Main Group Elements: VSEPR, Compounds, Boranes and Silicates, Allotropes, Industrial Chemistry and Acid–Base Concepts
1. Shapes of molecules by VSEPR
Valence-shell electron-pair repulsion places the bonding and lone pairs of a central atom as far apart as possible. The steric number (bonded atoms + lone pairs) fixes the electron geometry: 2 linear, 3 trigonal planar, 4 tetrahedral, 5 trigonal bipyramidal, 6 octahedral, 7 pentagonal bipyramidal. Repulsions fall in the order lone pair–lone pair > lone pair–bond pair > bond pair–bond pair, so lone pairs compress bond angles (CH₄ 109.5°, NH₃ 107°, H₂O 104.5°). In a trigonal bipyramid lone pairs go equatorial (fewer 90° contacts) and the more electronegative ligands go axial; in an octahedron two lone pairs go trans. Multiple bonds count as one region but repel more strongly.
| Steric number | Lone pairs | Shape | Examples |
|---|---|---|---|
| 4 | 1 | trigonal pyramidal | NH₃, XeO₃ |
| 4 | 2 | bent | H₂O, SCl₂ |
| 5 | 1 | see-saw | SF₄, TeCl₄ |
| 5 | 2 | T-shaped | ClF₃, BrF₃ |
| 5 | 3 | linear | XeF₂, I₃⁻ |
| 6 | 1 | square pyramidal | BrF₅, IF₅, XeOF₄ |
| 6 | 2 | square planar | XeF₄, ICl₄⁻ |
| 7 | 0 | pentagonal bipyramidal | IF₇ |
| 7 | 1 | distorted octahedral | XeF₆ |
2. Hydrides, halides, oxides, oxoacids, nitrides and sulfides
Hydrides are saline (ionic, with H⁻: LiH, CaH₂), covalent (molecular: CH₄, NH₃, H₂O, B₂H₆) or metallic/interstitial (PdHₓ). Electron-deficient diborane B₂H₆ has four terminal B–H bonds and two three-centre two-electron B–H–B bridges. Halides of the p-block hydrolyse when the central atom has a vacant low-lying orbital to accept water: SiCl₄ and PCl₅ hydrolyse readily, while CCl₄ and SF₆ are kinetically inert because carbon has no d orbitals and sulfur is sterically shielded. Oxides become more acidic across a period and with rising oxidation state (Na₂O basic, Al₂O₃ amphoteric, SiO₂ weakly acidic, P₄O₁₀ and SO₃ strongly acidic), and more basic down a group.
Oxoacids. The basicity of a phosphorus oxoacid equals its number of P–OH groups, not its number of hydrogens: H₃PO₄ is tribasic, H₃PO₃ (one P–H) dibasic and H₃PO₂ (two P–H) monobasic and a good reducing agent. For oxoacids of one element, acidity rises with the number of oxygens not bearing hydrogen (Pauling’s rule): HClO < HClO₂ < HClO₃ < HClO₄. Sulfur forms H₂SO₃, H₂SO₄, H₂S₂O₇ (disulfuric, oleum) and the peroxoacids H₂SO₅ and H₂S₂O₈, in which sulfur is still +6 and the extra oxidising power sits in the O–O bond. Nitrides range from ionic Li₃N and Mg₃N₂ (which give NH₃ with water) through covalent BN and Si₃N₄ to interstitial TiN; sulfides from ionic Na₂S to the layered MoS₂ and the cage molecules P₄S₃ and S₄N₄.
3. Boranes, carboranes, Wade’s rules and the isolobal analogy; silicones, silicates, BN, borazine, phosphazenes
Wade’s rules (polyhedral skeletal electron pair theory) count the electrons holding a borane cage together. Each B–H unit gives 2 skeletal electrons, each extra H 1, a C–H unit 3, and a negative charge adds its electrons. With n vertices, n + 1 skeletal pairs means a closo cage (a closed deltahedron: B₆H₆²⁻ octahedron, B₁₂H₁₂²⁻ and C₂B₁₀H₁₂ icosahedra); n + 2 a nido cage (one vertex missing: B₅H₉, B₁₀H₁₄); n + 3 an arachno cage (two missing: B₄H₁₀, B₅H₁₁). The styx numbers describe the same molecules in terms of 3c–2e B–H–B bridges (s), 3c–2e BBB bonds (t), 2c–2e B–B bonds (y) and extra BH₂ groups (x).
