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Chemistry · JEE / NEET syllabus

p-Block Elements (Groups 13 to 18)

The rationalised NCERT books no longer have a p-block chapter, but groups 13 to 18 are still part of the JEE Main and NEET syllabi. The syllabi put the weight on group trends and on the unusual behaviour of the first member of each group, so those come first here. The key compounds follow, since their structures apply the same ideas of size, electronegativity and available orbitals.

In this chapter: general trends across the p-block, inert pair effect, anomalous behaviour of B, C, N, O and F, group-wise notes on groups 13 to 18, diborane, boric acid, allotropes of carbon, silicones and silicates, ammonia and nitric acid, oxoacids of phosphorus and sulphur, ozone and sulphuric acid, chlorine and interhalogen compounds, and the fluorides and oxides of xenon.

The p-block has the outer configuration ns2np1−6 (helium, 1s2, is placed in group 18). It contains metals, metalloids and non-metals; non-metallic character is highest at the top right and metallic character increases down each group.

  • Oxidation states: the highest oxidation state equals the number of valence electrons (group number − 10). Down a group, the state two units lower becomes more stable. This inert pair effect arises because the ns2 electrons of heavy elements are poorly shielded by d and f electrons and are reluctant to take part in bonding. So Tl+ is more stable than Tl3+, Pb2+ more stable than Pb4+ (PbO2 is a strong oxidant), and Bi3+ more stable than Bi5+.
  • Anomalous first member: B, C, N, O and F differ from the rest of their groups because of their small size, high electronegativity, high ionisation enthalpy and absence of d orbitals. Their maximum covalency is 4, and they form pπ-pπ multiple bonds (C=C, C≡C, N≡N, C=O, O=O) that heavier members form far less readily.
  • Heavier members can use d orbitals to exceed a covalency of 4 (PCl5, SF6, IF7) and can form dπ-pπ bonds, as in the oxides and oxoacids of P and S.

Group 13: B, Al, Ga, In, Tl

  • Configuration ns2np1; common oxidation state +3, with +1 increasingly stable down the group.
  • Atomic radius increases down the group, except that Ga (135 pm) is smaller than Al (143 pm): the ten 3d electrons shield the nucleus poorly.
  • Boron is a non-metal, very hard, and forms only covalent compounds. BX3 molecules have only six electrons around boron, so they are electron deficient and act as Lewis acids (BF3 + F− → [BF4]−). Aluminium chloride exists as the dimer Al2Cl6.
  • Boric acid, B(OH)3, is a weak monobasic Lewis acid: it accepts OH− rather than releasing H+: B(OH)3 + 2H2O → [B(OH)4]− + H3O+. It has a layered structure of planar BO3 units held by hydrogen bonds. Borax is Na2[B4O5(OH)4]·8H2O, used in the borax bead test for transition metal ions.

Diborane, B2H6

4BF3 + 3LiAlH4 → 2B2H6 + 3LiF + 3AlF3 (in ether)2NaBH4 + I2 → B2H6 + 2NaI + H2 (laboratory)

Diborane is a colourless, highly toxic gas that catches fire in air (B2H6 + 3O2 → B2O3 + 3H2O) and is hydrolysed by water to boric acid and hydrogen. Heating with ammonia finally gives borazine, B3N3H6, called "inorganic benzene".

Each boron is roughly sp3 hybridised. The four terminal B-H bonds are ordinary two-centre two-electron bonds, and the two boron atoms and four terminal hydrogens lie in one plane. The two bridging hydrogens lie above and below this plane; each B-H-B bridge is a three-centre two-electron (3c-2e) bond, often called a banana bond. The bridge B-H bonds are longer than the terminal ones.

Structure of diborane with terminal and bridging hydrogenswww.iitmedicoguide.comBBHHHHHHterminal H2c-2e B-H bonds(4 in one plane)bridging H (above the plane)bridging H (below the plane)B-H-B bridges:3c-2e bondswww.iitmedicoguide.com
Diborane has 12 valence electrons: eight form the four terminal B-H bonds, and the remaining four hold the two B-H-B bridges, two electrons shared over three atoms in each.

