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VKC Notes · Main group chemistry · Civil Services P-I · Geo-Scientist P-I

14.4 Main Group Chemistry: Boranes, Borazine, Phosphazenes, Silicates, Silicones, Interhalogens, Xenon and S–N Compounds

★★★ 2021–2025: 16 parts, 145 marks (Civil Services 13 parts / 130 m, every year two or three 10-mark parts · Geo-Scientist 3 / 15 m on silicates and minerals). Mostly complete the reaction and draw the structure.

Syllabus lines. Civil Services: “Boranes, borazines, phosphazenes and cyclic phosphazene, silicates and silicones, interhalogen compounds; sulphur–nitrogen compounds, noble gas compounds”. Geo-Scientist asks silicates and framework structures under “Inorganic solids” (see 3.3). Liquid NH₃, HF and BrF₃ as solvents are in 14.2; cluster electron counting for metals in 13.5.

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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

14.4.1 Diborane and the three-centre two-electron bond ★★★

  • Preparation: ; in the laboratory . Colourless gas (bp −93 °C), ignites in air (), hydrolysed to boric acid: .SN
  • Electron deficient: 12 valence electrons, but eight atoms joined by seven ordinary bonds would need 14. Structure (electron diffraction, IR, ¹H NMR 4 : 2): two BH₂ units in one plane, two bridging H above and below it.SN
  • Bonding (hybridisation picture): each B is sp³. Two hybrids make normal 2c–2e terminal B–H bonds (8 e). In each bridge one hybrid of each B and the H 1s orbital combine into three MOs; only the bonding one is filled: a 3c–2e “banana” bond. Bridges are longer (133 pm) than terminal bonds (119 pm) CSE 2024 Q8(b)(ii) 10m.SN
  • Cleavage by Lewis bases decides the product CSE 2022 Q6(d) 10m CSE 2025 Q5(g) 10m: bulky bases (Me₃N, CO, THF) split each bridge symmetrically: . Small NH₃ attacks one boron twice: unsymmetrical cleavage to the ionic “diammoniate of diborane” . At higher temperature NH₃ gives borazine (14.4.3). Hydroboration of alkenes is in 11.1.SN
BBHHHHHHbridge B–H 133 pmterminal B–H 119 pm, B···B 177 pm3c–2eone bridge: B sp3 + H 1s + B sp3BBHψ3 antibondingψ2 non-bondingψ1 bonding↑↓2 e−
Fig. 14.4.1 Diborane and one 3c–2e bridge.
Draw it in 60 s: two B joined by two curved B–H–B bridges above and below; two H on each B, one wedge, one hash; beside it three lobes B, H, B over one filled level.

14.4.2 Higher boranes: Wade’s rules and styx numbers ★★

  • Boranes and borohydrides are deltahedral cages. Skeletal electron count: each B–H unit gives 2 e, each extra H (bridging or BH₂) 1 e, each negative charge 1 e. With n vertices: n + 1 pairs closo (closed deltahedron, BₙHₙ²⁻), n + 2 nido (one vertex missing, BₙHₙ₊₄), n + 3 arachno (two missing, BₙHₙ₊₆). The same counting works for carboranes (a CH unit gives 3 e) and metal clusters (13.5).SN
closo B6H62−6(2) + 2 = 14 e = 7 pairsn + 1: octahedronnido B5H95(2) + 4 = 14 e = 7 pairsn + 2: square pyramidarachno B4H104(2) + 6 = 14 e = 7 pairsn + 3: butterflyremove a vertex →remove another →open face carries 4 B–H–B2 BH2 + 4 B–H–B bridges
Fig. 14.4.2 Same 7 skeletal pairs: closo → nido → arachno.
Draw it in 60 s: an octahedron of six dots; rub out the bottom vertex (square pyramid); rub out one more (butterfly); write n + 1, n + 2, n + 3 under them.

Solved PYQ CSE 2023 Q6(e) 10m

Framework electrons and structures of B₇H₇²⁻, B₁₁H₁₃²⁻ and B₂H₇⁻.

