p-Block Elements JEE Notes, Download PDF & Formulas

p-Block Elements JEE Notes: Important Concepts

The p-block contains elements in which the last electron enters a p-orbital. Key features to keep in mind while revising:

• Electronic configuration: $$ns^2np^{1-6}$$ (increases across the block).
• Common oxidation states: maximum state = group number; inert pair effect generally stabilises lower states ($$ns^2$$ electrons remain non-bonding).
• Periodic trends: electronegativity and ionisation enthalpy rise from left to right, metallic character and atomic size drop, but anomalies appear due to d- and f-block contraction (Ga, In, Tl).
• Catenation: strong for C > Si > Ge; practically absent for Pb.
• Allotropy: multiple forms exist for C, P, S, O and Se; learn structural differences and stability order (diamond > graphite thermodynamically above 1500 K).
• Acid–base behaviour of oxides: basic → amphoteric → acidic across a period; down a group, acidity decreases.
• Maximum covalency: based on availability of vacant d-orbitals (e.g., $$PCl_5, SF_6, IF_7$$) except in C, N, O and F where it is limited to four.

Memorising these general ideas cuts down time spent on individual group trends because every exception in the exam is built on them.

Group 13 & 14 Highlights: Boron, Aluminium, Carbon and Silicon Families

Boron Family (Group 13)

• Anomalous Boron: non-metal, forms covalent hydrides $$\text{B}_n\text{H}_m$$, strong Lewis acid $$BF_3$$, $$BCl_3$$.
• Dimeric halides: $$Al_2Cl_6$$ exists in vapour and solid; degree of dimerisation falls down the group.
• Oxides: $$B_2O_3$$ acidic, $$Al_2O_3$$ amphoteric, $$Ga_2O_3$$ weakly basic, $$In_2O_3, Tl_2O$$ strongly basic.
• Inert pair effect: +1 state dominates in Tl (disproportionation of $$TlCl_3$$ → $$TlCl + Cl_2$$ in cold water).
• Borax bead test: $$Na_2B_4O_7·10H_2O$$ → on heating → $$NaBO_2 + B_2O_3$$; coloured beads with metal oxides help in qualitative analysis.

Carbon Family (Group 14)

• Allotropes of Carbon: diamond (sp³), graphite (sp²), fullerenes $$C_{60}$$, nanotubes.
• Catenation order: $$C \gg Si \gt Ge \gt Sn \gt Pb$$ (bond energy falls sharply after silicon).
• Oxidation states: +4 common; +2 gains stability down the group (pbCl₂ vs. pbCl₄).
• Inorganic chains: $$SiO_2$$ network, silicates, zeolites, silicones $$[–R_2SiO–]_n$$ important for polymer-type numericals.
• Important reaction: $$SiCl_4 + 2H_2O \rightarrow SiO_2 + 4HCl$$ (violent hydrolysis, contrasts with $$CCl_4$$ which resists hydrolysis).

When revising these two groups, practise concept application through mixed numericals. The concept modules in JEE Mains Online Coaching often begin with such cross-link questions where trends, structure and reactivity must be invoked together.

Group 15 & 16 Highlights: Nitrogen, Phosphorus, Oxygen and Sulphur Families

Nitrogen Family (Group 15)

• Molecular vs. metallic: $$N_2$$ forms $$p\pi–p\pi$$ triple bond (gives inertness), from P onward catenated polyatomic molecules $$P_4, As_4$$ appear; Bi shows metallic character.
• Anomalous Nitrogen: absence of d-orbitals restricts its covalency to four, hence $$NF_3$$ but not $$NF_5$$.
• Oxidation states: –3 to +5; stability of –3 drops down the group, +3 (inert pair) becomes more stable in Bi.
• Interconversion reactions: $$NH_3 + ClO^-$$ → $$N_2$$, $$HNO_2$$ ↔ $$NO + NO_2$$; learn redox balancing.
• Industrial processes: Ostwald (nitric acid) and Haber's (ammonia) – constants directly asked.

Oxygen Family (Group 16)

• Anomalous Oxygen: diatomic, gaseous, shows –2, –1 (peroxides) and –½ (superoxides) states; no vacant d-orbitals.
• Increasing metallic character: $$O < S < Se < Te < Po$$.
• Allotropes: $$S_8$$ puckered ring, $$S_6$$, plastic sulphur.
• Trends in acidity: $$H_2O < H_2S < H_2Se < H_2Te$$ (bond dissociation decreases despite lowering electronegativity).
• Important oxoacids: $$H_2SO_3, H_2SO_4, H_2S_2O_7$$; remember structure, basicity and oxidising power.

To strengthen these groups, attempt 15–20 mixed concept questions from the topic bank inside JEE Questions after each sub-topic. Tracking errors immediately improves recall of exceptions.

Group 17 & 18 Highlights: Halogens and Noble Gases

Halogens (Group 17)

• Oxidising ability: $$F_2 > Cl_2 > Br_2 > I_2$$ in aqueous medium (hydrate enthalpy dominates).
• Interhalogen compounds: general type $$AX_n$$ (n = 1,3,5,7); all are covalent, more reactive than parent halogens.
• Polyhalide ions: $$I_3^-$$ structure linear $$I–I–I$$, used in iodine tincture analysis.
• Oxoacids trend: acid strength increases with oxidation number: $$HOCl < HClO_2 < HClO_3 < HClO_4$$.
• Bleaching reactions: $$Cl_2 + H_2O \rightarrow HCl + HClO$$ (in situ nascent O forms bleaching action), $$SO_2$$ acts by reduction so reversible.

