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AITS 03Chemistry · Chapter 2 of 6

Chemistry · last-minute revision

The p-Block Elements, Groups 15–18

Group trends and the anomalous first member (NTA core), plus the compound-level facts that coaching papers still test: hydrides, oxoacids, Cl₂ reactions, interhalogens and xenon fluorides.

ns² np³⁻⁶one electron short, then filled

What this paper can askMust

The NTA NEET syllabus keeps groups 13–18 at the level of electronic configuration, group trends in physical and chemical properties, and the anomalous first member. Coaching papers still borrow classic compound questions (shapes, oxoacids, xenon fluorides, Cl2 reactions) from the older NCERT chapter. Both are covered here; compound-level material is marked Gap content.

GroupValence shellMembersOxidation statesNature down the group
15ns2 np3N P As Sb Bi−3, +3, +5N, P non-metal → As, Sb metalloid → Bi metal
16ns2 np4O S Se Te Po−2, +2, +4, +6O, S non-metal → Se, Te metalloid → Po metal (radioactive)
17ns2 np5F Cl Br I At−1 (F only −1); +1, +3, +5, +7 for Cl, Br, Iall non-metals; At radioactive
18ns2 np6 (He 1s2)He Ne Ar Kr Xe Rn0; Xe +2, +4, +6inert; Rn radioactive
Inert pair effect in one line

Down groups 15 and 16 the ns2 pair becomes reluctant to bond, so the lower state gains stability: Bi(+3) is stable, Bi(+5) is a strong oxidant (only BiF5 known); Te(+4) is more stable than Te(+6).

Group 15: trends and nitrogen's anomaliesMust

PropertyOrderNote
Ionisation enthalpyN > P > As > Sb > Bihalf-filled np3 → higher than group 16 neighbour
Atomic radiusN < P < As < Sb < Bionly small rise As → Bi (filled d and f shield poorly)
Melting pointrises up to As, then falls to Biboiling point rises steadily down
Hydride stability, EH3NH3 > PH3 > AsH3 > SbH3 > BiH3reducing power runs the opposite way
Basicity of hydridesNH3 > PH3 > AsH3 > SbH3 ≥ BiH3lone pair less available on large atom
Boiling point of hydridesPH3 < AsH3 < NH3 < SbH3 < BiH3NH3 raised by H-bonding but BiH3, SbH3 still higher (mass)
Oxides E2O3N, P acidic; As, Sb amphoteric; Bi basichigher oxidation state oxide is more acidic
Bond angle of group 15 and 16 hydrides decreasing down the grouplone pairENH₃ 107.8°PH₃ 93.6°AsH₃ 91.8°SbH₃ 91.3°BiH₃ 90.0°Group 15Group 16NH₃107.8°PH₃93.6°AsH₃91.8°SbH₃91.3°BiH₃90.0°H₂O104.5°H₂S92.1°H₂Se91.0°H₂Te90.0°Down the group: bond pairs move away from thesmaller-EN central atom, repulsion falls, angle → 90°
The two bonds close from 107.8° (NH₃) to 90° (BiH₃). The same logic gives H₂O 104.5° down to H₂Te 90°.

Why nitrogen is different

  • Small size, high electronegativity, high IE, no d orbitals → maximum covalency 4 (NH4+); cannot form NCl5, while P forms PCl5 and [PF6]−.
  • Forms pπ–pπ multiple bonds: N≡N (941.4 kJ mol−1), C≡N, N=O. Phosphorus prefers single bonds (P4).
  • N–N single bond is weaker than P–P (lone-pair repulsion at short bond length) → weaker catenation in N.
  • Heavier members form dπ–pπ bonds (R3P=O) and dπ–dπ bonds.
  • Halides: pentahalides more covalent than trihalides. All trihalides except those of N are stable; of nitrogen only NF3 is stable.

