Question
Why does boron show anomalous properties compared to other members of Group 13, particularly aluminium? Explain with reference to its size, ionisation enthalpy, and bonding behaviour.
Solution — Step by Step
Boron is the only non-metal in Group 13. Every other member — Al, Ga, In, Tl — is a metal. This single fact tells us boron is doing something structurally different, and the reason traces back to its atomic size.
Boron has an atomic radius of just 85 pm, far smaller than Al at 143 pm. A smaller atom with the same nuclear charge means a much higher charge density (charge per unit volume). This makes boron polarise anions heavily — it behaves more like a covalent species than an ionic one.
The first three ionisation enthalpies of boron sum to ~6880 kJ/mol. That’s enormous — releasing B³⁺ into solution is energetically not worth it for most reactions. So boron forms covalent compounds almost exclusively, while Al readily forms Al³⁺ in aqueous chemistry.
Boron’s valence shell is n=2, which has only s and p orbitals. This has two consequences: boron can never expand its octet (no d-orbitals available), and the empty 2p orbital makes BF₃, BCl₃ etc. strong Lewis acids. Al has access to 3d orbitals and can form 6-coordinate complexes — boron cannot.
Boron’s anomalous properties land it closer to silicon (its diagonal neighbour) than to aluminium. Both form covalent hydrides (B₂H₆ vs SiH₄), both form acidic oxides (B₂O₃ and SiO₂ are both glass-forming oxides), and both form halides that are Lewis acids prone to hydrolysis.
Why This Works
The root cause is the lithium effect scaled up: the first element in any group always has an anomalously small radius because it has no inner d or p electrons shielding the nucleus. For Group 13, this effect is dramatic enough to push boron across the metal/non-metal boundary entirely.
Covalent character increases when a cation is small and highly charged — this is Fajans’ Rule in action. Boron’s high charge density polarises surrounding electron clouds so strongly that ionic bonding becomes unfavourable. The result: boron behaves like a metalloid, forming network solids and covalent molecules rather than salts.
The empty 2p orbital is the other key detail examiners love. It explains why boron compounds are electron-deficient and form bridged structures like diborane (B₂H₆) — boron needs those 3-centre 2-electron bonds to compensate for having only 6 electrons in its valence shell.
Alternative Method — Fajans’ Rule Angle
If the question asks you to justify covalent character using Fajans’ Rule directly:
Covalent character increases when:
- Cation is small (B³⁺ radius ≈ 23 pm vs Al³⁺ ≈ 53 pm)
- Cation charge is high (both are 3+, but B³⁺ charge density >> Al³⁺)
- Anion is large and polarisable
B³⁺ has the highest charge density of any tripositive ion in the group. Apply Fajans’ criteria → boron compounds are covalent. Al³⁺ is larger → more ionic character → Al₂O₃ is amphoteric rather than purely acidic like B₂O₃.
This gives you a one-paragraph answer that directly scores the “justify” marks in JEE Main descriptive-type MCQs.
Common Mistake
Students write “boron cannot form B³⁺ because it doesn’t have enough electrons to lose” — that’s wrong. Boron has 3 valence electrons and can lose them in principle. The real reason is that the sum of its first three ionisation enthalpies is too high to be compensated by lattice/hydration energy. The thermodynamics don’t favour ionic bond formation, not the electron count.
For JEE Main, the diagonal relationship pair B–Si appears almost every year in some form — either directly asking for similarities, or asking why B₂O₃ is acidic (like SiO₂) while Al₂O₃ is amphoteric. Memorise three similarities: both form covalent hydrides, both have acidic oxides, both form halides that hydrolyse in water. This appeared as a direct 4-mark MCQ in JEE Main 2023 January Session.