Group 14: C, Si, Ge, Sn, Pb

  • Configuration ns2np2; oxidation states +4 and +2. The +2 state becomes more stable down the group: Sn2+ is a reducing agent (it tends to become Sn4+), while Pb4+ compounds are oxidising.
  • Carbon is unique: it shows the greatest catenation (strong C-C bonds), forms pπ-pπ multiple bonds with itself and with O and N, and has a maximum covalency of 4.
  • Allotropes of carbon: diamond (each C sp3, rigid three-dimensional network, hardest natural substance, non-conductor); graphite (each C sp2, hexagonal layers held by weak van der Waals forces, conducts electricity, soft and slippery, used as a dry lubricant); fullerenes (C60, sp2 carbons in a cage of 20 six-membered and 12 five-membered rings).
  • CO is poisonous because it binds haemoglobin to form carboxyhaemoglobin, about 300 times more stable than the oxygen-haemoglobin complex. CO2 is linear (C is sp hybridised).
  • SiO2 is a three-dimensional network in which each Si is bonded tetrahedrally to four O atoms and each O to two Si atoms.

Silicones and silicates

Silicones are organosilicon polymers with the repeating unit -(R2SiO)-. Methyl chloride reacts with silicon in the presence of copper at 573 K to give methyl-substituted chlorosilanes. Hydrolysis of dimethyldichlorosilane, (CH3)2SiCl2, gives (CH3)2Si(OH)2, which condenses into long chains; (CH3)3SiCl is added to cap the chain ends and control the chain length. Silicones repel water, resist heat and oxidation, are chemically inert and are good electrical insulators, so they are used in sealants, greases, electrical insulation and water-proofing of fabrics.

Silicates are built from SiO44− tetrahedra. When the tetrahedra share corners with each other they form chains, rings, sheets or three-dimensional networks, and the negative charge is balanced by metal ions. Feldspars and zeolites are three-dimensional aluminosilicates; the zeolite ZSM-5 is used to convert alcohols directly into gasoline.

Group 15: N, P, As, Sb, Bi

  • Configuration ns2np3, with a half-filled, extra stable p subshell. Oxidation states range from −3 to +5; the +3 state becomes more stable down the group.
  • Nitrogen is anomalous: it exists as N2 with a very strong triple bond (pπ-pπ), while phosphorus exists as P4; nitrogen cannot expand its covalency beyond 4; and its single N-N bond is weak because of repulsion between lone pairs on the small atoms.
  • Hydrides EH3: basicity and thermal stability decrease down the group (NH3 > PH3 > AsH3 > SbH3 > BiH3), while reducing character increases. The bond angle falls from NH3 (107.8°) to PH3 (93.6°), AsH3 (91.8°) and SbH3 (91.3°). NH3 has an unusually high boiling point for its size because of hydrogen bonding.
  • Phosphorus allotropes: white phosphorus (discrete tetrahedral P4 molecules with 60° bond angles, strained and very reactive, glows in the dark, stored under water); red phosphorus (polymeric chains of P4 units, much less reactive); black phosphorus (the most stable form, layered).
  • PCl5 is trigonal bipyramidal in the gas phase; its two axial bonds are longer, and weaker, than the three equatorial bonds. In the solid it exists as [PCl4]+[PCl6]−.

Ammonia

In the Haber process, N2(g) + 3H2(g) ⇌ 2NH3(g), ΔfH° = −46.1 kJ mol−1. Following Le Chatelier's principle, high pressure (about 200 × 105 Pa) favours ammonia; a temperature of about 700 K is a compromise between yield and rate, with iron oxide catalyst promoted by small amounts of K2O and Al2O3. The NH3 molecule is trigonal pyramidal, with three bond pairs and one lone pair on nitrogen. The lone pair makes it a Lewis base: with Cu2+ it forms the deep blue [Cu(NH3)4]2+.