Group 15 compounds you may meetGap content

N2 (lab)
NH4Cl + NaNO2 → N2 + 2H2O + NaCl; (NH4)2Cr2O7 →Δ N2 + 4H2O + Cr2O3; Ba(N3)2 → Ba + 3N2 (very pure)
NH3 (Haber)
N2 + 3H2 ⇌ 2NH3, ΔfH° = −46.1 kJ mol−1; ~200 bar, ~700 K, iron oxide with K2O and Al2O3
HNO3 (Ostwald)
4NH3 + 5O2 → 4NO + 6H2O (Pt/Rh gauze, 500 K, 9 bar); 2NO + O2 ⇌ 2NO2; 3NO2 + H2O → 2HNO3 + NO
Cu + HNO3
dilute: 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O; conc.: Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O
Zn + HNO3
dilute gives N2O; conc. gives NO2. Cr and Al are passive in conc. HNO3 (oxide film)
Brown ring test
[Fe(H2O)6]2+ + NO → [Fe(H2O)5(NO)]2+ + H2O
PH3
Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3; P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2 (disproportionation)
PCl3
P4 + 6Cl2 → 4PCl3; pyramidal, sp3; PCl3 + 3H2O → H3PO3 + 3HCl
PCl5
trigonal bipyramidal; axial P–Cl 240 pm longer than equatorial 202 pm (more repulsion); solid is [PCl4]+[PCl6]−; PCl5 + H2O → POCl3 + 2HCl → H3PO4
OxideN stateLookNature
N2O+1colourless gasneutral
NO+2colourless gas, paramagneticneutral
N2O3+3blue solidacidic
NO2+4brown gas, paramagnetic, dimerisesacidic
N2O4+4colourlessacidic
N2O5+5colourless solidacidic

Allotropes of P. White P4: tetrahedral with 60° angle strain, most reactive, glows in the dark, soluble in CS2, stored under water. Red: polymeric chains of P4 units, less reactive, does not glow. Black: most stable, layered, does not burn in air up to 673 K.

Structures of phosphorus oxoacids with P–H bonds highlightedOHHOHPOHOHOHPOHOOHOHPOHOOHPOOHOHPOH₃PO₂hypophosphorousP +1 · 1 P–OH → monobasic2 P–H → strong reductantH₃PO₃phosphorousP +3 · 2 P–OH → dibasic1 P–H → reducingH₃PO₄orthophosphoricP +5 · 3 P–OH → tribasicno P–HH₄P₂O₇pyrophosphoricP +5 · 4 P–OH → tetrabasicP–O–P bridge
Only P–OH hydrogens are ionisable. P–H bonds (brass) are not acidic but make the acid a reducing agent.
OxoacidP stateKey bondBasicity
H3PO2 hypophosphorous+1two P–H1
H3PO3 phosphorous+3one P–H2
H4P2O6 hypophosphoric+4P–P bond4
H3PO4 orthophosphoric+5three P–OH3
H4P2O7 pyrophosphoric+5P–O–P4
(HPO3)3 cyclotrimetaphosphoric+5ring of P–O–P3

4H3PO3 → 3H3PO4 + PH3 (disproportionation on heating). H3PO2 reduces AgNO3 to Ag.

Group 16: trends and key compoundsMust

PropertyOrder / factNote
Electron gain enthalpyO less negative than S; S most negative in the groupO is small: incoming electron repelled
Oxygen's states−2; +2 in OF2, +1 in O2F2F is more electronegative
+4 and +6 states+4 stability rises down, +6 fallsinert pair effect
Acid strength of H2EH2O < H2S < H2Se < H2TeE–H bond weakens down
Thermal stability of H2EH2O > H2S > H2Se > H2Tereducing character rises down (except H2O)
Bond angleH2O 104.5° > H2S 92.1° > H2Se 91° > H2Te 90°
HalidesSF6 exceptionally stable (steric crowding); SF4 see-saw; S2Cl2 dimeric monohalideonly hexafluorides exist as EX6
DioxidesSO2 reducing; SeO2 oxidisingreducing power of EO2 falls down

Anomalous oxygen: small size, high electronegativity → H-bonding in H2O; covalency limited to 4, whereas S reaches 6 in SF6 using d orbitals. O2 is a paramagnetic gas; S is a solid.

Ozone, sulphur, SO2, H2SO4

O3 formation
3O2 → 2O3, ΔH = +142 kJ mol−1; silent electric discharge through dry O2
O3 structure
angular, 117°; both O–O bonds 128 pm (resonance)
O3 as oxidant
PbS + 4O3 → PbSO4 + 4O2; 2I− + H2O + O3 → 2OH− + I2 + O2 (estimation of O3)
Ozone depletion
NO + O3 → NO2 + O2
Sulphur allotropes
rhombic (α, yellow) stable below 369 K; monoclinic (β) above 369 K; both S8 puckered crown rings; S2 vapour paramagnetic
SO2 reducing
2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO42− + 4H+; decolourises acidified KMnO4
Contact process
2SO2 + O2 ⇌ 2SO3 (V2O5, 720 K, 2 bar, ΔH = −196.6 kJ mol−1); SO3 + H2SO4 → H2S2O7 (oleum); oleum + H2O → H2SO4
H2SO4 behaviour
dehydrating: C12H22O11 → 12C + 11H2O; hot conc. oxidant: Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O; C + 2H2SO4 → CO2 + 2SO2 + 2H2O
Acid dissociation
Ka1 > 10 (very large), Ka2 = 1.2 × 10−2
Oxoacid of SS stateRemember
H2SO3 sulphurous+4two S–OH, one S=O
H2SO4 sulphuric+6two S–OH, two S=O
H2S2O7 pyrosulphuric (oleum)+6S–O–S bridge
H2SO5 peroxomonosulphuric (Caro's)+6one O–O
H2S2O8 peroxodisulphuric (Marshall's)+6S–O–O–S peroxide bridge

Oxides by nature: acidic SO2, SO3, Cl2O7, N2O5 · basic Na2O, CaO · amphoteric Al2O3 · neutral CO, NO, N2O.