| Cluster | Skeletal electrons | Pairs vs vertices n | Type |
|---|---|---|---|
| B₆H₆²⁻ | 6 × 2 + 2 = 14 | 7 = n + 1 | closo (octahedron) |
| C₂B₁₀H₁₂ | 2 × 3 + 10 × 2 = 26 | 13 = n + 1 | closo (icosahedron) |
| B₅H₉ | 5 × 2 + 4 = 14 | 7 = n + 2 | nido (square pyramid) |
| B₄H₁₀ | 4 × 2 + 6 = 14 | 7 = n + 3 | arachno (butterfly) |
The isolobal analogy (Hoffmann) says two fragments are isolobal when their frontier orbitals have the same number, symmetry, approximate energy and occupancy. CH₃ and Mn(CO)₅ each offer one singly occupied orbital pointing outward; CH₂ and Fe(CO)₄ offer two frontier orbitals with two electrons; CH and Co(CO)₃ offer three orbitals with three electrons. In the same sense a B–H vertex is isolobal with C–H⁺, which is why replacing BH⁻ by CH turns a borane anion into a neutral carborane. So Mn₂(CO)₁₀ is the metal analogue of ethane, and Co(CO)₃ units can replace CH vertices in tetrahedrane-like clusters — the bridge between borane, hydrocarbon and metal-cluster chemistry.
Silicates are built from SiO₄ tetrahedra sharing corners. The number of shared (bridging) oxygens per silicon sets the formula: 0 in orthosilicates SiO₄⁴⁻ (olivine), 1 in pyrosilicates Si₂O₇⁶⁻, 2 in rings (Si₆O₁₈¹²⁻ in beryl) and single chains (SiO₃)ₙ²ⁿ⁻ (pyroxenes), 2.5 in double chains Si₄O₁₁⁶⁻ (amphiboles, asbestos), 3 in sheets Si₂O₅²⁻ (micas, clays, talc) and 4 in three-dimensional frameworks SiO₂ (quartz), where replacing Si by Al gives aluminosilicate frameworks (feldspars, zeolites). Silicones (R₂SiO)ₙ come from hydrolysis and condensation of R₂SiCl₂; R₃SiCl caps chains and RSiCl₃ cross-links them, giving oils, rubbers and resins that are water-repellent and thermally stable. Boron nitride is isoelectronic with carbon: hexagonal BN is layered like graphite but an insulator (the B–N polarity localises the π electrons), and cubic BN, like diamond, is superhard. Borazine B₃N₃H₆, "inorganic benzene", is planar and isoelectronic with benzene but its polar B–N bonds make it add HCl or water readily. Phosphazenes (NPCl₂)ₙ, from PCl₅ and NH₄Cl, form rings (n = 3, 4) and high polymers with alternating P–N bonds; substituting Cl by OR gives stable elastomers.
4. Allotropes, industrial chemicals, noble gases and interhalogens
Carbon: diamond (sp³, three-dimensional, hardest, insulating), graphite (sp² layers 335 pm apart, conducting in-plane, thermodynamically stable at ambient conditions), fullerenes (C₆₀ with 12 pentagons and 20 hexagons), nanotubes and graphene. Phosphorus: white P₄ (tetrahedral, P–P–P angle 60°, strained, very reactive, glows in air and is stored under water), red (polymeric chains of P₄ units, less reactive) and black (puckered layers, the most stable, a semiconductor). Sulfur: rhombic α-S₈ (crown rings, stable below 95.5 °C), monoclinic β-S₈ (stable from 95.5 °C to the melting point) and plastic sulfur (long helical chains from quenching the melt).
