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p-Block Elements – JEE Main Formula Sheet & Class 12 Notes | Groups 13 to 18
This is the complete JEE Main Formula Sheet and Class 12 Formula Sheet for p-Block Elements — Chapter 20 from the Aakash Rapid Revision & Formula Bank. The p-block spans Groups 13 to 18 and contains the most chemically diverse set of elements in the periodic table — metals, non-metals, metalloids, and noble gases all coexist within these six groups. This chapter covers the Group 13 (Boron family), Group 14 (Carbon family), Group 15 (Nitrogen family), Group 16 (Oxygen family), Group 17 (Halogens), and Group 18 (Noble gases) — including allotropes, oxoacids, interhalogen compounds, anomalous properties of first elements, and the preparation and properties of key industrial compounds. p-Block chemistry contributes 6–10 questions in JEE Main every year — making it the single largest source of inorganic chemistry marks in the paper. Download the Free PDF below for all formulas, trends, and reaction details in one structured revision reference.
Topics Covered in This Formula Sheet
JEE Main Formula Sheet PDF Preview
Scroll to explore all p-Block Elements formulas — JEE Main & Class 12 Formula Sheet
Introduction: Why p-Block Is the Most Marks-Dense Chapter in JEE Main Inorganic Chemistry
Among all the chapters in Class 12 inorganic chemistry, the p-block is the broadest, the most detail-rich, and — for JEE Main — the highest-yield. Six groups, dozens of elements, hundreds of compounds, and an enormous range of reaction types are all housed under this single chapter heading. For a student who invests systematic preparation time here, the returns are disproportionately high: p-block questions in JEE Main appear in almost every session, covering allotropes, oxoacids, industrial processes, anomalous properties, and structural chemistry.
The chapter is best approached group by group rather than as a monolithic whole. Each group has its own internal logic — the anomalous behaviour of the first element (due to small size, inability to expand octet, and high electronegativity), the trend in oxidising/reducing power, the series of oxoacids and their relative acidic strength, and the industrial importance of key compounds. Once this group-by-group structure is in place, the connections across groups become apparent — the same charge-density logic that explains boron's anomaly explains nitrogen's anomaly; the same oxoacid acidic strength rule applies to sulphur, phosphorus, and chlorine oxoacids.
Download the Free PDF to access all p-block group summaries, reaction equations, and structural diagrams in one exam-ready reference.
Key Concepts Covered in This Chapter
Group 13 — Boron Family: Anomalous Boron, Borax, Diborane, and Aluminium
Why Group 13 Anomalies and Boron Compounds Are JEE Main Favourites
Group 13 elements (B, Al, Ga, In, Tl) have the outer configuration ns²np¹. They typically form +3 ions or covalent compounds with three bonds and one empty p-orbital — making them electron-deficient Lewis acids. Boron is anomalous compared to Al, Ga, In, Tl because of its very small size, high ionisation enthalpy, high electronegativity, and absence of d-orbitals. Boron forms only covalent compounds (B³⁺ ion does not exist in chemistry). BF₃ is a classic Lewis acid — it accepts an electron pair from donors (BF₃ + :NH₃ → F₃B←NH₃). Boron forms cluster (cage-like) structures and exhibits multicentre bonding (3c–2e bonds) unlike aluminium.
Borax (Na₂B₄O₇·10H₂O): The borax bead test is a classic qualitative test — borax fused with a metal salt on a platinum wire loop gives a characteristic coloured bead. Borax dissolves in water to give an alkaline solution: Na₂B₄O₇ + 7H₂O → 2NaOH + 4H₃BO₃ (boric acid). The tetraborate ion (B₄O₇²⁻) contains both sp² (trigonal planar, 3-coordinate) and sp³ (tetrahedral, 4-coordinate) boron atoms — a direct JEE Main structural question. Borax reacts with acids: Na₂B₄O₇ + H₂SO₄ + 5H₂O → Na₂SO₄ + 4H₃BO₃.
Diborane (B₂H₆): Prepared by: 2BF₃ + 6NaH → B₂H₆ + 6NaF; or 2BCl₃ + 6H₂ → B₂H₆ + 6HCl (with high pressure). Structure: contains two 3-centre 2-electron (banana) bonds — the two bridging H atoms are shared between the two B atoms (B–H–B bridges). Four terminal H atoms form normal 2c–2e bonds. The molecule is not electron-sufficient in the traditional sense — the bridging bonds are called banana bonds or 3c–2e bonds. B₂H₆ reacts with water: B₂H₆ + 6H₂O → 2B(OH)₃ + 6H₂; reacts with NH₃ at low temperature to give B₂H₆·2NH₃ (diborane diammoniate), on heating gives borazine (B₃N₃H₆, "inorganic benzene"). Orthoboric acid (H₃BO₃): a weak monobasic acid but acts as a Lewis acid (accepts OH⁻, not donates H⁺): H₃BO₃ + H₂O → [B(OH)₄]⁻ + H⁺. On heating gives metaboric acid (HBO₂) then tetraboric acid (H₂B₄O₇) then B₂O₃. Download the Free PDF for the diborane structure diagram and borax reactions.
