3.2.1 · D5p-Block
Question bank — Group 13 (Boron family) — anomaly of B, diagonal B-Si; BX₃ Lewis acidity; diborane B₂H₆ (3c-2e bond), borazine
Visual toolkit (build these pictures first)
Before the traps, lock down the three ideas the whole page leans on. Each has a figure — read the picture, then the words.
What a 3c–2e bond looks like. In an ordinary bond one electron pair glues two atoms. In diborane there aren't enough electrons for that everywhere, so one pair is smeared over three atoms — the "banana". Look at the figure: the shaded lobe touches both borons and the hydrogen at once.
What back-bonding (–) looks like. In boron has an empty orbital sticking up. A halogen with a full lone-pair orbital can tip that pair sideways into B's empty lobe — this sideways () donation is back-bonding. It matters how well the two lobes overlap.
Recall Where does the "6887 kJ/mol" come from?
It is the sum of boron's first three ionisation enthalpies (energy to strip 3 electrons one by one): kJ/mol. Forming an ionic solid or hydrated ion releases lattice/hydration energy of at most a couple of thousand kJ/mol for a small +3 cation — nowhere near enough to pay back ~6900. So the ionic route is a net energy loss and never forms.
True or false — justify
exists just like .
False. Boron has no -orbitals at any energy, so it cannot expand its octet beyond a covalency of 4 — the largest fluoride ion it forms is . Aluminium (period 3) can enlist its high-lying orbitals when surrounded by strongly electronegative F, reaching covalency 6.
A free ion is present in .
False. Stripping 3 electrons costs kJ/mol (sum of the first three ionisation enthalpies), never recovered by lattice or hydration energy, so B never forms a naked cation — every bond in is covalent.
is the strongest Lewis acid of the boron halides because F is most electronegative.
False. Electronegativity would help acidity, but it is overruled by strong back-bonding (see the overlap figure above): F donates its lone pair so well into B's empty that B is nearly satisfied, making the weakest acid. See Back-bonding (pπ–pπ).
Diborane contains a direct B–B σ-bond like the C–C bond in ethane.
False. The two borons are joined only through the two bridging hydrogens (two 3c–2e bonds shown in figure s01); with just 12 valence electrons an ethane-like skeleton (needing 16) is impossible.
Borazine and benzene are equally unreactive because both are aromatic.
False. Borazine's ring is polar (, — partial, not full, charges) because N is more electronegative, so it readily adds reagents like HCl, whereas benzene resists addition.
Boron is a metal because it sits in Group 13 with Al, Ga, In, Tl.
False. Boron is a hard, non-metallic metalloid; its tiny size and high ionisation enthalpy prevent it from losing electrons ionically, unlike the genuinely metallic members below it.
All B–H bonds in are identical in length.
False. The four terminal B–H bonds ( pm) are shorter and stronger than the two bridging B–H bonds ( pm), because a bridge spreads one electron pair over three atoms — thinner electron glue, longer bond (figure s01).
and are both acidic covalent oxides.
True. This is direct evidence of the B–Si diagonal relationship — both are glassy, weakly acidic giant-covalent oxides rather than basic ionic oxides.
Spot the error
"In there are 8 normal 2c–2e B–H bonds, so 16 valence electrons are used."
The molecule has only 12 valence electrons (). It cannot afford 8 ordinary bonds, which is exactly why it uses two 3c–2e bridges (4 terminal 2c–2e + 2 bridging 3c–2e = 12).
" (a stable hydrate)."
Wrong product — is fully hydrolysed: . B has an empty orbital that water's lone pair attacks, so it never survives as an intact hydrate. See Boric Acid H3BO3.
"Boron forms and then -hybridises to make ."
There is no step; B is always covalent. forms when neutral (a Lewis acid) accepts a fluoride lone pair into its empty .
"Because Cl is bigger than F, has stronger back-bonding than ."
Reversed. Bigger, more diffuse Cl orbitals overlap worse with B's compact (right panel of figure s02), so back-bonding is weaker in — which is precisely why is the stronger acid.
"Borazine is with three carbons swapped for boron."
Borazine is — carbons are replaced in alternating pairs by one B and one N, whose electrons average to 4 (like two carbons). It is isoelectronic with benzene but not a simple substitution.
"Diagonal relationship links B with the element directly below it, Al."
The diagonal partner is Si (one period down and one group right). Al is the group neighbour and behaves as a metal; the diagonal cancellation of trends is what makes B resemble Si instead.
Why questions
Why does Lewis acidity increase down the boron halides ()?
As the halogen gets larger (F→I), its lone-pair p-orbital grows more diffuse and overlaps the small B ever more poorly (figure s02), so back-donation weakens and B stays electron-hungry — a hungrier B is a stronger acid.
Why can boron never expand its octet while aluminium can?
Boron's valence shell is only with no -orbitals, capping covalency at 4; Al has orbitals which, though high in energy, drop low enough to bond when very electronegative atoms surround it, letting Al reach covalency 6.
Why is called "electron-deficient"?
It has fewer valence electrons (12) than it would need for a conventional 2c–2e bond on every link (16), forcing it to spread pairs over three atoms.
Why does B behave chemically like Si (diagonal relationship)?
Electronegativity rises across a period and falls down a group; along the B→Si diagonal these two effects cancel, giving both atoms a similar electronegativity and charge/size ratio (polarising power). See Ionisation Enthalpy Trends.
Why is still a Lewis acid at all, given that F back-donates so well?
Back-bonding only partially fills the empty ; boron remains net electron-deficient, so it still accepts a full lone pair (e.g. forming ) — just less eagerly than the heavier halides. See Boron Trihalides.
Why is borazine planar?
Each N's filled lone pair overlaps sideways with the empty of the neighbouring B, and this only works if all six p-orbitals stand parallel and perpendicular to the ring — which forces the ring flat. The result is a delocalised 6-electron π cloud (three N→B donations resonating around the ring), the same flat, aromatic-like arrangement as benzene.
Why does boron form covalent hydrides while its group neighbours' behaviour differs?
Its small size and high ionisation enthalpy make electron sharing (covalency), not electron loss (ionic bonding), energetically favourable — the same reason it resembles a non-metal.
Edge cases
Is possible but impossible? Explain the boundary.
Yes — 4 is boron's maximum covalency (four bonds fill its four valence orbitals ). A sixth bond would require -orbitals it does not possess at any energy.
What holds the two borons together in if there is no B–B bond?
Two bridging 3c–2e "banana" bonds (figure s01) — each single electron pair binds both borons and the hydrogen between them simultaneously.
In the symmetric cleavage , what happens to the bridge?
The two 3c–2e bridges break evenly, splitting the molecule into two identical units, each of which accepts a nitrogen lone pair to satisfy boron's empty orbital.
What is the limiting reason hydration cannot rescue the ionic route?
Even the very large hydration energy of a hypothetical tiny falls far short of the enormous kJ/mol needed to strip three electrons, so the ionic option is never thermodynamically viable.
If we replace both bridging H in diborane conceptually with lone pairs, does the structure still work?
No — the bridges are the only way to distribute the scarce 12 electrons; remove them and boron's octet collapses to 6 with no bonding between the two halves.
Does borazine's aromaticity make it as stable as benzene under attack by HCl?
No — its polar B–N π ring provides electrophilic () and nucleophilic () sites, so HCl adds across the ring, unlike inert benzene. Compare with p-Block Overview.