Industrial synthesis is an exercise in equilibrium and kinetics. Ammonia (Haber–Bosch): N₂ + 3H₂ ⇌ 2NH₃ is exothermic (ΔH° ≈ −92 kJ/mol) with Δn = −2, so high pressure (about 150–300 atm) raises the yield, and low temperature would too, but at a useless rate; a compromise near 400–450 °C is used with a promoted iron catalyst (K₂O, Al₂O₃), and NH₃ is condensed out and unreacted gas recycled. Sulfuric acid (contact process): S or sulfide ores are burned to SO₂; 2SO₂ + O₂ ⇌ 2SO₃ (exothermic) is run over V₂O₅ at about 400–450 °C; SO₃ is absorbed in concentrated H₂SO₄ to give oleum, H₂S₂O₇, which is diluted, because SO₃ with water directly forms a fine acid mist. Nitric acid (Ostwald process): 4NH₃ + 5O₂ → 4NO + 6H₂O over a Pt–Rh gauze at about 850 °C with very short contact time; NO is oxidised to NO₂, which is absorbed in water, 3NO₂ + H₂O → 2HNO₃ + NO, and the NO is recycled.
Noble gases. Xenon, with the lowest ionisation energy of the stable noble gases, forms compounds with the most electronegative elements. Heating Xe with F₂ gives XeF₂ (linear), XeF₄ (square planar) or XeF₆ (distorted octahedral) depending on the ratio and conditions — shapes that VSEPR predicts from three, two and one lone pairs. XeF₆ hydrolyses to XeO₃, a trigonal pyramidal molecule that is dangerously explosive when dry; XeF₂ is a clean fluorinating agent. Interhalogens XY, XY₃, XY₅ and XY₇ (X the larger halogen; IF₇ is the only XY₇) have shapes from VSEPR (ClF₃ T-shaped, BrF₅ square pyramidal, IF₇ pentagonal bipyramidal) and are more reactive than the parent halogens, except F₂, because the X–Y bond is weaker and polar; ClF₃ and BrF₃ are powerful fluorinating agents. Polyhalide ions such as I₃⁻ (linear) form when a halide coordinates a halogen molecule.
5. Acid–base concepts: Brønsted, Lewis, HSAB and acid–base catalysis
A Brønsted acid donates a proton and a base accepts one; every acid has a conjugate base, and the stronger the acid the weaker its conjugate base. In water the strongest acid that can exist is H₃O⁺ (the levelling effect), so HClO₄, HCl and HNO₃ all look equally strong; a less basic solvent such as acetic acid differentiates them. A Lewis acid accepts an electron pair and a base donates one, which covers metal ions, BF₃ and AlCl₃. Lewis acidity of the boron trihalides runs BF₃ < BCl₃ < BBr₃, the reverse of electronegativity, because the strong B–F π back-donation into boron’s empty p orbital must be lost when BF₃ pyramidalises to accept a base.
Pearson’s hard and soft acids and bases sorts Lewis acids and bases by polarisability. Hard acids are small, highly charged and weakly polarisable (H⁺, Li⁺, Na⁺, Mg²⁺, Al³⁺, Ti⁴⁺, Fe³⁺, BF₃); soft acids are large, low-charged and polarisable (Cu⁺, Ag⁺, Au⁺, Hg²⁺, Pd²⁺, Pt²⁺, zero-valent metals). Hard bases: F⁻, OH⁻, H₂O, NH₃, O²⁻, RCOO⁻; soft bases: I⁻, S²⁻, RS⁻, CN⁻, CO, PR₃, H⁻, SCN⁻ bound through S; borderline: Fe²⁺, Co²⁺, Ni²⁺, Cu²⁺, Zn²⁺, Br⁻, pyridine, N₃⁻. The principle — hard prefers hard, soft prefers soft — explains why Al and Ti occur as oxides and Hg, Pb and Cu as sulfides, why Ag⁺ binds I⁻ far more strongly than F⁻, and why thiocyanate binds hard metals through N and soft ones through S. Acid–base catalysis speeds reactions by protonating a substrate (ester hydrolysis, keto–enol tautomerism, dehydration) or by deprotonating it (aldol reactions); Lewis acids such as AlCl₃ catalyse Friedel–Crafts reactions by generating the electrophile.
Key takeaways
- VSEPR: lone pairs repel most, sit equatorial in a trigonal bipyramid and trans in an octahedron — ClF₃ T-shaped, SF₄ see-saw, XeF₂ linear, XeF₄ square planar.
- B₂H₆ has two 3c–2e bridges; phosphorus oxoacid basicity counts P–OH, not H; sulfur is +6 in H₂S₂O₈; oxides turn acidic across a period.
- Wade: n + 1 pairs closo, n + 2 nido, n + 3 arachno (BH 2, CH 3, extra H 1); CH₃ ≡ Mn(CO)₅ isolobal; bridging O per Si sets the silicate class.
