JEE Main & Advanced

p-Block Elements (Class 12)

p-Block Elements (Class 12) for JEE Main & Advanced

1
Module 1

Group 15 and Group 16

Group 15 — Nitrogen Family and CompoundsTopic 1

Group 15: N, P, As, Sb, Bi. Outer config: $ns^2\,np^3$.

  • Non-metals: N, P
  • Metalloids: As, Sb
  • Metal: Bi

Electronic Configuration:

Element$Z$Config
N7$[He]\,2s^2\,2p^3$
P15$[Ne]\,3s^2\,3p^3$
As33$[Ar]\,3d^{10}\,4s^2\,4p^3$
Sb51$[Kr]\,4d^{10}\,5s^2\,5p^3$
Bi83$[Xe]\,4f^{14}\,5d^{10}\,6s^2\,6p^3$

Trends:

PropertyTrend
Atomic radiusIncreases ↓
IEDecreases ↓
ENDecreases ↓
Metallic characterIncreases ↓
Oxidation states$-3$ to $+5$; $-3$ becomes less stable, $+3$ more stable down (inert pair)

Anomalous Behavior of Nitrogen:

  • Small size; no $d$ orbitals
  • Can form $p\pi-p\pi$ multiple bonds (N≡N)
  • Only $-3, 0, +1, +2, +3, +4, +5$ — many oxidation states
  • N₂ is gaseous, very stable due to triple bond ($945$ kJ/mol)

Ammonia (NH₃):

Preparation:

  • Lab: $NH_4Cl + Ca(OH)_2 \xrightarrow{\Delta} 2NH_3 + CaCl_2 + 2H_2O$
  • Industrial: Haber's process: $N_2 + 3H_2 \xrightarrow{Fe, 700K, 200atm} 2NH_3$

Properties:

  • Colorless, pungent gas
  • Highly soluble in water
  • Lewis base (lone pair on N): forms $NH_4^+$
  • Reduces $Cu^{2+}$ to deep blue $[Cu(NH_3)_4]^{2+}$
  • Reacts with HCl to give white fumes of $NH_4Cl$

Structure: Trigonal pyramidal, $sp^3$ with one lone pair on N. Bond angle $107°$.

Uses: Fertilizers ($NH_4NO_3$, urea), HNO₃ manufacture (Ostwald process), refrigerant.

Nitric Acid (HNO₃):

Preparation: Ostwald Process:

  1. $4NH_3 + 5O_2 \xrightarrow{Pt, 500K} 4NO + 6H_2O$
  2. $2NO + O_2 \to 2NO_2$
  3. $3NO_2 + H_2O \to 2HNO_3 + NO$ (NO recycled)

Properties:

  • Colorless when pure; yellow on standing (due to NO₂)
  • Strong oxidizing acid
  • With metals (except noble): gives various nitrogen oxides
  • Dilute HNO₃ + Cu → $Cu(NO_3)_2 + NO + H_2O$
  • Conc. HNO₃ + Cu → $Cu(NO_3)_2 + NO_2 + H_2O$
  • Passive with Fe, Al, Cr (forms protective oxide layer)
  • Reacts with non-metals: $C + 4HNO_3 \to CO_2 + 4NO_2 + 2H_2O$

Aqua regia: $3:1$ conc. HCl + HNO₃; dissolves Au, Pt.

Oxides of Nitrogen:

OxideNameHybridizationNotes
$N_2O$Nitrous oxide$sp$"Laughing gas"; anesthetic
NONitric oxide$sp$Colorless gas; brown on exposure to O₂ (forms NO₂)
$N_2O_3$Dinitrogen trioxide$sp^2$Unstable; blue
$NO_2$Nitrogen dioxide$sp^2$Brown gas; paramagnetic
$N_2O_4$Dinitrogen tetroxide$sp^2$Colorless; dimer of NO₂
$N_2O_5$Dinitrogen pentoxide$sp^2$White solid; anhydride of HNO₃

Phosphorus and its Compounds:

Allotropes:

  • White P (P₄): Tetrahedral; very reactive; glows in dark (phosphorescence); poisonous; stored under water
  • Red P: Polymeric; less reactive; non-toxic; used in matches
  • Black P: Most stable; layered structure; semi-conductor