Nitric acid

4NH3 + 5O2 → 4NO + 6H2O (Pt/Rh gauze, 500 K, 9 bar)2NO + O2 ⇌ 2NO2; 3NO2 + H2O → 2HNO3 + NOOstwald process. Distillation gives about 68% HNO3 by mass; dehydration with concentrated H2SO4 raises it to 98%.

HNO3 is a planar molecule and a strong oxidising agent. The product depends on concentration and on the metal:

3Cu + 8HNO3(dilute) → 3Cu(NO3)2 + 2NO + 4H2OCu + 4HNO3(conc.) → Cu(NO3)2 + 2NO2 + 2H2O4Zn + 10HNO3(dilute) → 4Zn(NO3)2 + N2O + 5H2OZn + 4HNO3(conc.) → Zn(NO3)2 + 2NO2 + 2H2O

Chromium and aluminium do not dissolve in concentrated HNO3 because a protective oxide film forms (passivity). Concentrated HNO3 oxidises non-metals: iodine to iodic acid, carbon to CO2, sulphur to H2SO4 and phosphorus to H3PO4. In the brown ring test for nitrate, Fe2+ reduces nitrate to NO, which forms the brown complex [Fe(H2O)5(NO)]2+ at the junction with concentrated H2SO4.

Oxoacids of phosphorus

AcidFormulaOxidation state of PP-OH / P-H bondsBasicity
Hypophosphorous (phosphinic)H3PO2+11 P-OH, 2 P-HMonobasic; strong reducing agent
Orthophosphorous (phosphonic)H3PO3+32 P-OH, 1 P-HDibasic; reducing agent
OrthophosphoricH3PO4+53 P-OH, no P-HTribasic
PyrophosphoricH4P2O7+54 P-OH, P-O-P linkTetrabasic
Metaphosphoric(HPO3)n+5cyclic or chain P-O-P linksOne per P

In all of them phosphorus is tetrahedral with one P=O bond. Only hydrogens attached to oxygen (P-OH) are ionisable; P-H hydrogens are not, but they give the acid reducing properties. On heating, H3PO3 disproportionates: 4H3PO3 → 3H3PO4 + PH3.

Group 16: O, S, Se, Te, Po

  • Configuration ns2np4. Oxygen shows mainly −2; S, Se and Te also show +2, +4 and +6, with +4 becoming more stable than +6 down the group.
  • Oxygen is anomalous: small size and high electronegativity allow strong hydrogen bonding, so H2O is a liquid while H2S is a gas. Oxygen's electron gain enthalpy is less negative than that of sulphur because the added electron enters a small, crowded 2p subshell.
  • Hydrides H2E: acidic character increases (H2O < H2S < H2Se < H2Te) and thermal stability decreases down the group, as the H-E bond becomes longer and weaker.
  • Ozone is a bent molecule (bond angle about 117°) with two equal O-O bonds of 128 pm, explained by resonance. It is a strong oxidising agent: it converts PbS to PbSO4 and iodide to iodine.
  • Sulphur exists as puckered S8 rings (crown shape). Rhombic (α) sulphur is stable below 369 K; above it, monoclinic (β) sulphur is stable.
  • SO2 is bent; in water it acts as a reducing agent, decolourising acidified KMnO4 and reducing Fe3+ to Fe2+.

Sulphuric acid and oxoacids of sulphur

In the contact process, sulphur or sulphide ore is burnt to SO2, which is oxidised: 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔrH° = −196.6 kJ mol−1, over V2O5 at about 720 K and 2 bar. SO3 is absorbed in H2SO4 to give oleum (H2S2O7), which is diluted with water to the desired concentration.

  • Dehydrating agent: it chars sugar (C12H22O11 → 12C + 11H2O) and removes water from many compounds.
  • Oxidising agent (hot, concentrated): Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O; C + 2H2SO4 → CO2 + 2SO2 + 2H2O.
  • Oxoacids: sulphurous acid H2SO3 (S +4), sulphuric acid H2SO4 (+6), peroxodisulphuric acid H2S2O8 (+6, contains an O-O peroxide link) and pyrosulphuric acid H2S2O7 (+6, contains an S-O-S link). In each, sulphur is tetrahedral.