- Haber: Fe catalyst, high P, ~400–450 °C; contact: V₂O₅, absorb SO₃ in H₂SO₄; Ostwald: Pt–Rh at ~850 °C. Black P and graphite are the stable allotropes.
- BF₃ < BCl₃ < BBr₃ in Lewis acidity (π back-donation); HSAB: hard–hard and soft–soft pairs are the stable ones.
Practice questions (19)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
What is the shape of the ClF₃ molecule?
Show answer
Answer: A — T-shaped
Cl has 7 valence electrons, three used in bonds, leaving two lone pairs: steric number 5. Both lone pairs go equatorial in the trigonal bipyramid, leaving the three F atoms in a T. Trigonal planar would need no lone pairs; see-saw is SF₄ with one.Which of the following species are linear?
Show answer
Answer: A — XeF₂; B — I₃⁻; D — BeCl₂ (gas phase)
XeF₂ and I₃⁻ have steric number 5 with three equatorial lone pairs, leaving the two ligands axial and linear; gaseous BeCl₂ has two bond pairs and no lone pairs. ClF₂⁺ has two bond pairs and two lone pairs, steric number 4, so it is bent.How many lone pairs are there on the xenon atom in XeF₄?
Numerical answer — type the value.
Show answer
Answer: 2
Xe has 8 valence electrons; four go into Xe–F bonds, leaving 4 electrons, i.e. 2 lone pairs. With steric number 6 they sit trans, so the molecule is square planar. XeF₂ has 3 lone pairs and XeF₆ 1.How many three-centre two-electron B–H–B bonds does diborane, B₂H₆, contain?
Numerical answer — type the value.
Show answer
Answer: 2
B₂H₆ has 12 valence electrons: four terminal 2c–2e B–H bonds use 8, and the remaining 4 form two 3c–2e B–H–B bridges above and below the B₂H₄ plane. Counting six ordinary bonds would need 14 electrons it does not have.What is the basicity of phosphorous acid, H₃PO₃?
Numerical answer — type the value.
Show answer
Answer: 2
H₃PO₃ is HPO(OH)₂: one hydrogen is bonded directly to phosphorus and is not ionisable, so only the two P–OH hydrogens are acidic. Counting all three hydrogens gives 3, the answer for H₃PO₄.What is the oxidation state of sulfur in peroxodisulfuric acid, H₂S₂O₈?
Numerical answer — type the value.
Show answer
Answer: 6
H₂S₂O₈ is (HO)(O)₂S–O–O–S(O)₂(OH): of its eight oxygens two form a peroxide bridge at −1 each and six are −2. So 2(+1) + 2x + 6(−2) + 2(−1) = 0 gives x = +6. Treating every oxygen as −2 gives the impossible +7.How many skeletal electron pairs does the closo-borane anion B₆H₆²⁻ have according to Wade’s rules?
Numerical answer — type the value.
Show answer
Answer: 7
Each of the six B–H units supplies 2 skeletal electrons (12) and the 2− charge adds 2, giving 14 electrons = 7 pairs = n + 1 for n = 6: a closo octahedron. Counting all valence electrons, including the B–H bonds, gives 13 pairs and the wrong class.According to Wade’s rules, pentaborane(9), B₅H₉, is
Show answer
Answer: A — nido
Five B–H units give 10 electrons and the four extra (bridging) H atoms 4, so 14 electrons = 7 pairs = n + 2 for n = 5: nido, a square pyramid derived from an octahedron with one vertex missing. B₄H₁₀, with the same 7 pairs on four vertices, is arachno.How many skeletal electron pairs does the carborane C₂B₁₀H₁₂ have according to Wade’s rules?
Numerical answer — type the value.
Show answer
Answer: 13
Each C–H unit contributes 3 skeletal electrons and each B–H unit 2: 2 × 3 + 10 × 2 = 26 electrons = 13 pairs = n + 1 for n = 12 — a closo icosahedron, isoelectronic with B₁₂H₁₂²⁻. Counting C–H as 2 gives 12 pairs, which would wrongly suggest a hypercloso cage.Which fragment is isolobal with CH₃?