Group 14 — Carbon Family: Allotropes, CO/CO₂, Carbides, and Silicates
Why Carbon Chemistry Is a Multi-Question JEE Main Section
Group 14 elements (C, Si, Ge, Sn, Pb) have configuration ns²np². Carbon's anomalous properties arise from its small size, high electronegativity, and absence of d-orbitals — it can form stable pπ–pπ multiple bonds, long chains (catenation), and has a maximum covalency of four. Allotropes of carbon: Diamond (sp³, tetrahedral, hardest natural substance, non-conductor — no delocalised electrons), Graphite (sp², planar layers, conductor — delocalised π electrons, used as lubricant and electrode), Fullerene C₆₀ (spherical cage, 20 hexagons + 12 pentagons, all sp² carbons — like a football). Coke, charcoal, and carbon black are amorphous forms.
Carbon monoxide (CO): prepared industrially from incomplete combustion of carbon (2C + O₂ → 2CO) or from water gas (C + H₂O → CO + H₂ at 1270 K). CO is colourless, odourless, toxic gas — combines irreversibly with haemoglobin to form carboxyhaemoglobin (affinity 200× higher than O₂). CO is a strong reducing agent and a ligand (forms metal carbonyls like Ni(CO)₄). CO burns in air: 2CO + O₂ → 2CO₂. Carbon dioxide (CO₂): linear molecule (O=C=O), sp hybridised carbon. Acidic oxide — dissolves in water to give H₂CO₃ (weak dibasic acid). Sublimes at 195 K (dry ice). Used in fire extinguishers, carbonated drinks, and supercritical CO₂ extraction. Greenhouse gas — absorbs IR radiation.
Carbides: ionic carbides — contain C²²⁻ (acetylide, gives C₂H₂ with water: CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂) or C⁴⁻ (methanide, gives CH₄: Al₄C₃ + 12H₂O → 4Al(OH)₃ + 3CH₄). Covalent carbides: SiC (carborundum, very hard), B₄C (boron carbide, extremely hard abrasive). Metallic/interstitial carbides: Fe₃C (cementite in steel), WC (tungsten carbide, cutting tools). Silicates: SiO₄⁴⁻ is the fundamental unit (tetrahedral). Orthosilicate (SiO₄⁴⁻), pyrosilicate (Si₂O₇⁶⁻), cyclic silicate (Si₃O₉⁶⁻), chain silicate (SiO₃²⁻), double chain, sheet silicate (Si₄O₁₀⁴⁻), 3D framework (SiO₂, quartz). Silicones are synthetic polymers of R₂SiO units — water repellent, thermally stable, biocompatible — used in sealants, lubricants, and medical implants. Download the Free PDF for allotrope comparison and silicate structure diagrams.
Group 15 — Nitrogen Family: NH₃, Oxides, HNO₃, and Phosphorus Oxoacids
Why Group 15 Chemistry Carries the Highest Marks in p-Block for JEE Main
Group 15 elements (N, P, As, Sb, Bi) have configuration ns²np³ — a half-filled p subshell gives extra stability. Nitrogen is anomalous: small size, high electronegativity, no d-orbitals (maximum covalency 4 as in NH₄⁺), ability to form stable pπ–pπ multiple bonds (N≡N triple bond in N₂ is very strong, 945 kJ/mol). Phosphorus (and heavier members) can expand their octet using d-orbitals, forming PCl₅, H₃PO₄, etc. — which nitrogen cannot do (NCl₅ does not exist).
Ammonia (NH₃): pyramidal structure, sp³ N (one lone pair). Lab preparation: Ca(OH)₂ + 2NH₄Cl → CaCl₂ + 2NH₃ + 2H₂O. Industrial: Haber process: N₂ + 3H₂ ⇌ 2NH₃ (450–500°C, 200–300 atm, Fe catalyst with Al₂O₃/K₂O as promoters). NH₃ is basic (lone pair on N), forms hydrogen bonds (high b.p. = –33°C for its molecular mass), is a good ligand (forms [Cu(NH₃)₄]²⁺ deep blue — used as a test for Cu²⁺). NH₃ burns in air with Pt catalyst: 4NH₃ + 5O₂ → 4NO + 6H₂O (Ostwald process, first step). Reacts with Lewis acids: NH₃ + BF₃ → F₃B←NH₃. NH₃ + excess Cl₂ → NCl₃ + 3HCl; Cl₂ + excess NH₃ → N₂ + 6HCl.
