AP EAPCET (Engineering) · Chemistry (JEE & NEET)
p-Block Elements
Group trends and chemistry of groups 13 to 18 including boron, carbon, nitrogen, oxygen, halogen and noble gas families.
Thirteen concepts covering JEE Main p-block depth for groups 13–18. Three mechanisms organise the chapter: the inert-pair effect down groups 13–15, anomalous first elements (B, C, N, O, F — small, no d-orbitals, strong multiple bonding), and oxidation-state / electron-count bookkeeping for oxoacids, diborane and xenon fluorides. Planar shapes (XeF₂, XeF₄, ClF₃) are drawn; octahedral XeF₆ and IF₇ are not. Rationalised NCERT dropped the Class 12 p-block PDF — content follows older NCERT/JEE syllabus expectations.
- AP EAPCET (Engineering)
- Medium level
- 13 concepts
- 5 practice questions
1Inert-pair effect and oxidation-state stability
Descending groups 13–15, the ns² electrons become increasingly reluctant to participate in bonding. That inert-pair effect stabilises the lower oxidation state: Tl⁺ over Tl³⁺, Pb²⁺ over Pb⁴⁺, Bi³⁺ over Bi⁵⁺.
The higher oxidation state remains accessible but becomes a stronger oxidiser down the group — PbO₂ and BiF₅ oxidise what their lighter congeners would not — because returning to the lower state is favoured.
Figure. Schematic preference for the higher oxidation state falling down groups 13–15. Lengths are ordinal, not energies — the figure teaches direction of the inert-pair trend, not a measured ΔG. Tl⁺, Pb²⁺ and Bi³⁺ win at the bottom.
Reading the group
- Name the group oxidation statesGroup 13: +3 and +1; group 14: +4 and +2; group 15: +5 and +3.
- Apply inert pair down the groupLower state gains relative stability; higher state becomes oxidising.
- Pick the exam consequenceTlCl is stable ionic; TlCl₃ is oxidising. Pb²⁺ salts are common; Pb⁴⁺ is not.
| Group | Higher OS | Lower OS | Heavier preference |
|---|---|---|---|
| 13 | +3 | +1 | Tl⁺ ≫ Tl³⁺ |
| 14 | +4 | +2 | Pb²⁺ ≫ Pb⁴⁺ |
| 15 | +5 | +3 | Bi³⁺ ≫ Bi⁵⁺ |
Which ion is more stable in aqueous chemistry?
- Pb⁴⁺
- Pb²⁺
- They are equally stable
The inert-pair effect stabilises Pb²⁺ over Pb⁴⁺. PbO₂ (Pb⁴⁺) is a strong oxidiser precisely because returning to Pb²⁺ is favoured.
2Group 13: boron anomaly, diborane, borax and AlCl₃
Boron is the anomalous first member of group 13: small size, high ionisation enthalpy, and a tendency to form covalent, electron-deficient compounds rather than B³⁺ salts. BF₃ and BCl₃ are planar Lewis acids with an incomplete octet — boron has only six electrons in the valence shell and accepts a lone pair readily.
Diborane (B₂H₆) is the textbook electron-deficient hydride: four terminal B–H bonds are ordinary two-centre two-electron bonds, while the two bridging B–H–B links are three-centre two-electron bonds. Aluminium chloride in the vapour and in non-polar solvents exists as the Al₂Cl₆ dimer with bridging chlorines, so each Al completes an octet; in water it ionises and the covalent dimer story no longer applies. Borax (Na₂B₄O₇·10H₂O) is the exam staple for qualitative analysis — the borax-bead test colours identify many metal ions.
Figure. Diborane is electron-deficient: six B–H connections would want twelve electrons for ordinary 2c–2e bonds, and that is exactly the valence count — the two bridging hydrogens must be 3c–2e banana bonds. Al₂Cl₆ bridges with chlorines for a related reason.
Reading group-13 staples
- Boron anomalySmall, high IE, covalent/electron-deficient — Lewis acids BF₃/BCl₃, not ionic B³⁺ salts.
- Diborane bookkeeping4 terminal 2c–2e B–H + 2 bridging 3c–2e B–H–B. Total valence electrons are short of an all-2c–2e count.
