JEE Main & Advanced

p-Block Elements (Class 11)

p-Block Elements (Class 11) for JEE Main & Advanced

1
Module 1

Group 13 (Boron Family)

Trends, Anomalies of BoronTopic 1

Group 13 (Boron family): B, Al, Ga, In, Tl. Outer config: $ns^2\,np^1$.

  • B is metalloid
  • Al, Ga, In, Tl are metals

Electronic Configurations:

Element$Z$Config
B5$[He]\,2s^2\,2p^1$
Al13$[Ne]\,3s^2\,3p^1$
Ga31$[Ar]\,3d^{10}\,4s^2\,4p^1$
In49$[Kr]\,4d^{10}\,5s^2\,5p^1$
Tl81$[Xe]\,4f^{14}\,5d^{10}\,6s^2\,6p^1$

General Trends:

PropertyTrend
Atomic radiusB < Al > Ga (Ga anomaly: filled $3d$ poorly shields) ≈ In < Tl
IEGenerally decreases down group; Ga > Al (anomaly)
Metallic characterIncreases ↓ (B is non-metal/metalloid)
Oxidation states$+3$ common; $+1$ becomes important from Ga downward (inert pair effect)

Inert Pair Effect: Tendency of $ns^2$ electrons to remain inert (not participate in bonding) in heavier elements due to poor shielding by $d$ and $f$ electrons. Tl shows stable $+1$ rather than $+3$.

Anomalous Properties of Boron:

  • Smallest size, highest IE in group
  • Non-metal (others metals)
  • Only covalent compounds (no $B^{3+}$ ion in compounds)
  • Maximum covalency = $4$ (no $d$-orbitals)
  • Highly stable boron-hydrogen compounds (boranes)

Differences B vs Al:

PropertyBoronAluminium
NatureNon-metalMetal
Reaction with acidsVery slow / passiveReactive
Halides Lewis acidityBX₃ are stronger Lewis acids (no back-bonding) than AlX₃AlCl₃ dimeric
Hydroxide nature$B(OH)_3$ weakly acidic$Al(OH)_3$ amphoteric
Oxide$B_2O_3$ acidic$Al_2O_3$ amphoteric

Lewis Acid Strength of BX₃: Surprisingly, despite F being most EN: $BF_3 < BCl_3 < BBr_3 < BI_3$. This is because back-bonding ($p\pi-p\pi$) from halogen to B partially fills B's empty $p$ orbital — strongest with F (best size match), weakest with I.

Aluminium:

  • Most abundant metal in Earth's crust ($\sim 8.3\%$)
  • Bauxite ($Al_2O_3 \cdot 2H_2O$) is the main ore
  • Hall-Héroult process: Electrolysis of molten $Al_2O_3$ in cryolite ($Na_3AlF_6$)
  • Amphoteric: dissolves in both acids and alkalis
  • Forms protective oxide layer (resistant to corrosion)
Worked Examples
1

Why is $BCl_3$ a stronger Lewis acid than $BF_3$?

Show solution

B has empty $2p$ orbital → can accept e⁻ pair (Lewis acid). In $BF_3$: F has filled $2p$ that overlaps well with B's empty $2p$ → strong back-bonding makes B less electron deficient. In $BCl_3$: Cl has $3p$, size mismatch with B's $2p$ → weak back-bonding. B more electron-deficient → stronger Lewis acid. Order: $BF_3 < BCl_3 < BBr_3 < BI_3$.

Final Answer: Weaker back-bonding in $BCl_3$ vs $BF_3$.

2

Write balanced equation for Al with NaOH.

Show solution

$2Al + 2NaOH + 2H_2O \to 2NaAlO_2 + 3H_2$ (sodium meta-aluminate + H₂)

In hot concentrated NaOH: $2Al + 6NaOH \to 2Na_3AlO_3 + 3H_2$.

Final Answer: Aluminate + H₂.

✎ Self-Check — 5 questions0 / 5
Q1.

Most metallic in Group 13:

Q2.

Inert pair effect is most pronounced in:

Q3.

$B(OH)_3$ in water is:

Q4.

Most stable oxidation state of Tl:

Q5.