Oxides of nitrogen — all five must be known for JEE Main: N₂O (laughing gas, dinitrogen oxide — linear, used as anaesthetic), NO (colourless, paramagnetic, brown fumes in air as it oxidises to NO₂ — 2NO + O₂ → 2NO₂), N₂O₃ (nitrogen sesquioxide, anhydride of HNO₂), NO₂ (brown gas, angular, paramagnetic, dimerises to N₂O₄ — 2NO₂ ⇌ N₂O₄), N₂O₅ (anhydride of HNO₃, white solid).
Nitric acid (HNO₃) — Ostwald process: Step 1: 4NH₃ + 5O₂ → 4NO + 6H₂O (Pt–Rh catalyst, 900°C). Step 2: 2NO + O₂ → 2NO₂. Step 3: 3NO₂ + H₂O → 2HNO₃ + NO (NO is recycled). HNO₃ is a strong oxidising agent: Cu + 4HNO₃(conc.) → Cu(NO₃)₂ + 2NO₂ + 2H₂O (brown fumes with conc.); 3Cu + 8HNO₃(dil.) → 3Cu(NO₃)₂ + 2NO + 4H₂O (colourless with dil.). P + 5HNO₃(conc.) → H₃PO₄ + 5NO₂ + H₂O. S + 2HNO₃(conc.) → H₂SO₄ + 2NO. Aqua regia = 3 parts conc. HCl + 1 part conc. HNO₃ — dissolves noble metals (Au, Pt).
Allotropes of phosphorus: White P (P₄ tetrahedral units, waxy solid, stored under water, very reactive, highly toxic, glows in dark — phosphorescence), Red P (polymeric P₄ units linked in chains, less reactive, non-toxic, used in safety matches), Black P (thermodynamically most stable, graphite-like layers, semiconductor). White P on prolonged heating at 250°C gives red P; at very high pressure gives black P. Oxoacids of phosphorus — key for JEE Main: H₃PO₂ (hypophosphorous acid, monobasic — only 1 ionisable OH group, two P–H bonds — a strong reducing agent), H₃PO₃ (phosphorous acid, dibasic — 2 ionisable OH groups, one P–H bond — reducing agent), H₃PO₄ (orthophosphoric acid, tribasic — 3 ionisable OH groups, no P–H bonds — not a reducing agent). The rule: P–H bonds in phosphorus oxoacids are non-ionisable and make the acid a reducing agent. Only P–OH groups are ionisable. Download the Free PDF for all nitrogen and phosphorus oxoacid structures.
Group 16 — Oxygen Family: Allotropes, H₂SO₄ Contact Process, and Sulphur Oxoacids
Why Group 16 Industrial Processes and Oxoacids Are High-Priority JEE Main Topics
Group 16 elements (O, S, Se, Te, Po) have configuration ns²np⁴ — they need 2 electrons to complete their octet and typically show –2 oxidation state in compounds, though heavier members show +2, +4, and +6 oxidation states as well. Oxygen is anomalous: very small size, highest electronegativity (after F), no d-orbitals (maximum covalency 2 as in H₂O, except in OF₂ where it shows +2), forms pπ–pπ multiple bonds (O=O in O₂, O=C=O in CO₂). Allotropes of oxygen: O₂ (dioxygen, paramagnetic — two unpaired electrons in degenerate π* orbitals, explained by MO theory) and O₃ (ozone, angular molecule, bond angle 117°, diamagnetic, sp² hybridised O, strong oxidising agent — turns moist KI paper blue by releasing [O]).
Allotropes of sulphur: Rhombic sulphur (α-S, stable below 369 K, yellow crystals, contains S₈ rings in crown form), Monoclinic sulphur (β-S, stable 369 K to 392 K, needle-shaped crystals, S₈ rings in boat form), Plastic sulphur (fibrous, formed by pouring boiling S into cold water — chains of S atoms), Sulphur vapour (S₂ at high T, paramagnetic). Transition temperature (rhombic ⇌ monoclinic) = 369 K (96°C). Sulphur dioxide (SO₂): Angular molecule (sp³ S, one lone pair), reducing agent (with KMnO₄, K₂Cr₂O₇) and oxidising agent (with H₂S: 2H₂S + SO₂ → 3S + 2H₂O). Bleaches by reduction (unlike Cl₂ which bleaches by oxidation) — SO₂ bleaching is temporary (colour returns on air oxidation). Acidic oxide: SO₂ + H₂O → H₂SO₃ (sulphurous acid).
