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Chemistry — Class 12 — L18: General Characteristics of p-Block Elements

NIOS Code 313 · Module 6 · Chemistry of Elements

Notes extracted from NIOS Chemistry Course (313), Lesson 18 — General Characteristics of p-Block Elements (313_Chemistry_Eng_Lesson18.pdf). Content covers sections 18.1–18.10.
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Overview — Module 6: General Characteristics of the p-Block Elements

The p-block comprises Groups 13, 14, 15, 16, 17 and 18 — elements characterised by filling of outermost p-orbitals. Outer configuration is ns² np¹–⁶. These elements and their compounds shape daily life: nitrogen in ammonia and fertilizers (and explosives like TNT); oxygen for respiration and combustion; carbon chains in carbohydrates, proteins and vitamins.

Vertical similarity is less marked in the p-block than in the s-block (especially Groups 13 and 15), but later groups show clearer family behaviour. Horizontal trends across a period are regular. This lesson covers occurrence, electronic configuration, atomic size, ionization enthalpy, electron gain enthalpy, electronegativity, metallic/non-metallic character, anomalous first members, inert pair effect, and general trends in hydrides, oxides and halides.

Section 1: Occurrence and Configuration (18.1–18.2)

p-Block elements do not follow one mode of occurrence. Some exist free and combined (O₂, N₂, C, S); noble gases only free; most others only combined. Abundance is uneven: O, Si, Al, N are plentiful; heavier members of each group are rarer.

Five rows of p-block correspond to filling 2p, 3p, 4p, 5p and 6p. Outer configuration: ns² np¹ (Group 13) through ns² np⁶ (Group 18, noble gases).

p-Block Groups 13–18 13ns²np¹ 14ns²np² 15ns²np³ 16ns²np⁴ 17ns²np⁵ 18ns²np⁶ Metal character ← left · Non-metal character → right Filling of outermost p-orbitals defines the block
Six groups of p-block with outer configurations ns²np¹ to ns²np⁶.

Section 2: Atomic Size and Ionization Enthalpy (18.3–18.4)

Atomic size decreases left to right across a period: electrons enter the same valence shell while nuclear charge rises, increasing effective nuclear charge (B 88 pm → F 64 pm). Down a group, size increases because new shells are added (B 88 → Tl 178 pm in Group 13) — the extra shell outweighs increased nuclear charge.

Size ↓ across period  |  Size ↑ down group
Same shell + ↑ Z_eff across · New shells down · Controls IE, EN, metallic character

First ionization enthalpy is energy to remove the most loosely bound electron from a gaseous atom (kJ mol⁻¹). It generally increases across a period (smaller atoms hold electrons more tightly) and decreases down a group (larger atoms, weaker attraction).

Important exception: Group 15 elements have higher first IE than Group 16 (N > O; P > S). Removing an electron from half-filled p³ is harder than from p⁴. Full/half-filled configurations are extra stable — classic exam trap.

IE Exception: N vs O N: 2s² 2p³ Half-filled · stable IE = 1403 kJ mol⁻¹ O: 2s² 2p⁴ Paired e⁻ in p IE = 1310 kJ mol⁻¹ Half-filled p³ → higher IE than next element
Nitrogen has higher first ionization enthalpy than oxygen despite smaller size of O.

Sample values (kJ mol⁻¹): B 801, C 1086, N 1403, O 1310, F 1681, Ne 2080. Al 577, Si 796, P 1062, S 999, Cl 1255, Ar 1521 — same pattern in the third period.

Intext 18.1 style comparisons: F is smaller than Cl; C smaller than Si; C smaller than B? No — B is larger than C across the period. Higher IE: Be > B (full 2s² of Be); Cl > S; He > Ne; O > S. Increasing IE order often Na < Be < N < He. Always compare size first, then half-filled/full-filled exceptions.

Section 3: Electron Gain Enthalpy and Electronegativity (18.5–18.6)

Electron gain enthalpyegH) is the energy change when an electron is added to a neutral gaseous atom: X(g) + e⁻ → X⁻(g). Usually negative (energy released). It becomes more negative across a period (smaller size, stronger nuclear attraction) and less negative down a group (larger size).

Famous exception: chlorine has more negative ΔegH than fluorine. The F atom is so small that adding an electron causes severe interelectronic repulsion; Cl’s larger size makes electron addition more favourable. Similar effects appear for first members of other groups.

ΔegH(Cl) more negative than ΔegH(F)
Small F → high e⁻–e⁻ repulsion · Cl larger → more favourable electron gain

Electronegativity measures ability of an atom in a covalent bond to attract the shared pair. It increases across a period and decreases down a group. Order of highest EN: F > O > N. Fluorine is the most electronegative element.