- AlCl₃ and boraxAl₂Cl₆ dimer (bridging Cl) in vapour/non-polar media; aqueous AlCl₃ ionises. Borax bead → metal identification.
| Compound | What to remember | Why it matters |
|---|---|---|
| BF₃ / BCl₃ | Planar, incomplete octet | Lewis acids |
| B₂H₆ | 4 terminal 2c–2e + 2 bridge 3c–2e | Electron-deficient hydride |
| Al₂Cl₆ | Cl-bridged dimer (vapour/non-polar) | Octet completion for Al |
| Borax | Na₂B₄O₇·10H₂O | Bead test / qualitative analysis |
Electron count in diborane
Show that B₂H₆ is electron-deficient if every B–H connection were a normal 2c–2e bond, and state how the bridging bonds resolve the deficit.
- Valence electrons: 2×B(3) + 6×H(1)12 e⁻
- Connections if all 2c–2e: 4 terminal + 2 bridges ×2 ends8 bonds would need 16 e⁻
- Deficit4 e⁻ short of an all-2c–2e picture
- Resolution2 bridges are 3c–2e (2 e⁻ each); terminals stay 2c–2e
Pro tip. Electron deficiency is a counting claim, not a metaphor: 12 electrons cannot fund eight ordinary 2c–2e bonds. The 3c–2e bridges are the bookkeeping fix NCERT teaches.
BF₃ is a Lewis acid primarily because
- Boron has a complete octet and a vacant d-orbital
- Boron has an incomplete octet and can accept a lone pair
- Boron is more electronegative than fluorine
Boron in BF₃ has only six valence electrons around it. It accepts an electron pair (e.g. from NH₃ or F⁻) to complete the octet — the definition of a Lewis acid here.
3Group 14: catenation, carbon allotropes, CO/CO₂ and silicates
Carbon catenates far more than silicon because the C–C bond is much stronger than the Si–Si bond; that single fact underwrites organic chemistry and the persistence of long carbon chains. Carbon's common allotropes at this level are diamond (sp³ network, hard, electrical insulator), graphite (sp² layers, soft, electrical conductor along the sheets), and fullerenes such as C₆₀ (cage molecules).
CO is a strong π-acceptor ligand and a reducing agent at high temperature (e.g. in metallurgy); CO₂ is a linear acidic oxide that forms carbonates with bases. Silicon chemistry is dominated by Si–O frameworks: silicates are built from SiO₄ tetrahedra that share corners, and silicones are polymeric −(R₂Si–O)− chains. Down the group the inert-pair effect (already covered) makes Pb²⁺ far more common than Pb⁴⁺.
Figure. Carbon allotropes differ by hybrid network: diamond is an sp³ 3D net, graphite stacked sp² sheets with mobile π electrons, C₆₀ a closed sp² cage. Silicon catenates far more weakly — that is the group anomaly the exam asks.
Group-14 forks
- CatenationC ≫ Si because bond enthalpy C–C ≫ Si–Si — long chains for C, short chains/rings for Si.
- AllotropesDiamond sp³ insulator/hard; graphite sp² conductor/soft layers; C₆₀ fullerene cages.
- Oxides and Si–OCO reducing / π-acid ligand; CO₂ acidic oxide. Silicates = linked SiO₄; silicones = R₂SiO polymers.
| Species | Key structure idea | Exam property |
|---|---|---|
| Diamond | sp³ 3D network | Hard, insulator |
| Graphite | sp² layers + π electrons | Soft, conducts in-plane |
| C₆₀ | Cage molecule | Fullerene allotrope |
| CO | Strong π-acceptor / reductant | Ligand; metallurgy |
| CO₂ | Linear acidic oxide | Carbonates with base |
| Silicates | Corner-sharing SiO₄ | Rock-forming frameworks |
| Silicones | −(R₂Si–O)− polymers | Thermal stability, water-repellent |
Diamond vs graphite from hybridisation
Relate the hybridisation of carbon in diamond and graphite to hardness and electrical conductivity.
- Diamond hybridisationsp³ — 3D covalent network
- Consequencevery hard; no mobile electrons → insulator
- Graphite hybridisationsp² — layers; p orbitals form π system
- Consequencelayers slide (soft); delocalised π electrons → conducts in-plane
Pro tip. Same element, two networks: the hybridisation decides whether electrons are locked in σ bonds (diamond) or free to move in a π system (graphite).
Carbon catenates more extensively than silicon mainly because
- Carbon has a higher atomic mass than silicon
- The C–C bond is much stronger than the Si–Si bond
- Silicon has no d-orbitals
Catenation tracks the strength of the element–element single bond. C–C is strong enough for long chains; Si–Si is weaker, so catenation is limited. Silicon does have accessible 3d orbitals — that is not the reason it catenates less.