Sulphuric acid (H₂SO₄) — Contact process: Step 1: S + O₂ → SO₂. Step 2: 2SO₂ + O₂ ⇌ 2SO₃ (V₂O₅ catalyst, 450–500°C, 1–2 atm — higher pressure thermodynamically favoured but too low for significant rate; temperature is a compromise). Step 3: SO₃ is absorbed in 98% H₂SO₄ (not water — would form acid mist): SO₃ + H₂SO₄ → H₂S₂O₇ (oleum/pyrosulphuric acid). Oleum is then diluted with calculated water: H₂S₂O₇ + H₂O → 2H₂SO₄. Properties of conc. H₂SO₄: strong dehydrating agent (C₁₂H₂₂O₁₁ → 12C + 11H₂O), strong oxidising agent (Cu + 2H₂SO₄(hot conc.) → CuSO₄ + SO₂ + 2H₂O; S + 2H₂SO₄(hot conc.) → 3SO₂ + 2H₂O), forms hydrates with water (highly exothermic — always add acid to water, not water to acid). Oxoacids of sulphur: H₂SO₃ (sulphurous acid, S=+4), H₂SO₄ (sulphuric acid, S=+6), H₂S₂O₇ (pyrosulphuric/oleum, S=+6), H₂S₂O₃ (thiosulphuric acid, S=+2 average), H₂S₂O₈ (peroxodisulphuric acid, S=+6 with O–O bridge), H₂SO₅ (peroxomonosulphuric/Caro's acid). Download the Free PDF for all sulphur oxoacid structures.
Group 17 — Halogens: Fluorine Anomaly, Interhalogen Compounds, and Oxoacids
Why Halogen Chemistry Is a Guaranteed Multi-Mark Section in JEE Main
Group 17 elements (F, Cl, Br, I, At) have configuration ns²np⁵ — one electron short of a noble gas configuration. They are the most electronegative and most reactive non-metals. All halogens are oxidising agents; oxidising power decreases F₂ > Cl₂ > Br₂ > I₂. Fluorine is anomalous: cannot show positive oxidation states (no d-orbitals, highest electronegativity — always shows –1 or 0), cannot act as central atom in compounds with oxygen (so no oxoacids of F), forms the strongest and shortest H–F bond among HX, and HF is a weak acid in dilute solution (due to strong H-bonding forming [HF₂]⁻ species) while HCl, HBr, HI are all strong acids. Thermal stability of HX: HF > HCl > HBr > HI (decreasing bond enthalpy). Acidic strength: HI > HBr > HCl > HF (HF is weakest due to strong H–F bond making it hard to ionise in dilute solution, though it has the highest electronegativity on F).
Halogen reactions: All halogens react with NaOH: Cl₂ + 2NaOH → NaCl + NaOCl + H₂O (cold dilute, gives bleaching powder equivalent — hypochlorite); 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O (hot conc., gives chlorate). F₂ + 2NaOH → 2NaF + ½O₂ + H₂O (F₂ too reactive, oxidises OH⁻). All halogens react with H₂: H₂ + X₂ → 2HX (reactivity F₂ > Cl₂ > Br₂ > I₂; the reaction of I₂ with H₂ is reversible and incomplete). Interhalogen compounds: formed between two different halogens. Types: XX' (ClF, BrF, BrCl, ICl, IBr), XX'₃ (ClF₃, BrF₃, IF₃, ICl₃), XX'₅ (BrF₅, IF₅, ClF₅), XX'₇ (IF₇). The heavier and larger halogen is always the central atom (it can accommodate more bonds using d-orbitals). Structures: XX'₃ → T-shaped (sp³d, 2 lone pairs); XX'₅ → square pyramidal (sp³d², 1 lone pair); XX'₇ → pentagonal bipyramidal. All interhalogen compounds are stronger oxidising agents than Cl₂ but weaker than F₂. They react with water to give HF and the oxyacid of the heavier halogen.
Oxoacids of halogens: Cl forms the most complete series: HClO (hypochlorous, Cl=+1), HClO₂ (chlorous, Cl=+3), HClO₃ (chloric, Cl=+5), HClO₄ (perchloric, Cl=+7). Acidic strength increases with increasing oxidation state of Cl: HClO < HClO₂ < HClO₃ < HClO₄ — because more electronegative oxygens pull electron density from O–H bond, making H⁺ easier to release. HClO₄ is the strongest known oxoacid. Fluorine does not form oxoacids (cannot show positive oxidation state). Bleaching powder (CaOCl₂) — prepared by passing Cl₂ over dry slaked lime: Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O. Active bleaching agent is HOCl (hypochlorous acid) formed when bleaching powder reacts with CO₂: CaOCl₂ + CO₂ → CaCO₃ + Cl₂ → HOCl → [O] (nascent oxygen bleaches). Download the Free PDF for all interhalogen compound structures and oxoacid tables.