Hall-Héroult process is used for:

Boron Compounds — Borax, Boric Acid, DiboraneTopic 2

Borax (Na₂B₄O₇·10H₂O):

Structure (in solid): $[B_4O_5(OH)_4]^{2-}$ ion with $2$ Na⁺.

Two of the four B atoms are $sp^2$ hybridized (planar, trigonal); two are $sp^3$ (tetrahedral). Anion has formula $[B_4O_5(OH)_4]^{2-}$.

Borax bead test: When heated, borax loses water → swells → forms transparent glassy bead of $Na_2B_4O_7 + 2NaBO_2 + B_2O_3$. With colored salts gives characteristic colors.

E.g., copper salt + borax bead → blue (Cu²⁺) → red (Cu in reducing flame).

Preparation of Borax:

  • From colemanite ($Ca_2B_6O_{11}$): boil with Na₂CO₃; CaCO₃ precipitates, borax remains in solution.

Boric Acid (H₃BO₃ or B(OH)₃):

Preparation:

  • From borax: $Na_2B_4O_7 + 2HCl + 5H_2O \to 4H_3BO_3 + 2NaCl$
  • Hydrolysis of $BX_3$ or boric oxide

Structure: Trigonal planar. H-bonded layered structure in solid (each B(OH)₃ unit H-bonded to 3 others).

Properties:

  • White crystalline solid; sparingly soluble in cold water; more soluble in hot water
  • Weak monoprotic Lewis acid: $B(OH)_3 + 2H_2O \to [B(OH)_4]^- + H_3O^+$
  • It does NOT donate H⁺ directly; instead accepts OH⁻ from water — Lewis acid behavior
  • $K_a \approx 6 \times 10^{-10}$

Heating Behavior: $$H_3BO_3 \xrightarrow{370 K} HBO_2 \xrightarrow{red\ heat} B_2O_3$$ (meta-boric acid → boric oxide)

Uses: Mild antiseptic, eye washes, manufacture of borosilicate glass, enamels.

Diborane (B₂H₆):

Preparation:

  • $2BF_3 + 6NaH \xrightarrow{450K} B_2H_6 + 6NaF$
  • $2NaBH_4 + I_2 \to B_2H_6 + 2NaI + H_2$

Structure: "Banana bonds" or 3-center-2-electron (3c-2e) bonds — two B-H-B bridges.

  • 2 H atoms are terminal (one per B), 4 are terminal end-on, 2 are bridging.
  • Actually: 4 terminal H + 2 bridging H.
  • Each B is $sp^3$ hybridized.
  • Both B share 2 bridging H atoms with delocalized 3c-2e bonds.
  • B-B distance ≈ $1.77$ Å; bridging B-H ≈ $1.33$ Å; terminal B-H ≈ $1.19$ Å.

Properties:

  • Colorless gas; toxic; spontaneously inflames in air
  • $B_2H_6 + 3O_2 \to B_2O_3 + 3H_2O + \text{heat}$
  • Hydrolysis: $B_2H_6 + 6H_2O \to 2B(OH)_3 + 6H_2$
  • Hydroboration of alkenes (Brown's reaction; used in synthesis of alcohols)
  • Reacts with NH₃ → Borazine ($B_3N_3H_6$, "inorganic benzene")

Higher Boranes: $B_5H_9, B_{10}H_{14}$ etc. Show similar 3c-2e bonding patterns.

Worked Examples
1

Why is the structure of diborane unique?

Show solution

$B_2H_6$ has only $12$ valence electrons but needs $14$ for normal Lewis structure (4 terminal + 2 bridge B-H + 1 B-B = 7 bonds × 2 = 14 e⁻). Solution: 2 bridging H atoms share their electrons with both B atoms simultaneously, forming 3-center-2-electron (3c-2e) banana bonds. Two bridge bonds + four normal terminal B-H bonds = $4 + 2 \times 2 = 8$ electron pairs, but only $6$ pairs of electrons are actually present (the bridge pairs are shared among 3 atoms).

Final Answer: Electron-deficient; uses 3c-2e bonds.

2

Why is boric acid considered a Lewis acid?

Show solution

$B(OH)_3$ has only $6$ valence e⁻ around B (octet incomplete). Accepts OH⁻ from water: $B(OH)_3 + H_2O \rightleftharpoons [B(OH)_4]^- + H^+$ Acidity comes from accepting electron pair, not directly donating H⁺.