4Group 15: nitrogen's oxidation states and NH₃
Nitrogen shows oxidation states from −3 to +5. Ammonia (N at −3) is a good base and a mild reducing agent; nitric acid (N at +5) is a strong oxidiser. The wide span is why nitrogen chemistry feels like several elements at once.
Nitrogen is the anomalous first element of group 15: small size, high electronegativity, strong N≡N triple bond, and no accessible d-orbitals. It cannot expand its octet, so NF₅ and NCl₅ do not exist while PCl₅ does. Down the group the E–H bond weakens, so hydride thermal stability falls and reducing power rises — BiH₃ is the strongest reducer among the group-15 hydrides. Bond angles fall NH₃ (107°) > PH₃ (93.5°) > AsH₃ (91.8°) as the central atom uses less hybridised p character.
Figure. Bond angles of the group-15 hydrides. NH₃ is well above the 90° p-orbital limit; PH₃ and AsH₃ sit close to 90° as hybridisation fades. Drawn as bars from a 90° baseline so the excess over pure p character is the bar length — not as a geometric angle figure, which aspect would distort.
Nitrogen checklist
- Read the oxidation state−3 in NH₃/NH₄⁺; +3 in HNO₂; +5 in HNO₃/NO₃⁻.
- Match the roleLow OS → base/reductant; high OS → oxidant.
- Anomaly and hydride trendN: no d-orbitals → no pentahalide. Down the group: thermal stability ↓, reducing power ↑ to BiH₃.
Oxidation state in three nitrogen compounds
Assign the oxidation number of N in NH₃, NO and HNO₃.
- NH₃: 3H at +1 → N + 3(+1) = 0N = −3
- NO: O at −2 → N + (−2) = 0N = +2
- HNO₃: H +1, 3O at −2 → N + 1 − 6 = 0N = +5
- Span check−3 → +5 covers the group-15 range named in the prose
Pro tip. HNO₃ as an oxidiser is N(+5) grabbing electrons to fall toward a lower state — the same inert-pair logic that makes Bi(+5) oxidising, applied at the top of the group to a different mechanism.
Among NH₃, PH₃, AsH₃ and BiH₃, the strongest reducing agent is
- NH₃
- PH₃
- BiH₃
E–H bond strength falls down the group, so the hydride becomes a better H-atom/electron donor. BiH₃ is the strongest reducer in group 15.
5Ammonia, HNO₃ and phosphorus oxoacid basicity
Ammonia is manufactured by Haber's process (N₂ + 3H₂ ⇌ 2NH₃, Fe catalyst, high P, moderate T). Nitric acid is made industrially by the Ostwald route: catalytic oxidation of NH₃ to NO, then to NO₂, then absorption in water. Concentrated HNO₃ is a strong oxidiser — it attacks most metals — but passivates Al, Fe and Cr by forming a coherent oxide film, so those metals appear unreactive toward cold concentrated acid.
Phosphorus oxoacids illustrate a counting rule that exams love: basicity equals the number of P–OH hydrogens, not the total hydrogen count. Hypophosphorous acid H₃PO₂ is monobasic (one P–OH, two P–H), phosphorous acid H₃PO₃ is dibasic (two P–OH, one P–H), and phosphoric acid H₃PO₄ is tribasic (three P–OH).
Figure. Basicity of the three common phosphorus oxoacids as the count of P–OH groups. Bar height is that integer (1, 2, 3), not Ka — the teaching is the counting rule, not acid strength order.
From NH₃ to oxoacid basicity
- RoutesHaber → NH₃; Ostwald → HNO₃ from NH₃.
- HNO₃ as oxidiserStrong oxidiser on most metals; cold conc. HNO₃ passivates Al/Fe/Cr via oxide film.
- P-oxoacid basicityCount P–OH only. H₃PO₂: 1; H₃PO₃: 2; H₃PO₄: 3. P–H hydrogens do not ionise.
Basicity of the phosphorus oxoacids
State the basicity of H₃PO₂, H₃PO₃ and H₃PO₄, and justify each from the number of ionisable hydrogens.
- H₃PO₂ structure count1 P–OH + 2 P–H → monobasic
- H₃PO₃ structure count2 P–OH + 1 P–H → dibasic
- H₃PO₄ structure count3 P–OH → tribasic
- Rulebasicity = n(P–OH), not n(H) total
Pro tip. If the formula is H₃POₓ, do not assume tribasic. Draw or recall P–OH versus P–H before quoting basicity.