Group 18 — Noble Gases: Properties and Xenon Compounds
Why Noble Gas Chemistry Including Xenon Fluorides Is a JEE Main One-Mark Section
Group 18 elements (He, Ne, Ar, Kr, Xe, Rn) have completely filled outer shells (ns²np⁶, except He = 1s²). They are monoatomic gases with very low boiling points (He has the lowest b.p. of any substance — 4.2 K). They were once called "inert gases" but this is now incorrect since several xenon compounds are known. Their ionisation enthalpies are the highest in their respective periods (filled shells, high Z*), and their electron gain enthalpies are positive (strongly resist adding electrons). Uses: He — filling balloons, cryogenics, deep-sea diving gas mixtures, MRI coolant. Ne — advertising signs (neon lights, glows red-orange). Ar — inert atmosphere for arc welding and metallurgical processes, filling incandescent light bulbs. Kr — filling fluorescent lamps. Xe — xenon lamps (high-intensity light source). Rn — radioactive, used in radiotherapy of cancer.
Xenon compounds: Xenon is the most chemically reactive noble gas due to its comparatively low ionisation enthalpy and available d-orbitals. Xenon fluorides: XeF₂ (linear, 3 lone pairs on Xe, sp³d, prepared by 1:1 mixture of Xe + F₂ at 400°C in sealed Ni tube at high pressure — or Xe + F₂ at 6 atm); XeF₄ (square planar, 2 lone pairs, sp³d², prepared at higher F₂ pressure, 20 atm); XeF₆ (distorted octahedral, 1 lone pair, sp³d³, prepared at 60–70 atm and 300°C). Hydrolysis: XeF₂ + H₂O → Xe + 2HF + ½O₂; 6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂; XeF₆ + H₂O → XeOF₄ + 2HF (partial), XeF₆ + 3H₂O → XeO₃ + 6HF (complete). Xenon oxides: XeO₃ (pyramidal, violent explosive oxidant), XeO₄ (tetrahedral, extremely powerful oxidiser, unstable). XeF₂, XeF₄, XeF₆ are all powerful fluorinating agents. Download the Free PDF for xenon compound preparation conditions and structures.
Download Free PDF — p-Block Elements Formula Sheet
All group-wise trend tables, anomalous property lists, oxoacid structures and acidic strength rankings, industrial process steps, allotrope descriptions, interhalogen compound structures, and xenon compound preparation conditions are compiled in the Aakash Rapid Revision & Formula Bank PDF — structured specifically for JEE Main, CBSE boards, and NEET.
Why This Chapter Is the Highest-Return Investment in JEE Main Chemistry
Four reasons make p-block chemistry the most marks-dense chapter for JEE Main preparation.
Volume of questions is unmatched. Six groups, six anomalous first elements, industrial processes for HNO₃, H₂SO₄, and NH₃, two entire series of oxoacids, six xenon compounds, a full table of interhalogen compounds — every piece of content here has appeared in JEE Main at some point. No other single chapter provides this many question opportunities.
Anomalous properties follow one unifying principle. Every first element in Groups 13 to 17 (B, C, N, O, F) is anomalous for the same set of reasons: small size, high electronegativity, no d-orbitals (maximum covalency limited to 4 or 2), and ability to form pπ–pπ multiple bonds. Students who understand this framework answer anomaly questions for any group without separate memorisation for each.
Oxoacid acidic strength questions are fast and formulaic. For any oxoacid, acidic strength increases with the number of terminal oxygen atoms (=O groups, not –OH groups). This one rule handles oxoacids of Cl (HClO < HClO₂ < HClO₃ < HClO₄), sulphur, nitrogen, and phosphorus simultaneously. For phosphorus specifically, basicity is counted by –OH groups only. These rules answer entire question types reliably.
Industrial processes are finite and highly structured. Haber process (NH₃), Ostwald process (HNO₃), Contact process (H₂SO₄) — each has specific reagents, catalysts, temperatures, pressures, and step-by-step reactions. Preparing these as structured summaries rather than narrative paragraphs makes revision efficient and exam recall fast. Download the Free PDF to access all three industrial processes in compact, exam-ready form.
Who Should Use This Formula Sheet?
Learning Outcomes After Completing This Chapter
After working through this chapter using the formula sheet and notes above, a student should be able to accomplish the following with confidence.