Final Answer: Accepts OH⁻ from water → acts as Lewis acid; not a Bronsted acid.

✎ Self-Check — 5 questions0 / 5
Q1.

Borax formula:

Q2.

Boric acid is:

Q3.

Number of 3c-2e bonds in $B_2H_6$:

Q4.

Hydrolysis of diborane gives:

Q5.

Borax bead test uses:

2
Module 2

Group 14 (Carbon Family)

Trends, Allotropes of CarbonTopic 1

Group 14: C, Si, Ge, Sn, Pb. Outer config: $ns^2\,np^2$.

  • C, Si: non-metals
  • Ge: metalloid
  • Sn, Pb: metals

Trends:

PropertyTrend
Atomic radiusIncreases ↓
IEDecreases ↓
Oxidation states$+4$ predominant; $+2$ becomes important down (inert pair)
Metallic characterIncreases ↓
Tendency for catenationC > Si >> Ge > Sn ≈ Pb
Stability of $+4$Decreases ↓
Stability of $+2$Increases ↓ (Pb²⁺ > Pb⁴⁺ stability)

Catenation: Property of an element to form chains/rings with itself. Most pronounced in carbon due to:

  • Small size (strong C-C bond, $\sim 348$ kJ/mol)
  • $sp^3, sp^2, sp$ hybridization
  • Tetravalency

Other elements show decreasing catenation; Si forms only short chains (silicates have Si-O-Si).

Carbon-Halogen Bonds:

  • CCl₄ is stable in water (no hydrolysis); SiCl₄ readily hydrolyzes (Si has $d$-orbitals)
  • $SnCl_4, PbCl_4$ hydrolyze easily

Allotropes of Carbon:

AllotropeStructureProperties
Diamond$sp^3$ tetrahedral 3D networkHardest natural substance; insulator; high refractive index
Graphite$sp^2$ planar hexagonal layersConducts electricity along layers; soft (slippery); used as lubricant, electrode
Fullerene (C₆₀)$sp^2$ hexagonal+pentagonal "soccer-ball" cageDiscrete molecule; "buckminsterfullerene"
Carbon nanotubesRolled-up graphite sheetsExtremely strong; conducting/semiconducting depending on geometry
Amorphous (coal, charcoal, soot)Disordered $sp^2$/$sp^3$Used for adsorption (activated charcoal)

Diamond Structure: Each C bonded to 4 others tetrahedrally. C-C bond length $1.54$ Å. Very high lattice energy.

Graphite Structure: Layers of hexagonal rings; each C bonded to 3 others ($sp^2$); C-C in layer $1.42$ Å; layers $3.35$ Å apart (van der Waals); 4th electron delocalized → conduction.

Fullerene: C₆₀ molecule shaped like soccer ball; $20$ hexagons + $12$ pentagons; $sp^2$ but slightly pyramidalized (curvature).

Worked Examples
1

Why does diamond have higher melting point than graphite (theoretically)? Actually graphite has slightly higher MP.

Show solution

Both have very strong covalent bonds, but in graphite, each layer is held by C-C ($\sim 1.42$ Å, very strong, slightly shorter than diamond's $1.54$ Å), and to break the structure, one must break covalent bonds. Diamond: $sp^3$ network, 3D. Graphite: $sp^2$ within layer; even higher bond strength due to partial $\pi$-bonding character. At very high T (sublimation), graphite sublimes slightly above diamond's melting point.

Final Answer: Both very high; graphite's MP > diamond due to stronger in-plane bonds.

2

Why does SiCl₄ react with water but CCl₄ does not?

Show solution

SiCl₄: Si has empty $3d$ orbitals → can accept lone pair from $H_2O$ → easy hydrolysis: $SiCl_4 + 4H_2O \to Si(OH)_4 + 4HCl$. CCl₄: C has no $d$-orbitals; cannot expand octet; no path for hydrolysis.

Final Answer: Availability of $d$-orbitals in Si enables hydrolysis.

✎ Self-Check — 5 questions0 / 5
Q1.

Most stable oxidation state of Pb:

Q2.

Hardest allotrope of carbon:

Q3.

Graphite conducts electricity due to:

Q4.

