Predict why it is more reactive than benzene.

Recall Solution

(a) $3 B_{2} H_{6} + 6 N H_{3} \to 2 B_{3} N_{3} H_{6} + 12 H_{2}$ Check: B $6 = 6$ ; N $6 = 6$ ; H left $3 \times 6 + 6 \times 3 = 18 + 18 = 36$ , right $2 \times 6 + 12 \times 2 = 12 + 24 = 36$ ✔. (b) A B–N unit carries $3 + 5 = 8$ valence electrons; a C–C unit carries $4 + 4 = 8$ . Averaging, each B–N pair matches two carbons → the ring is isoelectronic with $C_{6} H_{6}$ . Nitrogen donates its lone pair into boron's empty $p$ -orbital, creating a delocalised $π$ system across the planar ring (compare Benzene Aromaticity). (c) Because N is more electronegative than B, the ring is polarised: $N^{δ -}$ and $B^{δ +}$ . Benzene's ring is non-polar and evenly shared, but borazine's polar sites are reactive handles — e.g. it adds HCl across a B–N unit ( $H^{+}$ to N, $C l^{-}$ to B), something benzene resists.

Q10 (L4) — Thermochemistry of combustion

Diborane burns very exothermically: $B_{2} H_{6} + 3 O_{2} \to B_{2} O_{3} + 3 H_{2} O, Δ H = - 2165 kJ/mol .$ How much heat is released when 13.85 g of $B_{2} H_{6}$ is burned completely? (Molar mass $B_{2} H_{6} \approx 27.7$ g/mol.)

Recall Solution

Moles of diborane: $n = \frac{13.85 g}{27.7 g/mol} = 0.500 mol .$ Heat released: $q = n \times ∣Δ H ∣ = 0.500 \times 2165 = 1082.5 kJ .$ So burning 13.85 g of $B_{2} H_{6}$ releases about 1082.5 kJ — a huge output per gram, which is why boranes were studied as high-energy fuels.

Level 5 — Mastery

Q11 (L5) — Symmetric vs unsymmetric cleavage

$B_{2} H_{6}$ reacts with $N M e_{3}$ (a bulky base) to give $2 M e_{3} N \cdot B H_{3}$ (symmetric cleavage: the bridge splits into two equal $B H_{3}$ halves). With $N H_{3}$ (a small base) it instead gives the unsymmetric ionic product $[B H_{2} (N H_{3})_{2}]^{+} [B H_{4}]^{-}$ . Explain what governs which pathway occurs.

Recall Solution

The bridge $B - H - B$ can break in two ways:

Symmetric: each base grabs one $B H_{3}$ → $2 L \cdot B H_{3}$ . Needs the base to attack one boron each side evenly.
Unsymmetric: one boron ends up bonded to two base molecules ( $[B H_{2} L_{2}]^{+}$ ) while the other leaves as $[B H_{4}]^{-}$ . What decides: the size of the base.
A bulky base like $N M e_{3}$ cannot pile two molecules onto one small boron (steric crowding), so it settles for one-each → symmetric.
A small base like $N H_{3}$ can fit two onto one boron → unsymmetric ionic split. So: small base → unsymmetric; bulky base → symmetric. Sterics, not just basicity, choose the path.

Q12 (L5) — Predict the adduct that fails

$B F_{3}$ readily forms $F_{3} B \leftarrow N H_{3}$ . Would you expect the analogous back-bonding argument to make $B F_{3}$ a stronger or weaker acid toward $N H_{3}$ than $B C l_{3}$ ? Then explain the twist: experimentally, does adduct formation enthalpy always track the gas-phase acidity order? Reason it out.

Recall Solution

From back-bonding (Q3), intrinsic acidity is $B F_{3} < B C l_{3}$ , so $B C l_{3}$ binds $N H_{3}$ more strongly — this is confirmed experimentally. The twist / edge case: when $B X_{3}$ accepts a lone pair, boron rehybridises from planar $s p^{2}$ to pyramidal $s p^{3}$ , and the back-bonding that stabilised the free acid is lost on adduct formation. The molecule that had the strongest back-bonding ( $B F_{3}$ ) loses the most stabilisation upon binding. This "reorganisation penalty" reinforces the same order: $B F_{3}$ is the poorest acceptor. So both the intrinsic-deficiency view and the reorganisation view agree — $B F_{3} < B C l_{3} < B B r_{3} < B I_{3}$ holds robustly.

Recall Self-test checklist

Max covalency of B and why ::: 4; no d-orbitals, octet can't expand. Lewis acidity order of $B X_{3}$ ::: $B F_{3} < B C l_{3} < B B r_{3} < B I_{3}$ . Valence electrons in $B_{2} H_{6}$ ::: 12. Heat from 0.5 mol $B_{2} H_{6}$ combustion ::: ≈ 1082.5 kJ. Small base on diborane gives ::: unsymmetric ionic cleavage.