For Groups 13 and 14: explain six anomalous properties of boron compared to aluminium. Write the structure and bonding of diborane (3c–2e bonds). Write the borax–boric acid reaction. Compare diamond, graphite, and fullerene in terms of hybridisation, structure, and properties. Write reactions of CO as reducing agent and ligand. Classify carbides as ionic, covalent, or metallic with examples. Describe the basic silicate unit and name chain, sheet, and 3D framework silicates.
For Group 15: explain six anomalous properties of nitrogen. Write the Haber process conditions for ammonia. Name all five oxides of nitrogen with their oxidation states. Write the three steps of the Ostwald process for HNO₃. Write the reactions of Cu with dilute and concentrated HNO₃. Determine the basicity of H₃PO₂, H₃PO₃, and H₃PO₄ from their structures. Identify the reducing agent nature of phosphorus oxoacids.
For Group 16: compare rhombic and monoclinic sulphur. Write the three steps of the Contact process for H₂SO₄ with conditions. Explain why SO₃ is absorbed in oleum rather than water. Describe two reactions of SO₂ as both oxidant and reductant. Write the structures of XeF₂, XeF₄, and XeF₆ with shapes.
For Groups 17 and 18: explain six anomalous properties of fluorine. Name interhalogen compounds of types XX'₃, XX'₅, XX'₇ with examples and shapes. Rank HX acids in order of acidic strength and thermal stability with explanation. State the acidic strength order of chlorine oxoacids with reasoning. Write the preparation conditions for XeF₂, XeF₄, and XeF₆. Download the Free PDF to test all these outcomes before your exam.
Get the Free PDF for Quick Revision
Whether you are preparing for JEE Main, CBSE Class 12 boards, or NEET, having a focused formula sheet for p-block elements is essential — this is the chapter where marks are won and lost most decisively in inorganic chemistry. The Aakash Rapid Revision & Formula Bank PDF for p-Block Elements brings all six group summaries into one structured, exam-ready reference.
Conclusion — One Framework, Six Groups, Maximum Marks
The p-block chapter rewards students who see the pattern across groups rather than treating each group as a separate unit of memorisation. The anomalous behaviour of B, C, N, O, and F all stem from the same causes: small atomic size limits covalency to four or fewer (no d-orbital expansion), high electronegativity leads to unexpected bond character, and the ability to form pπ–pπ multiple bonds gives these elements unique structural chemistry not seen in their heavier congeners. Recognising this shared foundation reduces the memory load dramatically.
For industrial processes, the three major processes — Haber, Ostwald, Contact — each follow the same pattern: identify the equilibrium conditions that favour product formation, choose a catalyst to achieve acceptable rate, and handle the engineering constraints (SO₃ absorbed in oleum, not water; NH₃ oxidised at specific Pt–Rh catalyst conditions). Understanding the thermodynamic and kinetic logic makes these processes logically memorable rather than arbitrarily factual.
For oxoacids, the acidic strength rule based on terminal oxygen count handles all cases — and for phosphorus acids, the P–H bond rule for basicity and reducing power handles the most common question type for that sub-topic. Combine these frameworks with the Free PDF Download for rapid revision of specific compounds and reactions, and p-block chemistry becomes one of the most systematically conquerable sections of the entire JEE Main chemistry paper.
Frequently Asked Questions
Why can nitrogen not form compounds like PCl₅ or NCl₅?
Nitrogen cannot form compounds like NCl₅ because it does not have d-orbitals in its valence shell (Period 2, n=2 shell has only s and p orbitals). To form five bonds, an atom needs to promote electrons from its ground state configuration into additional available orbitals. Nitrogen's maximum possible valency is therefore 4 (as in NH₄⁺, where it uses its three p orbitals + one s orbital in the excited state model, or more accurately, forms four equivalent sp³ bonds). Phosphorus, in the Period 3 n=3 shell, has access to empty 3d orbitals and can expand its octet — PCl₅ is formed by promoting one 3s electron to a 3d orbital, allowing 5 bond formation. This octet expansion ability of Period 3 and beyond p-block elements is a direct consequence of d-orbital availability, and its absence in Period 2 explains the lower maximum covalency of N, O, and F compared to their heavier group members.
How do you determine the basicity of phosphorus oxoacids?