Phosphine (PH₃):

  • Colorless, smelly gas
  • $sp^3$ hybridized P, pyramidal
  • Weak base
  • Preparation: $Ca_3P_2 + 6H_2O \to 3Ca(OH)_2 + 2PH_3$
  • Used in match heads (with KClO₃)

Phosphorus Halides:

$PCl_3$: $P + 3Cl_2 \to PCl_3$ (limited Cl₂); pyramidal; reacts with water: $PCl_3 + 3H_2O \to H_3PO_3 + 3HCl$

$PCl_5$: Trigonal bipyramidal; only stable in gas phase; in solid: $[PCl_4]^+[PCl_6]^-$.

Oxyacids of P:

AcidFormulaBasicityOxidation state of P
Hypophosphorus$H_3PO_2$1 (mono)$+1$
Phosphorous$H_3PO_3$2 (di)$+3$
Phosphoric (Orthophosphoric)$H_3PO_4$3 (tri)$+5$
Pyrophosphoric$H_4P_2O_7$4$+5$
Metaphosphoric$(HPO_3)_n$1 per unit$+5$

Basicity = number of replaceable H atoms (i.e., $-OH$ groups attached to P, not H directly attached to P).

Worked Examples
1

Why doesn't N form $NF_5$ but P forms $PF_5$?

Show solution

N is in second period; only $2s$ and $2p$ orbitals available; max covalency $4$. Cannot expand octet. P is in third period; has empty $3d$ orbitals; can form $sp^3d$ hybridization (5 bonds) and $sp^3d^2$ (6 bonds). Hence $PF_5$, $PCl_5$ exist.

Final Answer: No $d$-orbitals in N, cannot expand octet beyond 4 bonds.

2

Why does $NH_3$ have higher BP than $PH_3$?

Show solution

$NH_3$ forms strong hydrogen bonds (N has high EN, small size, lone pair) — extensive H-bonding network. $PH_3$ doesn't form H-bonds (P has lower EN, doesn't satisfy H-bond requirements: H attached to F, O, N only). BP: $NH_3$ ($-33°$C) >> $PH_3$ ($-87°$C).

Final Answer: $NH_3$ has H-bonding; $PH_3$ does not.

✎ Self-Check — 5 questions0 / 5
Q1.

Most common oxidation states of Group 15:

Q2.

Pure HNO₃ is:

Q3.

Aqua regia is:

Q4.

Most stable allotrope of P:

Q5.

Basicity of $H_3PO_3$:

Group 16 — Oxygen Family and CompoundsTopic 2

Group 16 (Chalcogens): O, S, Se, Te, Po. Outer config: $ns^2\,np^4$.

  • Non-metals: O, S, Se
  • Metalloids: Te
  • Metal: Po (radioactive)

Electronic Configurations:

Element$Z$Config
O8$[He]\,2s^2\,2p^4$
S16$[Ne]\,3s^2\,3p^4$
Se34$[Ar]\,3d^{10}\,4s^2\,4p^4$
Te52$[Kr]\,4d^{10}\,5s^2\,5p^4$
Po84$[Xe]\,4f^{14}\,5d^{10}\,6s^2\,6p^4$

Trends:

PropertyTrend
Atomic radiusIncreases ↓
IEDecreases ↓
ENDecreases ↓
Metallic characterIncreases ↓
Oxidation states$-2, +2, +4, +6$; O mainly $-2$, others $+4, +6$ also

Oxygen:

  • Diatomic ($O_2$); paramagnetic (2 unpaired e⁻ — explained by MOT)
  • $O_3$ (ozone): allotrope, blue-grey gas, V-shaped
  • Most abundant element in Earth's crust (by mass)

Oxides:

TypePropertyExamples
AcidicReact with bases$CO_2, SO_2, P_4O_{10}, SO_3$
BasicReact with acids$Na_2O, MgO, CaO$
AmphotericReact with both$ZnO, Al_2O_3, BeO, PbO$
NeutralNo reaction$CO, NO, N_2O, H_2O$
PeroxideContains $O_2^{2-}$$H_2O_2, Na_2O_2$
SuperoxideContains $O_2^-$$KO_2$