The basicity of H₃PO₃ is
- 1
- 2
- 3
H₃PO₃ has two P–OH groups and one P–H. Only the P–OH hydrogens ionise, so it is dibasic — not tribasic despite three hydrogens in the formula.
6Group 16: oxygen's anomaly and sulphur
Oxygen is anomalous in group 16: small size, high electronegativity, and no accessible d-orbitals. It forms strong hydrogen bonds, prefers a −2 oxidation state, and does not form the expanded-octet compounds that sulphur does (SF₆ exists; OF₆ does not).
Sulphur shows allotropy (rhombic and monoclinic S₈ rings are the exam staples). SO₂ is both a reductant and a bleach in aqueous solution — it reduces many oxidising dyes and is itself oxidised toward sulphate. Ozone and the contact-process chemistry of H₂SO₄ get their own concept next; this one is the O-versus-S anomaly that makes those compounds possible for sulphur but not for oxygen in the same way.
Figure. Oxygen's anomaly is the weak O–O single bond and its preference for multiple bonds; sulphur catenates (S₈) and makes stronger single bonds. Compare the two rows before memorising allotrope names.
Oxygen vs sulphur
- Size and ENO is smaller and more electronegative → H-bonding, discrete O₂, limited catenation versus S.
- d-orbitalsS can expand its octet (SF₆, SO₄²⁻ resonance structures with expanded pictures); O cannot.
- SO₂ roleReducing/bleaching in water; oxidised to H₂SO₄ under forcing conditions.
| Property | Oxygen | Sulphur |
|---|---|---|
| H-bonding | Strong (H₂O) | Weak (H₂S) |
| Catenation | Limited (O₂, O₃, peroxides) | Extensive (S₈, polysulphides) |
| Expanded octet | No (n = 2) | Yes (SF₆, SO₄²⁻ pictures) |
| Common OS | −2 dominates | −2, +4, +6 all common |
SF₆ exists but OF₆ does not, mainly because
- Oxygen is more electronegative than sulphur
- Oxygen lacks accessible d-orbitals to expand its octet
- Fluorine cannot oxidise oxygen
Sulphur can accommodate six pairs using 3d orbitals in the valence picture taught at this level; oxygen's valence shell stops at n = 2 with no d-orbitals.
7Ozone and sulphuric acid (contact process)
Ozone (O₃) is an allotrope of oxygen: a bent molecule with resonance between two equivalent O–O–O pictures, and a strong oxidising agent. It bleaches by oxidation; unlike chlorine bleaching, ozone bleaching is often described as temporary in the older textile sense because atmospheric oxygen can re-oxidise some reduced dyes — exam stems still contrast O₃ with Cl₂ on that point. Ozone also cleaves alkenes (ozonolysis), which organic chemistry owns in detail.
Sulphuric acid is made by the contact process: SO₂ is oxidised to SO₃ over V₂O₅, SO₃ is absorbed in concentrated H₂SO₄ to give oleum (H₂S₂O₇), and oleum is diluted carefully to H₂SO₄. Concentrated H₂SO₄ is a powerful dehydrating agent (chars sugar, dries gases) and, when hot, an oxidiser — it can oxidise Cu, C and Br⁻/I⁻ under forcing conditions while itself being reduced toward SO₂.
Figure. Contact process: burn sulphur to SO₂, catalytically oxidise to SO₃ on V₂O₅, absorb into H₂SO₄ as oleum, then dilute. Ozone's bent resonance pair stays in the property table — the process flow is the drawable half.
O₃ and the contact process
- Ozone roleBent, resonance-stabilised oxidiser; bleaching and ozonolysis are the two common uses named in exams.
- Contact stepsSO₂ → SO₃ (V₂O₅) → absorb in conc. H₂SO₄ → oleum → dilute to H₂SO₄.
- H₂SO₄ dual characterCold conc.: dehydrating. Hot conc.: oxidising (Cu, C, halide ions → SO₂).
| Species | Role | Exam note |
|---|---|---|
| O₃ | Strong oxidiser / bleach | Also ozonolysis of alkenes |
| SO₂ → SO₃ | Contact oxidation | V₂O₅ catalyst |
| Oleum | H₂S₂O₇ | SO₃ absorbed in H₂SO₄ |
| Conc. H₂SO₄ | Dehydrating (± oxidising when hot) | Chars sugar; dries gases |
Contact-process sequence
Write the key stages that convert SO₂ into concentrated H₂SO₄ in the contact process, naming the catalyst.