Group 17: halogensMust

PropertyOrderWhy
Electron gain enthalpy (most negative)Cl > F > Br > IF is small: electron–electron repulsion
ElectronegativityF > Cl > Br > IF = 4.0, highest of all
Bond dissociation enthalpyCl2 > Br2 > F2 > I2F–F weakened by lone-pair repulsion
Oxidising powerF2 > Cl2 > Br2 > I2F2: low bond enthalpy, high hydration enthalpy of F−
Acid strength HXHF < HCl < HBr < HIH–X bond weakens down
Thermal stability HXHF > HCl > HBr > HIsame bond-strength reason
Boiling point HXHCl < HBr < HI < HFHF H-bonded
ColourF2 pale yellow, Cl2 greenish yellow, Br2 red-brown liquid, I2 violet-black solidvisible light excites outer electrons
Acid strength, oxoacids of ClHOCl < HClO2 < HClO3 < HClO4more O → more stable conjugate base; oxidising power runs the other way

Anomalous fluorine: shows only −1, forms only one oxoacid (HOF), strong H-bonding in HF, no d orbitals. OF2 and O2F2 are fluorides of oxygen, not oxides of fluorine.

F2 with water
2F2 + 2H2O → 4H+ + 4F− + O2
Cl2 with water
Cl2 + H2O → HCl + HOCl; HOCl gives nascent O → permanent bleaching by oxidation
Cl2 preparation
MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O; 2KMnO4 + 16HCl → 2KCl + 2MnCl2 + 8H2O + 5Cl2; Deacon: 4HCl + O2 → 2Cl2 + 2H2O (CuCl2, 723 K)
Cl2 + cold dilute NaOH
2NaOH + Cl2 → NaCl + NaOCl + H2O
Cl2 + hot conc. NaOH
6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O
Cl2 + dry slaked lime
2Ca(OH)2 + 2Cl2 → Ca(OCl)2 + CaCl2 + 2H2O (bleaching powder)
Cl2 + NH3
excess NH3: 8NH3 + 3Cl2 → 6NH4Cl + N2; excess Cl2: NH3 + 3Cl2 → NCl3 + 3HCl
Aqua regia
3 HCl : 1 HNO3; Au → [AuCl4]−, Pt → [PtCl6]2−

Interhalogens

TypeExamplesShapeHybridisation
XX′ClF, BrF, ICl, IBrlinear—
XX′3ClF3, BrF3, IF3, ICl3bent T-shapesp3d
XX′5ClF5, BrF5, IF5square pyramidalsp3d2
XX′7IF7pentagonal bipyramidalsp3d3

X is the larger halogen, X′ the smaller. Interhalogens are more reactive than the parent halogens (fluorine excepted) because X–X′ is weaker than X–X. Hydrolysis: XX′ + H2O → HX′ + HOX. ClF3 fluorinates U to UF6 for uranium enrichment.

Group 18: noble gases and xenon compoundsHigh yield

  • Very high IE, large positive electron gain enthalpy, only weak dispersion forces → very low boiling points that rise down the group; He has the lowest bp.
  • Bartlett made O2+[PtF6]−, noticed first IE of Xe (1170 kJ mol−1) ≈ O2 (1175 kJ mol−1), and made the first xenon compound, Xe+[PtF6]−.
CompoundMade fromShapeHydrolysis / use
XeF2Xe (excess) + F2, 673 K, 1 barlinear, sp3d, 3 lp2XeF2 + 2H2O → 2Xe + 4HF + O2
XeF4Xe + F2 (1 : 5), 873 K, 7 barsquare planar, sp3d2, 2 lp6XeF4 + 12H2O → 4Xe + 2XeO3 + 24HF + 3O2
XeF6Xe + F2 (1 : 20), 573 K, 60–70 bardistorted octahedral, sp3d3, 1 lpcomplete: XeF6 + 3H2O → XeO3 + 6HF
XeOF4partial hydrolysis: XeF6 + H2O → XeOF4 + 2HFsquare pyramidalcolourless volatile liquid
XeO2F2XeF6 + 2H2O → XeO2F2 + 4HFsee-saw—
XeO3hydrolysis of XeF4 / XeF6pyramidalcolourless explosive solid

Fluoride acceptor reactions: XeF2 + PF5 → [XeF]+[PF6]−; XeF4 + SbF5 → [XeF3]+[SbF6]−; XeF6 + MF → M+[XeF7]−.