Sulfur:

Allotropes:

  • Rhombic S (S₈): Stable at room T; yellow crystals; insoluble in water
  • Monoclinic S (S₈): Stable above $369$ K; needle-shaped
  • Plastic S, Engel's S, etc. (unstable)

Hydrogen Sulfide (H₂S):

  • Colorless, rotten egg smell, toxic gas
  • Slightly soluble in water; weak acid
  • Reducing agent: oxidized to S
  • $2H_2S + SO_2 \to 3S + 2H_2O$

Sulfur Dioxide ($SO_2$):

  • Colorless, pungent
  • $sp^2$, V-shaped (like ozone)
  • Acidic; forms $H_2SO_3$ with water
  • Reducing: $SO_2 + 2H_2O + Cl_2 \to H_2SO_4 + 2HCl$
  • Oxidizing: $SO_2 + 2H_2S \to 3S + 2H_2O$

Sulfuric Acid (H₂SO₄): "King of chemicals."

Preparation: Contact Process:

  1. Roasting: $S$ or sulfide ore $+ O_2 \to SO_2$
  2. Catalytic oxidation: $2SO_2 + O_2 \xrightarrow{V_2O_5, 720K} 2SO_3$
  3. Absorption (in conc. $H_2SO_4$ to avoid mist formation): $SO_3 + H_2SO_4 \to H_2S_2O_7$ (oleum)
  4. Dilution: $H_2S_2O_7 + H_2O \to 2H_2SO_4$

Properties:

  • Heavy, oily liquid; very viscous
  • Highly hygroscopic (strong dehydrating agent)
  • Strong oxidizing agent when conc. and hot
  • Strong dibasic acid
  • Reactions:
  • Dehydration: $C_{12}H_{22}O_{11} + 11H_2SO_4 \to 12C + 11H_2SO_4 \cdot H_2O$ (sugar → carbon)
  • Oxidation: $Cu + 2H_2SO_4(conc) \to CuSO_4 + SO_2 + 2H_2O$
  • With NaCl: $NaCl + H_2SO_4 \to NaHSO_4 + HCl$ (gas)

Structure of $SO_2, SO_3, H_2SO_4$:

  • $SO_2$: bent, $sp^2$, $\angle = 119.5°$
  • $SO_3$: trigonal planar, $sp^2$
  • $H_2SO_4$: tetrahedral around S, $sp^3$, two -OH groups
Worked Examples
1

Why is $O_2$ paramagnetic?

Show solution

Lewis structure $O=O$ suggests all paired; but experiment shows $O_2$ is paramagnetic (attracted by magnet). MOT explains: in $O_2$, the two $\pi^*$ orbitals are degenerate; Hund's rule places one electron each → 2 unpaired electrons → paramagnetic. Bond order = 2.

Final Answer: Two unpaired electrons in degenerate $\pi^*$ MOs (Molecular Orbital Theory).

2

Predict product: $Cu + dilute\,HNO_3$ vs $Cu + conc.\,HNO_3$.

Show solution

Dilute (cold, $\sim$ 6M): $3Cu + 8HNO_3(dil) \to 3Cu(NO_3)_2 + 2NO\uparrow + 4H_2O$ NO gas evolved (colorless; brown in air). Concentrated: $Cu + 4HNO_3(conc) \to Cu(NO_3)_2 + 2NO_2\uparrow + 2H_2O$ $NO_2$ (brown gas) evolved.

Final Answer: Dil HNO₃ → NO; conc. HNO₃ → NO₂.

✎ Self-Check — 5 questions0 / 5
Q1.

Allotrope of O₂:

Q2.

Most acidic oxide:

Q3.

$H_2S$ smells like:

Q4.

Industrial preparation of H₂SO₄ catalyst:

Q5.

$SO_2$ structure:

2
Module 2

Group 17 and Group 18

Group 17 — Halogens and InterhalogensTopic 1

Group 17 (Halogens): F, Cl, Br, I, At. Outer config: $ns^2\,np^5$.