Group 17: F, Cl, Br, I

  • Configuration ns2np5, one electron short of a noble gas. Halogens are strong oxidising agents; F2 is the strongest, and oxidising power decreases down the group.
  • Chlorine has the most negative electron gain enthalpy; fluorine's is less negative because of strong repulsion among electrons in its small 2p subshell.
  • Bond dissociation enthalpy: Cl2 > Br2 > F2 > I2. The F-F bond is unexpectedly weak because of lone pair repulsion between the small atoms.
  • Fluorine is anomalous: it shows only the −1 oxidation state, has no d orbitals, and HF is a liquid (hydrogen bonding) and a weak acid. Acid strength: HF < HCl < HBr < HI.
  • Chlorine: MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O (laboratory); Deacon's process, 4HCl + O2 → 2Cl2 + 2H2O (CuCl2, 723 K). With cold dilute NaOH it gives NaCl + NaOCl; with hot concentrated NaOH, 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O. With dry slaked lime it gives bleaching powder: 2Ca(OH)2 + 2Cl2 → Ca(OCl)2 + CaCl2 + 2H2O. Moist chlorine bleaches by oxidation (nascent oxygen), so the effect is permanent.
  • Oxoacids of chlorine: HOCl (+1), HClO2 (+3), HClO3 (+5), HClO4 (+7). Acid strength increases in the same order, HClO4 being the strongest, because the conjugate base is increasingly stabilised by resonance over more oxygen atoms.

Interhalogen compounds

Halogens combine with each other to give compounds of types XX′, XX′3, XX′5 and XX′7, where X is the larger, less electronegative halogen. They are covalent, diamagnetic and generally more reactive than the halogens themselves (except F2), because the X-X′ bond is weaker than the X-X bond. Their shapes follow VSEPR theory:

TypeExampleBond pairs + lone pairs on XHybridisationShape
XX′ClF, ICl1 + 3not needed (diatomic)Linear
XX′3ClF3, BrF33 + 2sp3dBent T-shape
XX′5BrF5, IF55 + 1sp3d2Square pyramidal
XX′7IF77 + 0sp3d3Pentagonal bipyramidal

Group 18: noble gases

Noble gases have closed-shell configurations (ns2np6; He 1s2), very high ionisation enthalpies and positive electron gain enthalpies, so they are very unreactive. Only xenon forms a well-known series of compounds. The first was made after it was noticed that the ionisation enthalpy of the O2 molecule (1175 kJ mol−1) is almost equal to that of xenon (1170 kJ mol−1): since O2+[PtF6]− was known, Xe+[PtF6]− was prepared.

Xe + F2 → XeF2 (xenon in excess, 673 K, 1 bar)Xe + 2F2 → XeF4 (Xe : F2 = 1 : 5, 873 K, 7 bar)Xe + 3F2 → XeF6 (Xe : F2 = 1 : 20, 573 K, 60 to 70 bar)
CompoundBond pairs + lone pairs on XeHybridisationShape
XeF22 + 3sp3dLinear (lone pairs in the equatorial plane)
XeF44 + 2sp3d2Square planar (lone pairs above and below the plane)
XeF66 + 1sp3d3Distorted octahedral
XeO33 σ bonds + 1 lone pairsp3Pyramidal
XeOF45 σ bonds + 1 lone pairsp3d2Square pyramidal
Shapes of XeF2 (linear) and XeF4 (square planar)www.iitmedicoguide.comXeFFF-Xe-F = 180°XeF₂: linearsp³d; 2 bond pairs + 3 lone pairsFXeFFFall five atomsin one planeXeF₄: square planarsp³d²; 4 bond pairs + 2 lone pairswww.iitmedicoguide.com
Lone pairs take the positions that keep them farthest from each other: three in the equatorial plane of XeF₂ and two opposite each other in XeF₄, so the atoms themselves end up linear and square planar.
Worked example: Use VSEPR theory to predict the shapes of XeF4 and ClF3.
Solution: Xe has 8 valence electrons; four are used in bonds with four F atoms, leaving 4 electrons = 2 lone pairs. Total 6 electron pairs, arranged octahedrally (sp3d2); the two lone pairs sit opposite each other, so XeF4 is square planar. Cl has 7 valence electrons; three form bonds with F, leaving 4 electrons = 2 lone pairs. Total 5 pairs, trigonal bipyramidal (sp3d); both lone pairs take equatorial positions, so ClF3 is bent T-shaped.