The basicity (number of ionisable protons, i.e., how many H⁺ a phosphorus oxoacid can donate) is determined only by counting the number of –OH groups directly bonded to the phosphorus atom. P–H bonds (where hydrogen is directly bonded to phosphorus without an oxygen bridge) are non-ionisable — these hydrogens cannot be released as H⁺. P=O bonds contribute to oxidising power but not to basicity. Using this rule: H₃PO₂ (hypophosphorous acid) has the structure H₂P(=O)(OH) — one P–OH group and two P–H groups — so it is monobasic. H₃PO₃ (phosphorous acid) has the structure HP(=O)(OH)₂ — two P–OH groups and one P–H group — so it is dibasic. H₃PO₄ (orthophosphoric acid) has the structure P(=O)(OH)₃ — three P–OH groups and no P–H groups — so it is tribasic. The P–H bonds also explain why H₃PO₂ and H₃PO₃ are reducing agents (the P–H bond is oxidised during reduction reactions), while H₃PO₄ is not a reducing agent.
Why is SO₃ absorbed in oleum rather than directly in water during the Contact process?
When SO₃ is absorbed directly into water, the reaction is extremely exothermic and produces a dense, corrosive mist of H₂SO₄ droplets (acid mist or acid fog): SO₃ + H₂O → H₂SO₄ + heat. This acid mist is very difficult to handle industrially — the droplets are too fine to settle or be easily collected, they contaminate the atmosphere, and they cause corrosion of equipment. To avoid this problem, SO₃ is instead absorbed into 98% concentrated H₂SO₄, forming oleum (pyrosulphuric acid, H₂S₂O₇): SO₃ + H₂SO₄ → H₂S₂O₇. This reaction occurs cleanly without mist formation. The oleum is then carefully diluted with a calculated amount of water to give the desired concentration of H₂SO₄: H₂S₂O₇ + H₂O → 2H₂SO₄. This two-step absorption-then-dilution approach eliminates the acid mist problem and is the standard industrial practice in the Contact process.
Why is the acidic strength of halogen hydracids in the order HI > HBr > HCl > HF?
The acidic strength of HX acids depends on how easily the H–X bond breaks to release H⁺ in aqueous solution. Two competing factors determine this: bond enthalpy (H–X bond dissociation energy) and hydration enthalpy of X⁻. Bond enthalpy is the dominant factor here. Going down Group 17 from F to I, the halogen atoms get larger and the H–X bond gets longer and weaker: H–F bond (567 kJ/mol) > H–Cl (432) > H–Br (366) > H–I (297 kJ/mol). The weaker the H–X bond, the easier it is to ionise, and the stronger the acid. HF is the weakest acid despite fluorine being the most electronegative — its strong H–F bond is the main reason for low ionisation. In dilute solution, HF also forms [HF₂]⁻ ions through hydrogen bonding, which reduces the concentration of free H⁺ further. Note that while this order holds in aqueous dilute solution, the intrinsic gas-phase acidity order is different.
What are the structures and shapes of XeF₂, XeF₄, and XeF₆?
XeF₂ has 2 bonding pairs + 3 lone pairs around the central Xe atom = 5 electron pairs total → sp³d hybridisation. The electron geometry is trigonal bipyramidal but the lone pairs occupy the equatorial positions, leaving the two F atoms in axial positions → the molecular shape is linear. XeF₄ has 4 bonding pairs + 2 lone pairs = 6 electron pairs → sp³d² hybridisation. The two lone pairs are placed opposite each other (trans) to minimise repulsion → the molecular shape is square planar. XeF₆ has 6 bonding pairs + 1 lone pair = 7 electron pairs → sp³d³ hybridisation. With one lone pair, the ideal octahedral geometry is distorted → the molecular shape is distorted octahedral (or pentagonal bipyramidal with one position occupied by a lone pair, depending on the isomeric form). XeO₃ has 3 bonding pairs + 1 lone pair → tetrahedral electron geometry → pyramidal molecular shape (like NH₃). XeO₄ has 4 bonding pairs + 0 lone pairs → regular tetrahedral shape.
Why is fluorine anomalous compared to the other halogens?
Fluorine differs from Cl, Br, and I in several important ways, all traceable to its unique combination of properties — smallest size, highest electronegativity, absence of d-orbitals in n=2 shell, and exceptionally strong H–F bond. Key anomalies: Fluorine cannot show positive oxidation states (no d-orbitals for octet expansion, and its extreme electronegativity means it always attracts electrons). Other halogens show +1, +3, +5, +7 states. Fluorine forms no oxoacids (cannot show positive OS to bond with O). HF is the weakest HX acid in dilute solution despite F being most electronegative — the very strong H–F bond (567 kJ/mol) resists ionisation. Fluorine forms the strongest element-element bond (F–F bond paradoxically weaker than Cl–Cl because of lone pair repulsion in the very compact F atom, but H–F is extremely strong). F₂ is the strongest oxidising agent and can react even with noble gases (XeF₂). F₂ reacts with glass (SiO₂), which no other halogen does. These anomalies make fluorine one of the most JEE Main-tested elements in Group 17.