  • All non-metals; diatomic molecules
  • F, Cl: gases; Br: liquid; I, At: solids
  • Very reactive; want one more electron → strong oxidizing agents

Electronic Configurations:

Element$Z$Config
F9$[He]\,2s^2\,2p^5$
Cl17$[Ne]\,3s^2\,3p^5$
Br35$[Ar]\,3d^{10}\,4s^2\,4p^5$
I53$[Kr]\,4d^{10}\,5s^2\,5p^5$
At85$[Xe]\,4f^{14}\,5d^{10}\,6s^2\,6p^5$ (radioactive)

Trends:

PropertyTrend
Atomic radiusIncreases ↓
IEDecreases ↓
ENDecreases ↓ (F highest)
Oxidizing powerDecreases ↓ (F strongest oxidizing agent)
Bond energy of X-XF-F is anomalously low; otherwise decreases
ColorIncreasing intensity ↓ (Cl: yellow-green, Br: red-brown, I: violet)

Anomalous behavior of F:

  • Highest EN (most reactive non-metal)
  • F-F bond is unusually weak due to small size, lone pair-lone pair repulsion
  • No oxidation state > $-1$ in compounds (cannot expand octet)
  • Forms only one oxoacid (HOF, hypofluorous acid; unstable)

Reactions of Halogens:

1. With Metals: Forms ionic/covalent halides. 2. With Hydrogen: $X_2 + H_2 \to 2HX$. Decreasing reactivity: F (explosive) > Cl > Br > I. 3. With Water: Disproportionation in alkali:

  • Cold: $Cl_2 + 2OH^- \to Cl^- + ClO^- + H_2O$
  • Hot, concentrated: $3Cl_2 + 6OH^- \to 5Cl^- + ClO_3^- + 3H_2O$

Hydrohalic Acids (HX):

  • Strength: HF < HCl < HBr < HI (longer, weaker H-X bond → easier H⁺ release)
  • BP: HF >> HCl < HBr < HI (HF has H-bonding)

Oxoacids of Cl:

AcidFormulaCl ONStrength
HypochlorousHClO$+1$Weakest
Chlorous$HClO_2$$+3$–
Chloric$HClO_3$$+5$–
Perchloric$HClO_4$$+7$Strongest

Strength increases with oxidation state of Cl (more O = more electron withdrawal from -OH).

Bleaching Powder: $CaOCl_2$ or $Ca(OCl)(Cl)$. Preparation: $Ca(OH)_2 + Cl_2 \to CaOCl_2 + H_2O$.

Interhalogen Compounds: Between two halogens. General formula $XY_n$ ($n = 1, 3, 5, 7$). $X$ is less electronegative (larger).

TypeFormulaExampleShape
AB$XY$ClF, BrF, BrCl, IFLinear
$AB_3$$XY_3$$ClF_3, BrF_3, IF_3, ICl_3$T-shaped
$AB_5$$XY_5$$BrF_5, IF_5$Square pyramidal
$AB_7$$XY_7$$IF_7$Pentagonal bipyramidal

Reactivity: $IF_7 > IF_5 > BrF_5 > IF_3 > BrF_3$.

Pseudohalogens: Behave like halogens. Examples: $CN^-, OCN^-, SCN^-$. Form $(CN)_2$ (cyanogen) and similar molecules.

Worked Examples
1

Why is bond energy of F-F lower than Cl-Cl?

Show solution

F atoms are very small; in F-F bond, lone pairs on each F atom are very close → strong lone pair-lone pair repulsion. This weakens the F-F bond. In Cl-Cl, atoms are larger; lp-lp repulsion is less. Bond energies: F-F = $158$ kJ/mol, Cl-Cl = $242$ kJ/mol.

Final Answer: F is small; high lp-lp repulsion weakens F-F bond.

2

Why is HClO₄ strongest of oxoacids of Cl?

Show solution

HClO₄ has $4$ O atoms attached to Cl ($+7$ oxidation state). Conjugate base $ClO_4^-$ has $4$ equivalent resonance structures (negative charge delocalized over $4$ O); highly stable conjugate base. Hence HClO₄ is strongest acid. Strength: $HClO_4 > HClO_3 > HClO_2 > HClO$.

Final Answer: Most O atoms → most stable conjugate base → strongest acid.

✎ Self-Check — 5 questions0 / 5
Q1.

Strongest oxidizing agent in Group 17:

Q2.