Xenon fluorides are strong fluorinating agents. XeF6 is hydrolysed completely to XeO3 (XeF6 + 3H2O → XeO3 + 6HF); partial hydrolysis gives XeOF4 and XeO2F2. Uses of noble gases: helium in weather balloons, in the gas mixture for deep-sea divers (it is much less soluble in blood than nitrogen) and in cryogenic work; neon in discharge tubes and signs; argon to provide an inert atmosphere in metallurgy and to fill electric bulbs.

Common mistakes: (1) Calling boric acid a protonic acid or a tribasic acid; it is a monobasic Lewis acid. (2) Counting P-H hydrogens when finding basicity; H3PO3 is dibasic and H3PO2 monobasic. (3) Taking F as having the most negative electron gain enthalpy; chlorine does. (4) Drawing XeF4 as tetrahedral or ClF3 as trigonal planar, forgetting lone pairs. (5) Explaining the smaller radius of Ga with the lanthanoid contraction; here it is the poor shielding by 3d electrons.

JEE and NEET focus

  • Trends in radius, ionisation enthalpy, electron gain enthalpy and oxidation states, and the inert pair effect.
  • Reasons for the anomalous behaviour of B, C, N, O and F.
  • Structures: diborane, boric acid, silicones, PCl5, oxoacids of P and S, ozone.
  • Trends in hydrides of groups 15 and 16 and acid strength of hydrogen halides and oxoacids of chlorine.
  • VSEPR shapes and hybridisation of interhalogens and xenon compounds.

Practice questions

The basicity of orthophosphorous acid, H3PO3, is:

  1. 1
  2. 2
  3. 3
  4. 4
Show answer
B. It has two P-OH groups and one non-ionisable P-H bond.

The shape of XeF4 is:

  1. Tetrahedral
  2. Square planar
  3. See-saw
  4. Square pyramidal
Show answer
B. Six electron pairs (4 bond + 2 lone), with the lone pairs trans to each other.

The B-H-B bridge bonds in diborane are:

  1. Ionic bonds
  2. Two-centre two-electron bonds
  3. Three-centre two-electron bonds
  4. Hydrogen bonds
Show answer
C. Two electrons hold three atoms (B, H, B) together in each bridge.

The most basic hydride of group 15 is:

  1. NH3
  2. PH3
  3. AsH3
  4. BiH3
Show answer
A. The lone pair on the small nitrogen atom is most available; basicity falls down the group.

The strongest acid among the oxoacids of chlorine is:

  1. HOCl
  2. HClO2
  3. HClO3
  4. HClO4
Show answer
D. ClO4− spreads its charge over four oxygen atoms, so it is the most stable conjugate base.

Copper reacts with dilute nitric acid to give mainly:

  1. NO2
  2. NO
  3. N2O
  4. H2
Show answer
B. 3Cu + 8HNO3(dilute) → 3Cu(NO3)2 + 2NO + 4H2O. Concentrated acid gives NO2.

The hybridisation and shape of ClF3 are:

  1. sp2, trigonal planar
  2. sp3, pyramidal
  3. sp3d, bent T-shape
  4. sp3d2, square planar
Show answer
C. Three bond pairs and two lone pairs on Cl; the lone pairs are equatorial.

Gallium has a smaller atomic radius than aluminium because:

  1. Of the lanthanoid contraction
  2. The 3d electrons shield the nuclear charge poorly
  3. Gallium is a liquid
  4. Gallium has fewer protons
Show answer
B. The extra nuclear charge after the 3d series is not fully shielded, so the outer electrons are pulled in.
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