Uses: He — meteorological balloons, diluent for O2 in diving apparatus (low solubility in blood), liquid He as cryogen for MRI magnets. Ne — discharge tubes and advertisement lights. Ar — inert atmosphere in arc welding and metallurgy, filling bulbs. Kr, Xe — special-purpose lamps.

Shapes galleryMust

Shapes of p-block molecules with lone pairs highlightedClClClClClPPCl₅trigonal bipyramidalsp³d · 5 bpFFFFSSF₄see-sawsp³d · 4 bp + 1 lpFFFClClF₃bent T-shapesp³d · 3 bp + 2 lpFFXeXeF₂linearsp³d · 2 bp + 3 lpFFFFFBrBrF₅square pyramidalsp³d² · 5 bp + 1 lpFFFFXeXeF₄square planarsp³d² · 4 bp + 2 lpFFFFFFXeXeF₆distorted octahedralsp³d³ · 6 bp + 1 lpOOOXeXeO₃pyramidalsp³ · 3 bp + 1 lp
Lone pairs (pulsing brass) take equatorial positions in trigonal bipyramids and opposite positions in octahedra. Count bp + lp first, then name the shape from bond positions only.
Steric numberArrangement0 lp1 lp2 lp3 lp
4tetrahedralCH4 tetrahedralNH3, XeO3 pyramidalH2O bent—
5trigonal bipyramidalPCl5SF4 see-sawClF3 T-shapeXeF2, I3− linear
6octahedralSF6BrF5, XeOF4 square pyramidalXeF4 square planar—
7pentagonal bipyramidalIF7XeF6 distorted octahedral——

Standard question patternsMust

Basicity of a phosphorus oxoacid
Count P–OH groups only. H3PO2 → 1, H3PO3 → 2, H3PO4 → 3, H4P2O7 → 4.
Which acid is a reducing agent
Look for P–H bonds: H3PO2 and H3PO3.
Shape / hybridisation of Xe or interhalogen
Steric number = bonded atoms + lone pairs on centre; lone pairs = (valence e− − bonds used)/2.
Orders within a group
Separate the property: bond enthalpy (Cl2 > Br2 > F2 > I2) is not the same order as electron gain enthalpy (Cl > F > Br > I) or electronegativity (F > Cl).
Anomalous first member
Small size + high EN + no d orbitals. N: no NCl5; O: no OF6; F: only −1, one oxoacid.
Products of Cl2 + NaOH
Cold dilute → NaOCl; hot concentrated → NaClO3.
Hydrolysis of XeF6
Complete → XeO3; one H2O → XeOF4; two H2O → XeO2F2.

Traps that cost marksMust

Trap · Ratio / direction reversal

Bond angle and basicity fall down group 15 (NH3 highest), but acid strength of hydrides rises down group 16 and group 17 (HI strongest, HF weakest). Write an arrow on the rough sheet before picking an order.

Trap · Two orders that look alike

Electron gain enthalpy: Cl > F > Br > I. Bond dissociation enthalpy: Cl2 > Br2 > F2 > I2. F sits in second place in one and third in the other.

Trap · Right content, wrong arrangement

In shape-matching questions ClF3 (T-shape) and SF4 (see-saw) are both sp3d; BrF5 (square pyramidal) and XeF4 (square planar) are both sp3d2. Decide lone-pair count for each row before looking at the option codes.

Trap · Basicity of H3PO3

Three H atoms, but only two are on oxygen. Answer 2, not 3.

Trap · Attribution

First noble-gas compound: Neil Bartlett, Xe+[PtF6]−, prompted by O2+[PtF6]−.

Ten-question checkMust

Attempt all ten before opening any answer. Aim for 9/10 in under 8 minutes.

Q1.The shape of XeF4 molecule is

  1. (A)tetrahedral
  2. (B)see-saw
  3. (C)square planar
  4. (D)square pyramidal
Show answer

Answer (C). Xe has 8 valence electrons; 4 used in bonds, 2 lone pairs remain. Steric number 6 (sp3d2) with lone pairs opposite each other → square planar.