Bond energy of F-F vs Cl-Cl:

Q3.

Most basic hydrogen halide:

Q4.

Strongest acid among oxoacids of Cl:

Q5.

Shape of $ClF_3$:

Group 18 — Noble Gases and Xenon CompoundsTopic 2

Group 18 (Noble Gases / Inert Gases): He, Ne, Ar, Kr, Xe, Rn. Outer config: $ns^2\,np^6$ (except He: $1s^2$).

  • All gases at room T
  • Monatomic; no bonding (high IE, no valence requirement)
  • Inertness due to filled valence shell

Electronic Configurations:

Element$Z$Config
He2$1s^2$
Ne10$[He]\,2s^2\,2p^6$
Ar18$[Ne]\,3s^2\,3p^6$
Kr36$[Ar]\,3d^{10}\,4s^2\,4p^6$
Xe54$[Kr]\,4d^{10}\,5s^2\,5p^6$
Rn86$[Xe]\,4f^{14}\,5d^{10}\,6s^2\,6p^6$ (radioactive)

Trends:

PropertyTrend
Atomic radiusIncreases ↓
IEHighest in period; decreases ↓ (Rn lowest in group)
BP, MPVery low (Van der Waals only); increase ↓
ReactivityAll very low; reactivity increases ↓ from Xe

Occurrence: $\sim 0.0124\%$ in air; mostly Ar.

Uses:

  • He: balloons, deep-sea diving, MRI
  • Ne: discharge lamps, advertising signs
  • Ar: inert atmosphere in welding, light bulbs
  • Kr: photography, lasers
  • Xe: photographic flash bulbs, lasers
  • Rn: tracer in radiation therapy

Xenon Compounds (Discovered 1962 by Bartlett):

Xe can form compounds due to large size, low IE (relatively), and availability of $d$ orbitals.

Fluorides:

CompoundHybridizationShapeNotes
$XeF_2$$sp^3d$Linear (3 lp at equatorial)Made from Xe + F₂
$XeF_4$$sp^3d^2$Square planar (2 lp axial)–
$XeF_6$$sp^3d^3$Distorted octahedral (1 lp)–

Other Compounds:

  • $XeOF_4$: pyramidal (with 1 lp)
  • $XeO_3$: pyramidal
  • $XeO_4$: tetrahedral
  • $XePtF_6$: first Xe compound (1962, by Bartlett)

Hydrolysis of XeF compounds:

  • $XeF_2 + H_2O \to Xe + HF + O_2$
  • $6XeF_4 + 12H_2O \to 4Xe + 2XeO_3 + 24HF + 3O_2$
  • $XeF_6 + 3H_2O \to XeO_3 + 6HF$

KrF₂ also exists. No compounds of He, Ne, Ar known.

Worked Examples
1

Why are noble gases monoatomic?

Show solution

Noble gases have completely filled valence shells ($ns^2\,np^6$ or $1s^2$ for He). Already have stable noble gas configuration. No tendency to gain/lose/share electrons to form bonds. At very low T, weak London forces allow them to condense to liquid/solid, but no chemical bonding occurs in monoatomic gas form.

Final Answer: Filled valence shells; no bonding tendency.

2

Predict hybridization and shape of $XeF_4$.

Show solution

Xe in $XeF_4$: Xe central atom. Electrons around Xe = 8 (its 8 valence e⁻) + 4 from F − 4 shared = unclear directly.

Better: Xe has 8 valence e⁻. Forms 4 Xe-F bonds, using 4 e⁻; leaving 4 e⁻ as 2 lone pairs. Total electron pairs around Xe = $4 + 2 = 6$ → $sp^3d^2$. Geometry: octahedral with 2 lp; lp's at axial positions → square planar shape.

Final Answer: $sp^3d^2$ hybridization; square planar shape.

✎ Self-Check — 5 questions0 / 5
Q1.

Noble gases have valence config:

Q2.

First Xe compound by Bartlett:

Q3.

Hybridization of Xe in $XeF_2$:

Q4.

Shape of $XeO_4$:

Q5.

Noble gas used in welding (inert atmosphere):

Ready to test yourself?

Attempt the full timed mock tests — Main & Advanced level.

Start Mock Test 1 →