p-Block Elements (Groups 13 to 18)
Inorganic Chemistry
Weightage: 3-4 Questions (12-16 Marks)
JEE Unit 10
“Welcome back! You already know p-block descriptive chemistry from NCERT — the elements, their configurations, and the standard reactions. This guide clubs Groups 13 to 18 into ONE unit exactly the way JEE Main syllabus frames it: general trends across periods and groups, plus the unique behaviour of the first element in each group (B, C, N, O, F). We keep only the compound facts that JEE Main has actually tested — the borax bead test, BCl3/BF3 Lewis acid chemistry, the Sn/Pb inert-pair oxide flip, Nessler's reagent, why sulfur exists as S8 while oxygen stays O2, hydride/hydracid trends, and xenon compound structures — and skip the JEE-Advanced-level depth (full interhalogen hydrolysis sets, silicate classification, phosphorus oxoacid bond-counting) that Main does not probe.”
— SCORECHEM ACADEMIC TEAM
1. General p-Block Trends and First-Row Anomalies
- Location and configuration: Groups 13–18, valence configuration ns2np1–6. Only block containing metals, metalloids and non-metals together. Maximum oxidation state = (Group number − 10): Group 13 → +3, Group 14 → +4, Group 15 → +5, Group 16 → +6, Group 17 → +7.
- Inert pair effect: down Groups 13–15, the ns2 pair increasingly resists bonding (poor shielding by intervening d/f electrons), so the (Group − 2) oxidation state becomes progressively MORE stable in heavier members: Tl+ > In+ > Ga+; Pb2+ > Sn2+ > Ge2+; Bi3+ > Sb3+ > As3+.
- Why the first element of each group (B, C, N, O, F) is anomalous: (1) exceptionally small atomic size, (2) high electronegativity, (3) high ionization enthalpy, (4) NO vacant d-orbitals in the valence shell, so maximum covalency is capped at 4 (N never shows covalency 6; nitrogen belongs to Period 2 and simply cannot expand its octet). Consequences you will see repeatedly below: N forms no pentahalide (NCl5 doesn't exist, NF3 is the only stable trihalide), CCl4 resists hydrolysis while SiCl4 hydrolyses readily, and O/N/C form strong pπ–pπ multiple bonds WITH THEMSELVES (O=O, N≡N, C=C) that heavier congeners cannot — S8, not S=S, is sulfur's stable form because 3pπ–3pπ self-bonding is too weak. Sulfur CAN still form pπ–pπ bonds with a small, electronegative partner like C or N (C=S, or the resonance-stabilised thiocyanate S=C=N− ↔ S⊖–C≡N) — it is only S–S self-bonding that fails, not S bonding to small atoms in general.
- Diagonal relationship carries over from s-block logic: B–Si is the Group-13/14-relevant diagonal pair (both metalloid-like, both form covalent halides and complex anions with F−: [BF4]−, [SiF6]2−).
⚠️ JEE Trap: "Maximum oxidation state" and "most stable oxidation state" are different questions. The Group-number-minus-10 rule gives the HIGHEST possible state (e.g. +5 for N, +3 for Bi); the inert pair effect then tells you which state is actually most STABLE for a given heavy element (Bi3+, not Bi5+).
2. Group 13: The Boron Family
Fig. 1: Group 13 does NOT vary smoothly — Ga and Tl break the expected trend at every property.
- Elements: B (non-metal), Al, Ga, In, Tl (metals) — ns2np1. B has an exceptionally high melting point (2453 K, rigid B12 lattice); Ga has an unusually LOW melting point (303 K, liquid in summer) but a very high boiling point, so it is used in high-temperature thermometers.
- The four Group 13 anomalies (all driven by poor 3d shielding in Ga and poor 4f shielding in Tl): atomic radius B < Ga < Al < In < Tl; IE1 order B > Tl > Ga > Al > In; electronegativity B > Tl > In > Ga > Al; stability of +1 state rises Al+ < Ga+ < In+ < Tl+ (Tl+ is the everyday Tl species; Tl3+ is a strong oxidizer).
- Oxidation states and oxides: B, Al show only +3. Ga, In, Tl show both +1 and +3. Oxide basicity rises down the group: B2O3 acidic, Al2O3/Ga2O3 amphoteric, In2O3/Tl2O3 basic.