Q2.The basicity of phosphorous acid, H3PO3, is

  1. (A)2
  2. (B)3
  3. (C)1
  4. (D)zero
Show answer

Answer (A). Structure: one P=O, one P–H and two P–OH. Only P–OH hydrogens ionise, so it is dibasic.

Q3.The correct order of H–E–H bond angle is

  1. (A)NH3 > PH3 > AsH3 > SbH3
  2. (B)SbH3 > AsH3 > PH3 > NH3
  3. (C)PH3 > NH3 > AsH3 > SbH3
  4. (D)NH3 > AsH3 > PH3 > SbH3
Show answer

Answer (A). As the central atom grows and its electronegativity falls, bond pairs lie farther from it, repulsion drops, angle approaches 90°: 107.8° > 93.6° > 91.8° > 91.3°.

Q4.Which of these has the most negative electron gain enthalpy?

  1. (A)F
  2. (B)Br
  3. (C)Cl
  4. (D)I
Show answer

Answer (C). F is so small that the added electron feels strong repulsion, so its value is less negative than chlorine's.

Q5.Assertion (A): Nitrogen does not form pentahalides.
Reason (R): Nitrogen has no d orbitals in its valence shell and cannot expand its covalency beyond four.

  1. (A)Both A and R are true, and R is the correct explanation of A
  2. (B)Both A and R are true, but R is not the correct explanation of A
  3. (C)A is true, but R is false
  4. (D)A is false, but R is true
Show answer

Answer (A). Both true; the lack of d orbitals is exactly why NX5 cannot form while PCl5 can.

Q6.Complete hydrolysis of XeF6 gives

  1. (A)XeOF4 and HF
  2. (B)XeO2F2 and HF
  3. (C)XeO4 and HF
  4. (D)XeO3 and HF
Show answer

Answer (D). XeF6 + 3H2O → XeO3 + 6HF. XeOF4 and XeO2F2 are partial-hydrolysis products with one and two water molecules.

Q7.The correct order of bond dissociation enthalpy of halogens is

  1. (A)F2 > Cl2 > Br2 > I2
  2. (B)Cl2 > F2 > Br2 > I2
  3. (C)I2 > Br2 > Cl2 > F2
  4. (D)Cl2 > Br2 > F2 > I2
Show answer

Answer (D). F–F is unusually weak because lone pairs on the two small F atoms repel. Cl > F > Br > I is the electron-gain order, not this one.

Q8.The shape of ClF3 is

  1. (A)trigonal planar
  2. (B)trigonal pyramidal
  3. (C)bent T-shape
  4. (D)see-saw
Show answer

Answer (C). Cl: 7 valence electrons, 3 bonds, 2 lone pairs → steric number 5 (sp3d). Both lone pairs equatorial → T-shape. See-saw needs 4 bonds + 1 lone pair.

Q9.Which oxoacid of phosphorus contains a P–P bond?

  1. (A)H4P2O7
  2. (B)H4P2O6
  3. (C)H3PO3
  4. (D)(HPO3)3
Show answer

Answer (B). Hypophosphoric acid has a direct P–P bond (P +4). Pyrophosphoric and cyclotrimetaphosphoric acids have P–O–P bridges.

Q10.Chlorine reacts with hot and concentrated NaOH to give

  1. (A)NaCl and NaOCl
  2. (B)NaCl and NaClO3
  3. (C)NaOCl and NaClO3
  4. (D)NaClO4 and NaCl
Show answer

Answer (B). 6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O. NaOCl is the product with cold dilute NaOH.

60-second recap before the paperMust

  1. Inert pair effect: lower state more stable down groups 15–16 (Bi +3, Te +4).
  2. Group 15 hydrides: stability, basicity and bond angle all fall NH3 → BiH3; reducing power rises. bp: PH3 lowest.
  3. Group 16 and 17 hydrides: acid strength rises down (HI, H2Te strongest); stability falls.
  4. N, O, F anomalies: small size, high EN, no d orbitals → no NCl5, H-bonding, F only −1, HOF only.
  5. Electron gain: Cl > F > Br > I, and S more negative than O. Bond enthalpy: Cl2 > Br2 > F2 > I2.
  6. Basicity = P–OH count; P–H makes reductant. H3PO3 dibasic, H3PO2 monobasic.
  7. Cl2 + cold dil NaOH → NaOCl; hot conc → NaClO3. Excess NH3 → N2; excess Cl2 → NCl3.
  8. XeF2 linear, XeF4 square planar, XeF6 distorted octahedral, XeO3 pyramidal, XeOF4 square pyramidal.
  9. ClF3 T-shape, BrF5 square pyramidal, IF7 pentagonal bipyramidal, SF4 see-saw, PCl5 trigonal bipyramidal (axial longer).