- 2SO₂ + O₂ ⇌ 2SO₃V₂O₅ catalyst
- SO₃ + H₂SO₄H₂S₂O₇ (oleum)
- H₂S₂O₇ + H₂O2H₂SO₄
- Why not SO₃ into water directly?fog of H₂SO₄ — absorption in acid avoids it
Pro tip. The industrial trick is absorbing SO₃ in acid, not dumping it into water. Oleum is the intermediate that makes dilution controllable.
In the contact process the catalyst for SO₂ → SO₃ is
- Fe
- V₂O₅
- Pt only — V₂O₅ is obsolete
V₂O₅ is the standard contact-process catalyst taught at this level. Iron is Haber's catalyst for ammonia; platinum can catalyse the oxidation but is not the named industrial staple in NCERT/JEE.
8Halogens: oxidising power and reactivity
The halogens are the most reactive non-metals. Oxidising power falls F₂ > Cl₂ > Br₂ > I₂, tracking their standard reduction potentials. Fluorine is reactive enough to oxidise water; iodine is a mild oxidiser by comparison.
Fluorine is the anomalous first halogen: highest electronegativity, strongest H-bonding in HF, and almost no positive oxidation states in oxoacids of the HOX type that chlorine enjoys. Interhalogen shapes (ClF₃, BrF₅, IF₇) are a separate concept — oxidising power here is only the X₂ ranking.
Figure. Standard reduction potentials E°(X₂/X⁻) in volts. Drawn lengths follow the values: F₂ at 2.87 V dwarfs I₂ at 0.54 V. Oxidising power is exactly this order.
Comparing halogen oxidisers
- Read E°More positive E°(X₂/X⁻) → stronger oxidiser.
- Predict displacementA halogen higher in the series displaces a lower one from its halide: Cl₂ liberates Br₂ from Br⁻.
- Separate from HX acidityOxidising power of X₂ and acidity of HX are different trends — do not merge them.
Displacement from E°
E°(Cl₂/Cl⁻) = 1.36 V and E°(Br₂/Br⁻) = 1.07 V. Can Cl₂ oxidise Br⁻ under standard conditions? Find E°_cell.
- Higher E° couple: Cl₂/Cl⁻Cl₂ reduced
- Lower E° couple: Br₂/Br⁻Br⁻ oxidised
- E°_cell = 1.36 − 1.070.29 V > 0
- Net reactionCl₂ + 2Br⁻ → 2Cl⁻ + Br₂
Pro tip. This is the same 'higher E° oxidises the reduced form of the lower couple' rule from electrochemistry — the halogen series is just that rule with four named points.
Which halogen can displace iodide from KI but not fluoride from KF?
- F₂
- Cl₂
- None of them
Cl₂ oxidises I⁻ (E° gap favours it) but cannot oxidise F⁻ — fluorine's couple sits above chlorine. F₂ would oxidise both, but the question asks for selective displacement of iodide.
9Hydrogen halide acidity vs hydrogen bonding
Acid strength of the hydrogen halides rises HF < HCl < HBr < HI. The H–X bond weakens down the group faster than electronegativity effects can compensate, so HI ionises most readily in water.
HF is the anomaly students conflate: it is the weakest hydrohalic acid yet forms the strongest hydrogen bonds, because acidity tracks bond strength while H-bonding tracks fluorine's high electronegativity and small size — two different questions.
Figure. Two rankings of the same four molecules: acidity rises HF < HCl < HBr < HI, while hydrogen-bond strength falls HF >> HCl > HBr > HI. HF's high boiling point is the H-bond story, not the acidity story.
Separating the two trends
- AcidityGoverned mainly by H–X bond enthalpy down the group → HF weakest acid.
- Hydrogen bondingGoverned by EN and size of X → HF strongest H-bonds (and abnormal boiling point).
- Do not mergeStrong H-bonding does not make HF a strong acid; it makes HF associated.
Acidity vs hydrogen bonding in HF
Why is HF the weakest hydrohalic acid yet forms the strongest hydrogen bonds?
- Acid strength ↔ ease of H–X cleavageH–F is the strongest H–X bond
- Consequence for acidityHF ionises least → weakest acid
- H-bonding ↔ EN and size of XF smallest and most electronegative
- Consequence for H-bondsHF forms the strongest hydrogen bonds
Pro tip. Bond strength governs acidity down the group; electronegativity governs hydrogen-bond strength. Same atom, two properties, opposite rankings against the rest of the group.