- Lewis acidity of BX3: electron-deficient (only 6 valence electrons around B), so BX3 compounds act as Lewis acids. Strength order BI3 > BBr3 > BCl3 > BF3, opposite to what halogen electronegativity alone predicts, because back-bonding (halogen lone pair → boron's empty 2p orbital) is strongest for the small, close-fitting F, which most effectively quenches B's electron deficiency. BF3 forms a stable adduct with NH3 (BF3·NH3) and is trigonal planar (sp2).
- BCl3 hydrolysis: BCl3 is covalent and hydrolyses completely in neutral water to B(OH)3 + 3HCl (the tetrahedral [B(OH)4]− ion forms only in BASIC medium, and boron never forms an aquo-cation like [B(H2O)6]3+ since it is not a transition metal).
- Diborane B2H6: electron-deficient, held together by two 3-centre-2-electron ("banana") B–H–B bridge bonds plus four normal 2-centre-2-electron terminal B–H bonds.
Fig. 2: Left: Lewis acid strength BI3>BBr3>BCl3>BF3. Right: diborane's electron-deficient bridge bonding.
- Borax and the borax bead test: borax is Na2B4O7·10H2O. On heating it loses water and melts into a glassy bead of sodium metaborate + boric anhydride: Na2B4O7 → 2NaBO2 + B2O3. B2O3 then combines with transition-metal-oxide impurities to give characteristic coloured beads: Cu → blue Cu(BO2)2 in the non-luminous (oxidizing) flame, which is reduced by carbon in the luminous flame to colourless/red Cu(I) metaborate CuBO2 (Cu goes +2 → +1); Co gives a dark blue bead, Cr gives a green bead.
- Alums: general formula M2SO4·M′2(SO4)3·24H2O. M+ = Na+, K+, NH4+ (Li+ is too small and cannot form an alum); M′3+ = Al3+, Fe3+, Cr3+.
⚠️ JEE Trap: Trichlorides and triiodides of ALL Group 13 elements are covalent (BCl3, AlCl3, GaCl3... and BI3, AlI3, GaI3...) — even Al's, despite Al being a metal, because Al3+'s high charge density polarizes the halide heavily (Fajans' rule). Only ionic radii of the M3+ cations genuinely increase down the group.
3. Group 14: The Carbon Family
Fig. 3: Down Group 14 the LOWER (+2) state grows more stable; Pb2+ beats Pb4+, reversing C/Si's preference.
- Elements: C, Si, Ge, Sn, Pb — ns2np2. Covalent radius rises sharply C→Si, then only slightly further down (poor d/f shielding).
- +4 vs +2 stability (inert pair effect, the single most-tested Group 14 fact): for C and Si only +4 is stable (no stable +2 state). Down the group +4 stability falls and +2 rises: Ge4+ > Ge2+, Sn4+ > Sn2+, but the trend flips at Pb: Pb2+ > Pb4+. Consequently PbO2 (Pb4+) is a strong oxidizing agent (readily falls back to Pb2+) while SnCl2 (Sn2+) is a strong reducing agent (readily rises to Sn4+): 2HgCl2 + SnCl2 → Hg2Cl2↓ (white) + SnCl4, and excess SnCl2 reduces this further to black Hg.
- ΔrG° sign rule for these reactions: PbO2 + Pb → 2PbO is spontaneous (ΔrG° < 0) because it converts the unstable Pb4+/Pb0 pair into the stable Pb2+ product. SnO2 + Sn → 2SnO is non-spontaneous (ΔrG° > 0) because it converts stable Sn4+ into the less-stable Sn2+.
- Amphoteric oxide pairs: among [SiO2, CO2] (both acidic), [GeO, GeO2] (both amphoteric), [SnO, SnO2] (both amphoteric) and [PbO, PbO2] (PbO amphoteric, but PbO2 is oxidizing, not simply amphoteric) — exactly 2 pairs have BOTH oxides amphoteric.
- Hydrolysis of tetrahalides: CCl4 is completely inert to water (carbon has no d-orbitals to expand its octet and accept the incoming OH−/H2O lone pair); SiCl4, GeCl4, SnCl4 all hydrolyse readily using empty d-orbitals.