  1. B₇H₇²⁻: 7(2) + 2 = 16 e = 8 pairs = n + 1: closo, pentagonal bipyramid.
  2. B₁₁H₁₃²⁻: 11(2) + 2 + 2 = 26 e = 13 pairs = n + 2: nido, icosahedron minus one vertex; the two extra H bridge the open pentagonal face.
  3. B₂H₇⁻: 2(2) + 5 + 1 = 10 e = 5 pairs = n + 3: arachno, H₃B–H–BH₃⁻ with one bent B–H–B bridge.SN
  • Lipscomb’s styx code counts the four bond types: s = B–H–B bridges, t = closed BBB 3c–2e bonds, y = 2c B–B bonds, x = extra terminal H (BH₂ groups). B₂H₆ 2002, B₄H₁₀ 4012, B₅H₉ 4120.SN

14.4.3 Borazine, the “inorganic benzene” ★★

  • Preparation: CSE 2023 Q6(a) 10m CSE 2025 Q5(g) 10m; better: , then reduction of the B-trichloroborazine with NaBH₄.SN
  • Planar ring of alternate B and N, all B–N 144 pm, isoelectronic with benzene (bp 55 °C, smells like benzene). But the N lone pairs donate only partly into the empty B 2p orbitals: the ring is polar (δ+ on B, δ− on N) and only weakly aromatic. So, unlike benzene, it adds HCl, HBr or H₂O across B–N (Cl to B, H to N): , and hydrolyses slowly to B(OH)₃ + NH₃.SN
benzene: complete pπ–pπ ringall C–C 139 pm, aromaticBHδ+NHδ−BHδ+NHδ−BHδ+NHδ−borazine B3N3H6: N lone pair → empty B 2ppartial, polar π; adds HCl across B–Nisoelectronic,both 6 π e
Fig. 14.4.3 Benzene and borazine.
Draw it in 60 s: two hexagons; full circle in benzene; in borazine B and N alternate, H on every atom, δ+ on B, δ− on N, a dashed circle.

14.4.4 Phosphazenes ★★★

  • Preparation: ; n = 3 and 4 rings are separated by distillation. Heating the trimer to 250 °C opens it to the linear polymer (NPCl₂)ₙ, “inorganic rubber”, which hydrolyses in air; replacing Cl by OCH₂CF₃ or OPh gives stable, flame-resistant elastomers and biomedical polymers.SN
  • Cl on P is easily substituted (NaOR, amines, RMgX); a second substituent goes to the same P (geminal) or the next P (non-geminal) depending on the group.
PClClNPClClNPClClN(NPCl2)3: planar ring, all P–N 158 pmbenzeneC 2p + C 2p (pπ–pπ) all round the ring: aromaticborazineN 2p → B 2p (pπ–pπ): polar, partialphosphazeneN 2p + P 3d (dπ–pπ): a node at every P, soπ stops: three P–N–P “islands” (Dewar)no 4n + 2 rule: (NPCl2)4 as stable as the trimer
Fig. 14.4.4 (NPCl₂)₃ and the island model.
Draw it in 60 s: hexagon P–N–P–N–P–N, two Cl on each P; shade three P–N–P islands, one centred on each N; all P–N equal, 158 pm.

Solved PYQ CSE 2021 Q6(a) 10m

How does the bonding in cyclic phosphazene differ from that in benzene and borazine?

  1. Benzene: six C 2p orbitals overlap all round the ring (pπ–pπ); complete delocalisation, aromatic, all C–C 139 pm.
  2. Borazine: filled N 2p donates into empty B 2p (pπ–pπ); polar and partial, so it adds HCl.
  3. Phosphazene: P uses 3s and 3p for four σ bonds; N 2p overlaps P 3d (dπ–pπ). The d orbitals change sign at P, so π delocalisation stops at each P: three P–N–P islands (Dewar), no ring current and no 4n + 2 rule; all P–N bonds are equal and short (158 pm vs 177 pm single). The modern view adds a polar P⁺–N⁻ σ bond strengthened by negative hyperconjugation into P–Cl σ*.SN

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