Show answer
Answer: A — Mn(CO)₅
CH₃ has one singly occupied frontier orbital, as does the 17-electron Mn(CO)₅ (a d⁷ octahedral fragment missing one ligand); that is why Mn₂(CO)₁₀ mirrors ethane. Fe(CO)₄ is isolobal with CH₂ and Co(CO)₃ with CH.In a sheet silicate with the repeating unit (Si₂O₅)²⁻, how many oxygen atoms of each SiO₄ tetrahedron are shared with neighbouring tetrahedra?
Numerical answer — type the value.
Show answer
Answer: 3
With b bridging oxygens per Si, each tetrahedron owns 4 − b/2 oxygens. Si₂O₅ means 2.5 O per Si, so 4 − b/2 = 2.5 and b = 3: three corners shared in a sheet, one oxygen left pointing out. Single chains (SiO₃) share 2 and frameworks (SiO₂) all 4.The mineral beryl, Be₃Al₂Si₆O₁₈, contains which silicate unit?
Show answer
Answer: A — A cyclic Si₆O₁₈¹²⁻ ring
Six tetrahedra each share two corners to close a ring: Si₆O₁₈ with charge 6 × 4 − 18 × 2 = −12, balanced by 3Be²⁺ + 2Al³⁺ = +12. Isolated tetrahedra are olivine’s motif, double chains the amphiboles’.In the P₄O₁₀ molecule, how many oxygen atoms bridge two phosphorus atoms?
Numerical answer — type the value.
Show answer
Answer: 6
P₄O₁₀ keeps the P₄ tetrahedron but inserts an oxygen into each of its six P–P edges (6 bridging O), and each phosphorus carries one terminal P=O (4 terminal O): 6 + 4 = 10. P₄O₆ has the same six bridges and no terminal oxygens.Which statements about allotropes are correct?
Show answer
Answer: A — White phosphorus consists of P₄ tetrahedra with P–P–P angles of 60°; B — Black phosphorus is the most thermodynamically stable form of phosphorus
The 60° angles make white P strained and reactive, and black P (puckered layers) is the stable allotrope. Diamond’s electrons are all in localised sp³ bonds, so it insulates; rhombic sulfur is the room-temperature form, with monoclinic stable only above 95.5 °C.Which catalyst is used for the oxidation of SO₂ to SO₃ in the contact process?
Show answer
Answer: A — V₂O₅
Vanadium pentoxide on a silica support catalyses 2SO₂ + O₂ ⇌ 2SO₃ at about 400–450 °C. Promoted iron is the Haber catalyst and Pt–Rh gauze the Ostwald catalyst for oxidising ammonia.For the Haber–Bosch synthesis N₂ + 3H₂ ⇌ 2NH₃ (ΔH° ≈ −92 kJ/mol), which statements are correct?
Show answer
Answer: A — High pressure increases the equilibrium yield of NH₃; B — Lower temperature would increase the equilibrium yield but slow the reaction
Four moles of gas become two, so pressure pushes the equilibrium to NH₃; the reaction is exothermic, so K falls with temperature and 400–450 °C is a compromise with rate. A catalyst speeds both directions equally and leaves K unchanged.Borazine, B₃N₃H₆, is isoelectronic with benzene. Why does it add HCl readily while benzene does not?
Show answer
Answer: A — Its B–N bonds are polar, with electrophilic B and nucleophilic N, so π delocalisation is weaker
Borazine is planar with six π electrons, but they sit mostly on nitrogen, so the ring is only weakly aromatic: H⁺ adds to N and Cl⁻ to B, giving B₃N₃H₉Cl₃. Boron carries the empty p orbital, not a lone pair.What is the correct order of Lewis acid strength of the boron trihalides towards a strong base such as NH₃?
Show answer
Answer: A — BF₃ < BCl₃ < BBr₃
Electronegativity alone would make BF₃ the strongest, but the small F atoms donate strongly into boron’s empty 2p orbital, and that π bonding is lost when BF₃ pyramidalises to bind a base. The weaker B–Br π overlap costs less, so BBr₃ is the strongest acid.By the HSAB principle, which ligand forms the most stable complex with Hg²⁺?
Show answer
Answer: A — I⁻
Hg²⁺ is a large, polarisable soft acid and pairs best with the soft base I⁻; stability falls through Br⁻ and Cl⁻ to the hard F⁻ and H₂O. A hard acid such as Al³⁺ would show the reverse order.