- Non-existence of PbI4: Pb4+ would oxidize the strongly reducing I− before a stable tetraiodide could form, and the Pb–I bond energy released is insufficient to promote a 6s2 electron into 6p; so PbI4 does not exist, even though PbCl4 and PbBr4 do.
- Catenation: C >> Si > Ge ≈ Sn; Pb shows essentially none. Governed by C–C bond energy (348 kJ/mol), the strongest single bond among Group 14 self-links.
⚠️ JEE Trap: The inert-pair flip is between Sn and Pb, not Ge and Sn. Both Ge and Sn still prefer +4 over +2; only Pb flips to preferring +2. Don't assume the flip starts as soon as the inert pair effect is "switched on."
4. Group 15: The Nitrogen Family
- Elements: N, P, As, Sb, Bi — ns2np3 (extra-stable half-filled configuration → unusually high IE1 for the group, especially N at 1402 kJ/mol).
- N is anomalous: absence of d-orbitals caps N's covalency at 4 ([NH4]+), so N forms no pentahalide (unlike P, As, Sb which form both EX3 and EX5). Among nitrogen trihalides only NF3 is stable; NCl3, NBr3, NI3 are unstable/explosive and hydrolyse readily.
- +5 vs +3 stability down the group: common oxidation states of Group 15 are −3, +3 and +5, but (inert pair effect again) the stability of the +5 state FALLS down the group while +3 rises — Bi5+ compounds (BiF5 only) are strong oxidizers and Bi3+ is Bismuth's stable everyday state, mirroring Tl/Pb in Groups 13/14.
- Oxide acidity: acidity of E2O3 falls down the group as the element becomes more metallic — N2O3, P2O3 acidic; As2O3, Sb2O3 amphoteric; Bi2O3 basic. Electronegativity likewise falls N > P > As > Sb, so among N/P/As/Sb, N is most electronegative (its oxide most acidic) and Sb is least electronegative (its oxide amphoteric).
Fig. 4: The first hydride (NH3, H2O) always breaks the smooth boiling-point trend via hydrogen bonding.
- EH3 hydride trends (N→Bi): thermal stability falls down the group (E–H bond weakens); reducing character rises (BiH3 strongest reducer); basicity falls (lone-pair density is highest on the small N atom and gets diffused in larger orbitals lower down); bond angle falls sharply from NH3 (107.8°) to PH3 (93.6°) onward (heavier hydrides use nearly pure p-orbitals for bonding). Boiling point is the one exception that does NOT rise smoothly down the group: PH3 < AsH3 < NH3 < SbH3 < BiH3, because NH3 is lifted anomalously high by hydrogen bonding, but not high enough to beat SbH3/BiH3's much larger van der Waals mass.
- Ammonia (NH3): made industrially by the Haber process (N2 + 3H2 ⇌ 2NH3, ~200 atm, 700 K, Fe catalyst). Cannot be dried over conc. H2SO4, P4O10, or anhydrous CaCl2 (all react chemically with it) — only CaO (quicklime) is used. Forms deep-blue [Cu(NH3)4]2+ with Cu2+, and a soluble [Ag(NH3)2]+ complex that dissolves AgCl.
- Nessler's reagent test (a favourite PYQ identification tool): NH4+/NH3 + alkaline K2[HgI4] gives a characteristic brown precipitate of Millon's base (H2N·HgO·HgI). A "p-block element E forming a binary cation EH4+ that gives this brown precipitate, with IE1 = 1402 kJ/mol" identifies E = Nitrogen.
- Nitric acid and metals: conc. HNO3 passivates Fe, Cr and Al (protective oxide layer forms, no visible reaction) — the same passivation behaviour you saw for Al in Group 13.
⚠️ JEE Trap: Group 15 hydride boiling point is NOT monotonic. Don't assume "boiling point rises down the group" applies uniformly — only stability, reducing character (rising) and basicity (falling) are clean monotonic trends; boiling point has the NH3-anomaly kink.
5. Group 16: The Oxygen Family
- Elements: O, S, Se, Te, Po — ns2np4. IE1 of Group 16 is LOWER than the corresponding Group 15 element in the same period (O < N, S < P) despite higher nuclear charge, because Group 15's half-filled np3 is extra stable — the same logic you used for Period 2/3 ionization-energy anomalies in Unit 9.