The strongest acid among the hydrogen halides is
- HF
- HCl
- HI
HI has the weakest H–X bond and ionises most readily. HF is the weakest acid despite fluorine's electronegativity.
10Interhalogens: XX′ₙ shapes from ClF₃ to IF₇
Interhalogens have the formula XX′ₙ with n = 1, 3, 5 or 7. The central atom X is the larger, less electronegative halogen; the terminal atoms X′ are usually fluorine when n is large, because only F is small and electronegative enough to pack around I or Br at high coordination. So IF₇ exists while ICl₇ does not appear in the standard set.
Shapes follow VSEPR once you count bonds and lone pairs on the central atom: ClF₃ is T-shaped (steric number 5, two equatorial lone pairs), BrF₅ is square pyramidal (steric number 6, one lone pair), and IF₇ is pentagonal bipyramidal (steric number 7, no lone pair). Only the planar T of ClF₃ is drawn here; the higher-coordinate shapes need honest 3D.
Figure. ClF₃ as a planar T: two axial F atoms and one equatorial F. The two lone pairs sit in the equatorial belt opposite the drawn F and are named rather than drawn as dots — dots would claim angles the top view cannot certify beside the axial pair.
From formula to shape
- Read XX′ₙX = larger central halogen; X′ = more electronegative terminals (often F). n = 1,3,5,7.
- Steric number on XClF₃: 3 bonds + 2 LP → 5 → T-shaped. BrF₅: 5 bonds + 1 LP → 6 → square pyramidal. IF₇: 7 bonds + 0 LP → 7 → pentagonal bipyramidal.
- What not to inventDo not flatten BrF₅ or IF₇ into a hexagon of bonds — decline those figures.
| Molecule | Steric no. | Shape |
|---|---|---|
| ClF / BrF / IF | 2 (linear diatomic) | Linear |
| ClF₃ | 5 | T-shaped |
| BrF₅ | 6 | Square pyramidal (not drawn) |
| IF₇ | 7 | Pentagonal bipyramidal (not drawn) |
Shape of ClF₃
Determine the hybridisation and molecular shape of ClF₃.
- Cl valence electrons7
- 3 Cl–F bonds; leftover electrons4 e⁻ = 2 lone pairs
- Steric number = 3 bonds + 2 LP5 → sp³d
- Two LP equatorial in TBP setT-shaped molecule
Pro tip. In a trigonal bipyramid, lone pairs prefer the equatorial belt. Two equatorial lone pairs leave three fluorines in a T — the same logic that makes XeF₂ linear with three equatorial lone pairs.
The shape of ClF₃ is
- Trigonal planar
- T-shaped
- Square planar
Steric number 5 with two lone pairs equatorial gives a T-shaped molecular geometry. Square planar is the XeF₄ / ICl₄⁻ story (steric number 6, two LP trans).
11Oxoacid strength and oxidation state
For oxoacids of the same central atom, more oxygen atoms — equivalently, a higher oxidation state — mean a stronger acid: HClO < HClO₂ < HClO₃ < HClO₄. Extra oxygen atoms withdraw electron density from the O–H bond and stabilise the conjugate base by delocalisation.
Across a period at the same oxidation state, higher electronegativity of the central atom also strengthens the acid (H₂SO₄ > H₃PO₄ in the usual comparison), but the chlorine oxoacid series is the clean exam pattern.
Figure. Chlorine oxidation number in the four oxoacids — the quantity the strength order tracks. Bar height is the oxidation number (+1 to +7), not Ka; Ka rises monotonically with that number in this series.
Ordering oxoacids
- Same central atom?Count O atoms / read the oxidation state — more is stronger.
- Stabilise the anionMore O → more delocalisation of the negative charge after H⁺ leaves.
- Name the extremeHClO₄ is the strongest chlorine oxoacid and one of the strongest common acids.
Ordering chlorine oxoacids
Arrange HClO, HClO₂, HClO₃ and HClO₄ in increasing acid strength, and state the oxidation number of Cl in each.
- Oxidation numbers of Cl+1, +3, +5, +7
- More O / higher OS → stronger acidstrength rises with OS
- Increasing strengthHClO < HClO₂ < HClO₃ < HClO₄
- Conjugate basesClO⁻ < ClO₂⁻ < ClO₃⁻ < ClO₄⁻ in stability
Pro tip. If two oxoacids share the central atom, count oxygen atoms before reaching for electronegativity arguments.