Property of catenation is best shown by:

Q5.

CCl₄ in water:

Carbon Compounds, Silicates, SiliconesTopic 2

Oxides of Carbon:

OxidePropertiesNotes
COColorless, odorless, very toxicForms strong complex with hemoglobin (200× O₂'s affinity)
CO₂Colorless, slightly solubleGreenhouse gas; in dry ice form is solid
C₃O₂Carbon suboxideRare; OCCCO structure

Carbon Monoxide (CO):

  • Bonding: $:C\equiv O:$ — triple bond with both lone pairs
  • Excellent ligand (forms metal carbonyls); $\sigma$-donor from C, $\pi$-acceptor
  • Strong reducing agent

Carbon Dioxide (CO₂):

  • $O=C=O$ linear, $sp$ hybridized C
  • Forms carbonic acid: $CO_2 + H_2O \rightleftharpoons H_2CO_3$
  • Solid form ("dry ice") sublimes; used as refrigerant
  • Green house gas; rising levels causing global warming

Silicon Compounds:

Silica ($SiO_2$):

  • 3D giant covalent network with $sp^3$ Si bonded to 4 O atoms (tetrahedrally)
  • Quartz, sand, glass forms
  • High MP ($\sim 1710°$C); insoluble; chemically inert
  • Reacts with HF: $SiO_2 + 4HF \to SiF_4 + 2H_2O$
  • Reacts with NaOH: $SiO_2 + 2NaOH \to Na_2SiO_3 + H_2O$

Silicates: Compounds containing $SiO_4^{4-}$ tetrahedra. Building unit; can polymerize.

TypeStructureExample
OrthosilicateIsolated $SiO_4^{4-}$$Mg_2SiO_4$ (olivine)
Pyrosilicate$Si_2O_7^{6-}$ — 2 tetrahedra sharing 1 OThortveitite
CyclicRings of tetrahedraBeryl $Be_3Al_2Si_6O_{18}$
ChainSingle chainPyroxenes
Double chainTwo chains linkedAsbestos (amphiboles)
Sheet2D layersTalc, mica, clays
3D frameworkAll four O sharedQuartz, feldspars, zeolites

Silicones: Synthetic polymers with general formula $R_2SiO$, alternating Si-O-Si backbone with R = alkyl/aryl groups.

Preparation: $R_2SiCl_2 + 2H_2O \to R_2Si(OH)_2 + 2HCl$ Condensation: $nR_2Si(OH)_2 \to (R_2SiO)_n + nH_2O$

Properties:

  • Stable to heat and chemicals
  • Water-repellent
  • Excellent electrical insulators
  • Used in lubricants, water-repellent coatings, surgical implants, sealants

Important Silicon Compounds:

  • SiC (Silicon Carbide / Carborundum): Very hard ($\sim 9.5$ on Mohs); abrasive
  • Si₃N₄: Refractory ceramic
  • Silanes (SiH₄ etc.): Analogous to alkanes but reactive (less stable than alkanes)

Zeolites: Crystalline aluminosilicates with porous structure. Used as:

  • Molecular sieves (separate molecules by size)
  • Catalysts in petroleum cracking
  • Water softeners (ion exchange)
Worked Examples
1

Why is CO poisonous?

Show solution

CO binds to hemoglobin's iron with ~$200\times$ greater affinity than O₂. This blocks O₂ transport in blood; cells suffocate (hypoxia). Severe poisoning leads to death.

Final Answer: Forms stable carboxyhemoglobin; blocks O₂ uptake.

2

Write structures of: $SiO_4^{4-}$, $Si_2O_7^{6-}$.

Show solution

$SiO_4^{4-}$: tetrahedral, Si bonded to 4 O atoms with charge $-4$. $Si_2O_7^{6-}$: two $SiO_4$ tetrahedra sharing one O corner; ${}_5O = ⏐$ structure with $1$ shared O between two Si.

Final Answer: Orthosilicate (isolated tetrahedron); pyrosilicate (two tetrahedra sharing one O).

✎ Self-Check — 5 questions0 / 5
Q1.

CO is dangerous because it:

Q2.

Carborundum is:

Q3.

Silicones are:

Q4.

Zeolites are used as:

Q5.

Diamond structure has:

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