- Electron gain enthalpy anomaly: S has a MORE negative EGA than O (S > Se > Te > O in magnitude), because O's very compact 2p subshell suffers strong inter-electronic repulsion when accepting an extra electron — the exact same reasoning as Cl > F in Group 17 below.
- H2E hydride trends (O→Te): acidic character rises down the group (H2O < H2S < H2Se < H2Te); thermal stability falls; reducing power rises (H2Te strongest). Boiling point exception: H2S < H2Se < H2Te < H2O — water's extensive hydrogen bonding lifts its boiling point (373 K) far above even the much heavier H2Te.
- Hexafluorides: only the hexaFLUORIDES (EF6) are stable across the group; SF6 is exceptionally inert because six tightly packed F atoms sterically shield the central S from attack.
⚠️ JEE Trap: Group 15 and Group 16 hydride boiling-point anomalies look similar but the culprit element differs. Group 15: NH3 is the anomaly (still below SbH3, BiH3). Group 16: H2O is the anomaly (and is the HIGHEST of its group, not just elevated) — O's hydrogen bonding is strong enough to beat even H2Te.
6. Group 17: The Halogen Family
Fig. 5: Left: four halogen property anomalies driven by F's small size. Right: more O atoms (higher O.S.) on the same halogen = stronger acid.
- Elements: F, Cl, Br, I — ns2np5, smallest radii in their respective periods (highest Zeff).
- Four classic Group 17 anomalies, all traced to F's small size: electron gain enthalpy magnitude Cl > F > Br > I (compact 2p subshell of F suffers strong electron-electron repulsion on accepting one more electron); bond dissociation energy Cl2 > Br2 > F2 > I2 (high lone-pair/lone-pair repulsion weakens the very short F–F bond); hydration energy F− > Cl− > Br− > I− (smallest ion, most charge density); standard reduction potential F2 > Cl2 > Br2 > I2 (F2 remains the strongest oxidizer overall — its weak F–F bond is more than compensated by F−'s huge hydration energy).
- Acid strength trends (HX): thermal stability HF > HCl > HBr > HI (falls with weakening bond); acidic/reducing strength HF < HCl < HBr < HI (rises as bond weakens, HI most easily ionized/oxidized). Boiling point: HCl < HBr < HI < HF (H-bonding lifts HF above even the heavier HI). Melting point: HCl < HBr < HF < HI — here HI (largest, most polarizable, strongest lattice packing) edges out HF in the solid state, even though HF wins on boiling point. Both orders were directly tested together in a 2026 PYQ as a two-statement question.
- Cannot prepare HBr/HI using conc. H2SO4: it oxidizes the strongly reducing HBr/HI to Br2/I2; use non-oxidizing conc. H3PO4 instead.
- Oxoacid acidity (same halogen, rising oxidation state): HClO < HClO2 < HClO3 < HClO4 — more O atoms on the same central halogen stabilize the conjugate base further by resonance, so acidity rises with oxidation state.
Fig. 6: Count Xe's valence electrons + bonded F/O to get hybridisation, then remove lone pairs for the molecular shape.
⚠️ JEE Trap: F2 is the strongest oxidizer despite having a WEAK F-F bond and a LOW-magnitude EGA compared to Cl. Oxidizing power (SRP) depends on the whole thermodynamic cycle (bond energy + electron affinity + hydration energy), not on any single step; F's huge hydration energy dominates.
7. Group 18: Noble Gases
- Configuration: ns2np6 (He: 1s2), closed shell → chemical inertness, very high IE1, near-zero/positive electron gain enthalpy.
- Bartlett's discovery (1962): IE1 of Xe (1170 kJ/mol) is close to that of O2 (1175 kJ/mol); since O2+[PtF6]− was already known, Bartlett successfully made Xe+[PtF6]−, the first noble-gas compound. Only Xe (and to a lesser extent Kr) form a significant range of compounds, exclusively with the most electronegative elements F and O.
- Xenon fluorides — structures (JEE Main tests this table directly, not the reaction mechanisms):
| Compound | Hybridisation | Lone pairs on Xe | Molecular shape |
|---|---|---|---|
| XeF2 | sp3d | 3 | Linear |
| XeF4 | sp3d2 | 2 | Square planar |
| XeF6 | sp3d3 | 1 | Distorted octahedral |
| XeO3 | sp3 | 1 | Pyramidal |
| XeOF4 | sp3d2 | 1 | Square pyramidal |
- How they're made (formula-level only): XeF2 from Xe + F2 (2:1); XeF4 from Xe + F2 (1:5); XeF6 from Xe + F2 (1:20, higher pressure). Complete hydrolysis of XeF6 gives pyramidal XeO3; partial hydrolysis (1 H2O) gives square-pyramidal XeOF4.