The strongest acid among the following is
- HClO
- HClO₂
- HClO₄
HClO₄ has Cl in the +7 state with four oxygens — maximum electron withdrawal and conjugate-base delocalisation in the series.
12Xenon fluorides: XeF₂, XeF₄ and XeF₆
Xenon forms fluorides because it is large and relatively ionisable for a noble gas. The three exam shapes are XeF₂ (linear, sp³d), XeF₄ (square planar, sp³d²) and XeF₆ (distorted octahedral, sp³d³).
Electron-pair geometry follows the steric number: count Xe’s eight valence electrons, use two per Xe–F bond, and place the leftover pairs to minimise repulsion. Even numbers of lone pairs in an octahedral set sit trans to each other — that is why XeF₄ is square planar rather than see-saw.
Figure. XeF₂ as a linear F–Xe–F unit. The three lone pairs occupy the equatorial belt of the trigonal bipyramid and are named rather than drawn — dots at 120° would claim a geometry the top view cannot show honestly beside the square-planar figure.
Steric number to shape
- Count valence electrons on XeXe has 8. Each F uses 1 of them in a bond (or count 2 per bond from Xe if you prefer the expanded picture — stay consistent).
- Steric numberBonds + lone pairs on Xe. XeF₂: 2 bonds + 3 LP → 5 (linear). XeF₄: 4 bonds + 2 LP → 6 (square planar). XeF₆: 6 bonds + 1 LP → 7 (distorted octahedral).
- Place lone pairsIn an octahedral set, two lone pairs go trans.
Shape of XeF₄
Determine the hybridisation and molecular shape of XeF₄.
- Xe valence electrons8
- 4 Xe–F bonds use 4e⁻; leftover4e⁻ = 2 lone pairs
- Steric number = 4 bonds + 2 LP6 → sp³d²
- Two LP trans in octahedral setsquare planar molecule
Pro tip. Even numbers of lone pairs in an octahedral set always take opposite positions to minimise repulsion — the same rule makes ICl₄⁻ square planar.
The shape of XeF₂ is
- Bent
- Linear
- Trigonal planar
Steric number 5 with three lone pairs in the equatorial belt of a trigonal bipyramid leaves the two F atoms axial — molecular shape linear.
13XeF₄ square planar (and why XeF₆ is not drawn)
XeF₄ has steric number 6: four bonds and two lone pairs. The electron-pair geometry is octahedral; the two lone pairs sit trans, so the four fluorines occupy a square — molecular shape square planar, hybridisation sp³d².
XeF₆ has steric number 7 and a distorted octahedral structure. That distortion is a 3D object; this vocabulary has no wedge/dash or true octahedron, so XeF₆ is carried in words and on the formula sheet rather than as a fake flat sketch.
Figure. XeF₄ in top view: four coplanar Xe–F bonds at 90°. The two lone pairs point above and below the page and are named rather than faked with wedge/dash. Aspect 1.0 keeps the square's angles honest.
From XeF₄ to XeF₆
- XeF₄ recallsp³d², square planar, LP trans.
- XeF₆ count6 bonds + 1 LP → steric number 7 → distorted octahedral.
- What not to drawA flat hexagon of bonds is not XeF₆. Decline the figure rather than invent one.
In XeF₄ the two lone pairs are
- Adjacent (90°), giving a see-saw shape
- Opposite (180°), giving square planar
- Absent — XeF₄ has no lone pairs
In an octahedral electron-pair set, two lone pairs minimise repulsion by sitting trans. The four F atoms then form a square.
Notes
- Group trends: descending a group, the inert-pair effect stabilises the lower oxidation state (e.g. Pb^{2+}>Pb^{4+}, Tl^+>Tl^{3+}).
- Group 15: NH_3 is a good base and reducing agent; nitrogen shows oxidation states from −3 to +5, and HNO_3 is a strong oxidiser.
- Group 16: sulphur shows allotropy; SO_2 is reducing and bleaching, and oxygen behaves anomalously due to its small size and absence of d-orbitals.
- Group 17 (halogens): the most reactive non-metals; oxidising power is F_2>Cl_2>Br_2>I_2, and they form interhalogens and oxoacids such as HOCl and HClO_4.
- Group 18 (noble gases): xenon forms fluorides — XeF_2 (linear), XeF_4 (square planar), XeF_6 (distorted octahedral) — due to its large size and low ionisation energy.