Noble Gases (Group 18)

• Compounds of Xe: $$XeF_2, XeF_4, XeF_6, XeO_3, XeO_4$$ – VSEPR shapes often tested (linear, square planar, octahedral with distortion).
• Clathrate hydrates: $$Ar·6H_2O, Kr·6H_2O$$ formed at low T, high P; release gas on warming.
• Separation methods: fractional distillation of liquid air, adsorption on coconut charcoal (temperature swing).
• Uses: He in cryogenics/Supersonic, Ne in neon signs, Ar in inert welding atmosphere.

Advanced examiners frequently design assertion–reason or multi-concept numericals combining VSEPR, hybridisation and oxidation states of xenon. Reviewing those in the JEE Advanced Previous Papers archive clarifies pattern recognition.

Important Formulas and Results at a Glance

$$\text{Electronegativity trend (across period)}: \uparrow \quad \quad \text{Electronegativity trend (down group)}: \downarrow$$

$$\text{Acid strength of hydrides}: HX \uparrow \;\; \text{with} \;\; \text{decreasing} \; H–X \; \text{bond energy}$$

$$\text{Bond order of } N_2 = 3 \quad ; \quad \mu = 0 \; (\text{diamagnetic})$$

For a clean, printable collection of JEE Formula Sheets, clip this table to your study diary and mark uncertain entries in red for quick recall.

Solved Example 1: Trend Application

Arrange the following oxides in increasing acidic character: Al₂O₃, Ga₂O₃, In₂O₃, B₂O₃.

Solution: Across the period acidity increases, down the group basicity increases. Therefore

In₂O₃ (most basic) < Ga₂O₃ < Al₂O₃ < B₂O₃ (most acidic).
Answer: In₂O₃ < Ga₂O₃ < Al₂O₃ < B₂O₃

Solved Example 2: Calculation of Percentage Purity

5.0 g of impure CaCO₃ is treated with excess dilute HCl. The CO₂ produced is passed into 500 mL of 0.1 M Ba(OH)₂. The residual Ba(OH)₂ requires 20 mL of 0.2 M HCl for neutralisation. Calculate percentage purity of CaCO₃.

Solution:

1. Ba(OH)₂ moles initially = 0.5 L × 0.1 M = 0.05 mol.
2. HCl unused Ba(OH)₂ moles = (0.02 L × 0.2 M)/2 = 0.002 mol (because Ba(OH)₂:HCl = 1:2).
3. Ba(OH)₂ consumed by CO₂ = 0.05 – 0.002 = 0.048 mol (stoichiometry 1:1).
4. Thus CO₂ moles = 0.048 mol → CaCO₃ moles = 0.048 mol.
5. Pure CaCO₃ mass = 0.048 × 100 g mol⁻¹ = 4.8 g.
6. Percentage purity = (4.8 / 5.0) × 100 = 96 %.

Solved Example 3: VSEPR & Hybridisation

Predict the geometry and hybridisation of XeF₄.

Solution: Xe has 8 valence electrons, F contributes 4 × 1 = 4. Total 12 electrons = 6 electron pairs. Out of these, 4 are bond pairs and 2 are lone pairs: AX₄E₂ type. Hybridisation = $$sp^3d^2$$ (octahedral basis) and lone pairs occupy axial positions leading to square planar geometry.

Solved Example 4: Redox Disproportionation

Write the balanced equation for disproportionation of H₃PO₃ in alkaline medium.

Solution: Oxidation numbers: P = +3. Products are H₂PHO₂ (PH₃O₂) where P = +1 and HPO₄²⁻ where P = +5.

Balanced equation:
$$2H_3PO_3 \rightarrow H_2PHO_2 + HPO_4^{2-} + 2H^+ + 2e^-$$ adjusted with OH⁻ and H₂O gives final alkaline form:
$$3H_3PO_3 \rightarrow H_2PHO_2 + 2HPO_4^{2-} + 2H^+ + 2e^-$$ (students may simplify by half-reaction method).

JEE Important Points, Common Mistakes and Quick Revision

• Watch lower oxidation state stability: Students often forget that +1 for Tl, +2 for Pb and +3 for Bi dominate due to inert pair effect. Examiner likes to frame MCQ on the ease of reduction or disproportionation.
• Bleaching powder misconception: The active ingredient is $$Ca(OCl)_2$$, not $$CaCl_2$$. Reaction with dil. acid releases $$Cl_2$$, not nascent oxygen directly.
• Incorrect bond angles in hydrides: NH₃ > PH₃ > AsH₃ because of decreasing electronegativity causing larger bp–lp repulsion drop.
• Hybridisation short-cut errors: For molecules with odd number of electrons (e.g., IF₇) always compute steric number; do not equate directly with bonds.
• Physical vs chemical bleaching: SO₂ bleaches by reduction (reversible), Cl₂ by oxidation (irreversible) – a favourite assertion–reason pair.

After revising, solve the last 10 years of JEE Mains Previous Papers for this chapter in one sitting. Set a stopwatch for 45 minutes and mark any question you solve in over 2 minutes for second-round practice.