- Clathrates: noble-gas atoms trapped inside ice/quinol crystal cages (~6H2O : 1 gas atom). He and Ne are too small and escape the cavities, so they do NOT form clathrates.
⚠️ JEE Trap: Hybridisation is fixed by TOTAL electron pairs (bond + lone) around Xe, not by the final visible shape. XeF4's octahedral electron geometry (sp3d2, 2 lone pairs) collapses to a square PLANAR molecular shape once the two lone pairs are placed axially — always count electron pairs first, name the shape second.
8. How JEE Frames p-Block Questions
- Two-statement / assertion-reason MCQs: the dominant p-block format — one statement states a trend, the other an exception or a compound fact (borax bead test, BCl3 hydrolysis, hydride boiling points). Verify each statement independently before combining.
- Identify-the-element questions: give a reaction/test (Nessler's reagent, a characteristic precipitate) plus a numeric IE1/radius value, and ask you to name the element — always cross-check the qualitative test AND the number.
- Amphoteric-oxide-pair counting: list several [MO, MO2]-type oxide pairs across Group 14 and ask how many pairs are both amphoteric — work each pair individually (SiO2/CO2 are NOT amphoteric, they're acidic).
- Reaction-sequence identification (Pb chemistry): a short chain of reactions (PbCl2 + K2CrO4 → A, A + NaOH → B, etc.) testing whether you know Pb2+'s common salts (PbCrO4, [Pb(OH)4]2−) rather than any exotic chemistry.
- ΔrG° sign questions: given two comproportionation-type reactions (e.g. PbO2 + Pb → 2PbO vs SnO2 + Sn → 2SnO), decide spontaneity purely from which oxidation state is more stable for that element — no thermodynamic data needed.
⚠️ JEE Trap: p-Block questions reward memorised EXCEPTIONS, not smooth group trends. Every high-frequency JEE Main p-block question (Ga/Al radius, Pb/Sn stability flip, BF3 weakest Lewis acid, HX boiling vs melting point) is built around the one element or property that breaks the "obvious" trend.
9. Quick Sheet and Checklist
| Idea | Rule |
|---|---|
| Max oxidation state | Group number − 10 |
| Inert pair stability | Tl+ > In+ > Ga+; Pb2+ > Sn2+ > Ge2+; Bi3+ > Sb3+ > As3+ |
| BX3 Lewis acidity | BI3 > BBr3 > BCl3 > BF3 (back-bonding weakest acidity for BF3) |
| Group 14 +4/+2 flip | Ge, Sn prefer +4; Pb alone prefers +2 |
| Group 15/16 hydride BP anomaly | NH3 mid-pack high; H2O highest of its whole group |
| Group 17 EGA / BDE / SRP | Cl>F>Br>I (EGA); Cl2>Br2>F2>I2 (BDE); F2>Cl2>Br2>I2 (SRP, still) |
| Oxoacid acidity (same halogen) | Rises with O.S.: HClO < HClO2 < HClO3 < HClO4 |
| XeF2 / XeF4 / XeF6 | sp3d/Linear; sp3d2/Square planar; sp3d3/Distorted octahedral |
Before the exam, check you can:
- State the Group-number-minus-10 rule and separate "maximum" from "most stable" oxidation state using the inert pair effect.
- Explain why BF3 is the weakest Lewis acid of the BX3 series, and describe the borax bead test's colour and oxidation-state changes for Cu.
- Predict which Group 14 oxide pairs are amphoteric and identify the Sn/Pb +4-vs-+2 stability flip.
- Recall the Group 15 and Group 16 hydride property trends, including which one (boiling point) is NOT monotonic and why.
- Rank halogen EGA, BDE, hydration energy and SRP, and explain why F2 stays the strongest oxidizer despite a weak F-F bond.
- Read off hybridisation, lone pairs and shape for XeF2, XeF4, XeF6, XeO3 and XeOF4 from the structures table.