Formulas
- Acidic strength of hydrogen halides: HF<HCl<HBr<HI
- Oxidising power of halogens: F_2>Cl_2>Br_2>I_2
- Bond angle: NH_3(107^\circ)>PH_3(93.5^\circ)>AsH_3(91.8^\circ)
- Oxoacid strength: HClO<HClO_2<HClO_3<HClO_4
- Xe fluorides: XeF_2 (sp^3d), XeF_4 (sp^3d^2), XeF_6 (sp^3d^3)
Exam traps & shortcuts
- The inert-pair effect means the stability of the higher oxidation state falls down groups 13-15.
- Thermal stability of group-15/16/17 hydrides falls down the group while their reducing power rises (BiH_3 is the strongest reducer in group 15).
- More oxygen atoms on the central atom (higher oxidation state) means a stronger oxoacid, so HClO_4 is the strongest chlorine oxoacid.
Reference tables
Each row is one exam-ready ordering. Anomalies are named in the last column.
| Trend | Order | Anomaly / note |
|---|---|---|
| HX acid strength | HF < HCl < HBr < HI | HF weakest acid, strongest H-bonds |
| X₂ oxidising power | F₂ > Cl₂ > Br₂ > I₂ | Tracks E° |
| Group-15 bond angle | NH₃ > PH₃ > AsH₃ | Approaches 90° down the group |
| Cl oxoacid strength | HClO < HClO₂ < HClO₃ < HClO₄ | Tracks Cl oxidation state |
| Hydride reducing power (15) | NH₃ < … < BiH₃ | E–H bond weakens down the group |
| P-oxoacid basicity | H₃PO₂ < H₃PO₃ < H₃PO₄ (as basicity 1,2,3) | Count P–OH, not total H |
Steric number is bonds + lone pairs on the central atom.
| Molecule | Steric no. | Hybridisation | Shape |
|---|---|---|---|
| XeF₂ | 5 | sp³d | Linear |
| XeF₄ | 6 | sp³d² | Square planar |
| XeF₆ | 7 | sp³d³ | Distorted octahedral (not drawn) |
| ClF₃ | 5 | sp³d | T-shaped |
Reconstructible from the concepts above.
| Relation | Reads as | Watch for |
|---|---|---|
| Inert-pair effect | Lower OS stabilised down groups 13–15 | Higher OS becomes oxidising |
| First-element anomaly | B, C, N, O, F differ from congeners | No d-orbitals; strong multiple bonds |
| HF < HCl < HBr < HI | Hydrohalic acid strength | Not the H-bond order |
| F₂ > Cl₂ > Br₂ > I₂ | Oxidising power | E° order |
| HClO < … < HClO₄ | Oxoacid strength | Count oxygen / OS |
| Basicity of H₃POₓ | n(P–OH) | P–H does not ionise |
| Contact process | SO₂ → SO₃ (V₂O₅) → oleum → H₂SO₄ | Do not absorb SO₃ in water |
Recap
Read only this the night before.
- Inert pair
- Down groups 13–15 the lower OS wins: Tl⁺, Pb²⁺, Bi³⁺. The higher OS becomes the oxidiser.
- Group 13
- BF₃ Lewis acid (incomplete octet). B₂H₆ has 3c–2e bridges. Al₂Cl₆ dimer; borax for bead tests.
- Group 14
- Catenation C ≫ Si. Diamond sp³ hard/insulator; graphite sp² soft/conductor. Silicates = SiO₄ networks.
- Nitrogen span
- N from −3 (NH₃) to +5 (HNO₃). No NF₅ (no d-orbitals). Hydride reducing power rises to BiH₃.
- P oxoacids
- Basicity = P–OH count: H₃PO₂ 1, H₃PO₃ 2, H₃PO₄ 3.
- O₃ / H₂SO₄
- Ozone oxidises/bleaches. Contact process: V₂O₅, oleum intermediate. Conc. H₂SO₄ dehydrates; hot, oxidises.
- Halogens
- Oxidising power F₂ > Cl₂ > Br₂ > I₂. HX acidity the other way: HF weakest, HI strongest.
- Interhalogens
- ClF₃ T-shaped; BrF₅ square pyramidal; IF₇ pentagonal bipyramidal. Central atom is the larger halogen.
- Oxoacids / Xe
- More O on central atom → stronger acid (HClO₄). XeF₂ linear, XeF₄ square planar, XeF₆ distorted octahedral.
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