What is the structure of diborane (B₂H₆) and what are 3-centre 2-electron bonds?
Diborane (B₂H₆) has an unusual "banana-shaped" or "bridged" structure. It cannot have the same structure as ethane (C₂H₆) because boron has only 3 valence electrons per atom — there are not enough electrons for 7 conventional 2-centre 2-electron bonds. The actual structure has four terminal B–H bonds (two on each B) which are normal 2-centre 2-electron bonds, plus two bridging H atoms that connect the two B atoms. Each bridging bond involves one H atom and two B atoms sharing just 2 electrons among three atoms — this is a 3-centre 2-electron (3c–2e) bond, also called a banana bond or half-bond. The geometry gives the molecule a butterfly or double-triangular shape with the two B atoms and the two bridging H atoms forming a planar diamond (rhombus) in the centre, and the four terminal H atoms projecting above and below. The two B atoms are sp³ hybridised. B₂H₆ hydrolyses readily: B₂H₆ + 6H₂O → 2B(OH)₃ + 6H₂ and reacts with NH₃ to give borazine at higher temperatures.
Why does the acidic strength of chlorine oxoacids increase in the order HClO < HClO₂ < HClO₃ < HClO₄?
The acidic strength of oxoacids increases with the number of terminal (non-hydroxyl) oxygen atoms attached to the central atom. Terminal oxygen atoms (=O) are highly electronegative and withdraw electron density from the O–H bond through inductive effect. The more terminal oxygens present, the more electron density is pulled away from the O–H bond, making it easier for H⁺ to dissociate. In HClO (Cl=+1, zero terminal O atoms), the inductive effect from terminal oxygen is absent — weakest acid. In HClO₂ (Cl=+3, one =O group), mild electron withdrawal — stronger. In HClO₃ (Cl=+5, two =O groups), more withdrawal — stronger still. In HClO₄ (Cl=+7, three =O groups), maximum electron withdrawal from three terminal oxygens — HClO₄ is the strongest known oxoacid. This same logic applies to oxoacids of sulphur (H₂SO₃ < H₂SO₄), nitrogen, and phosphorus — the number of terminal =O atoms, not the oxidation state alone, is the determining factor for acidic strength.
What are interhalogen compounds and why is the heavier halogen always the central atom?
Interhalogen compounds are binary compounds formed between two different halogen elements. They are represented as XX'ₙ where X is the larger (heavier) halogen, X' is the smaller (lighter) halogen, and n = 1, 3, 5, or 7. Examples: ClF, BrF₃, IF₅, IF₇. The heavier halogen is always the central atom because: it has a larger atomic size which allows it to accommodate more surrounding atoms without excessive steric strain; it has lower electronegativity, so it is the less electronegative partner and acts as the central positive-oxidation-state atom; and it has available d-orbitals (being in Period 3 or below) to expand its octet and accommodate 3, 5, or 7 bonds. The lighter halogen (F, Cl) acts as the terminal atom, always in the –1 oxidation state (or lower). Interhalogen compounds are generally more reactive than the individual halogens (except F₂) because the X–X' bond is weaker and more polar than the X–X or X'–X' homonuclear bonds. They are strong fluorinating or halogenating agents used in organic synthesis and nuclear fuel reprocessing.
What are the three allotropes of carbon and how do they differ in structure and properties?
Diamond: each carbon is sp³ hybridised and forms four strong σ bonds to four adjacent carbons in a giant covalent lattice extending in all three dimensions. No delocalised electrons are present. Properties: hardest natural substance (Mohs hardness 10), very high melting point (3823 K), electrical insulator, transparent, high refractive index (used in jewellery and cutting tools). Graphite: each carbon is sp² hybridised, forming three σ bonds in a planar hexagonal layer. The fourth electron of each carbon is in a p orbital perpendicular to the plane, forming a delocalised π system across each layer. The layers are held together by weak van der Waals forces and can slide over each other. Properties: soft, slippery (lubricant), electrical conductor (delocalised electrons), thermodynamically more stable than diamond at room temperature and pressure. Buckminsterfullerene C₆₀: spherical cage of 60 carbon atoms, each sp² hybridised, forming 20 hexagonal and 12 pentagonal faces (like a football). All bond angles are close to 120° but the curved surface introduces some sp³ character. Properties: molecular solid, dissolves in benzene (purple solution), semiconductor, can be doped to give superconductors (K₃C₆₀ is a superconductor at 18 K). C₆₀ was discovered by Kroto, Curl, and Smalley in 1985 (Nobel Prize in Chemistry 1996).
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