Section 4: Metallic and Non-Metallic Behaviour (18.7)

Metals form positive ions (lose e⁻); non-metals form negative ions (gain e⁻). Across a period: metallic character decreases, non-metallic increases (size ↓, IE ↑). Down a group: metallic character increases, non-metallic decreases (size ↑, IE ↓). Thus carbon is a non-metal, lead is metallic; fluorine is the most non-metallic element, while thallium is metallic.

Noble gases have near-zero electron affinity and very high IE — little tendency to gain or lose electrons under normal conditions — explaining their chemical inertness (with exceptions for heavier noble gas compounds under special conditions).

Section 5: Anomalous First Elements (18.8)

The first element of each p-block group (B, C, N, O, F) differs sharply from heavier congeners. Causes:

  • Small size and compact 2p orbitals — strong interelectronic repulsion; N–N, O–O, F–F single bonds weaker than P–P, S–S, Cl–Cl.
  • High electronegativity — strong H-bonding in X–H···Y where X, Y = N, O, F.
  • pπ–pπ multiple bonds — C=C, C≡C, N≡N, O=O effective with 2p; Si, P, S 3p orbitals too large for good π-overlap.
  • Maximum coordination number 4 — no d-orbitals in valence shell. BF₄⁻ and NH₄⁺ contrast with [AlF₆]³⁻ and [PCl₆]⁻ for heavier elements that use d-orbitals.
Period 2 vs Heavier Congeners Period 2 pπ–pπ: O=O, N≡N Max CN = 4 H-bonds (N,O,F) Period 3+ Weak pπ–pπ CN 6 with d-orbitals S₈ solid · P₄ · Cl₂ O₂ gas vs S solid: multiple bonding vs catenation
Oxygen is a gas (O=O); sulphur is a solid (S₈) because 3p π-overlap is poor.

This explains why oxygen is a diatomic gas while sulphur is a yellow solid of S₈ rings, and why carbon chemistry is so rich in multiple bonds compared with silicon.

Section 6: Inert Pair Effect (18.9)

In Groups 13, 14 and 15, higher oxidation states become less stable down the group. B and Al are almost always +3; Tl is stable as +1. C is tetravalent; Ge, Sn, Pb show +2 (Pb²⁺ very stable). Sb and Bi prefer +3 over +5.

Outer configs ns²np¹, ns²np², ns²np³ suggest +3, +4, +5 — but heavy elements often leave the ns² pair non-bonding. This reluctance of s-electrons to bond is the inert pair effect.

Inert pair: ns² reluctant to bond → lower OS stable (Tl⁺, Pb²⁺, Bi³⁺)
Causes: high promotion energy ground→valence state · poorer orbital overlap · weaker bonds for large atoms

Two physical causes: (1) high promotion energy from ground state (e.g. ns²np¹) to valence state (ns¹np²); (2) poorer overlap of large orbitals → lower bond energy. Once energies are considered carefully, “inert pair” is a convenient name rather than a mysterious force — but NIOS expects the term and its consequences for Tl, Pb and Bi.

Exam phrasing: “Is there an inert pair present or is it a misnomer?” Answer: the pair is not truly inert in every sense — the effect is energetic (promotion cost + weak bonds). Consequence remains clear: lower oxidation states become more stable for heavier Group 13–15 elements. Tl⁺ compounds are common; PbO₂ is a strong oxidant because Pb⁴⁺ wants to become Pb²⁺; Bi⁵⁺ is strongly oxidising for the same reason.

Section 7: Hydrides, Oxides and Halides (18.10)

p-Block elements (except noble gases) form hydrides, oxides and halides with fairly regular group trends.

18.10.1 Hydrides

Covalent molecules; bond angles follow VSEPR: CH₄ 109.5°, NH₃ ~107°, H₂O ~104°. Volatile. Acid strength generally increases left to right and top to bottom — HI is more acidic than HCl; H₂Te more acidic than H₂O. Group 13 hydrides include B₂H₆; Group 14 CH₄ to PbH₄; Group 15 NH₃ to BiH₃; Group 16 H₂O to H₂Po; Group 17 HF to HI.

CH₄ 109.5° · NH₃ ~107° · H₂O ~104°
VSEPR angles · Acid strength of hydrides ↑ down Group 17 (HI strongest)

18.10.2 Oxides

Many oxides form: NO, NO₂, N₂O₃, N₂O₅; P₄O₆, P₄O₁₀; XeO₃, XeO₄. Trends: (i) basic character of oxides (same oxidation state) increases down a group; (ii) acidity increases with oxidation state of the element in a period. SO₂ is more acidic than Al₂O₃ or CO₂ among common exam comparisons.

18.10.3 Halides

Mostly covalent. Covalent character decreases down a group; increases with higher oxidation state of the central atom — PbCl₄ is more covalent than PbCl₂; BCl₃ more covalent than AlCl₃. Fluorides often stabilize higher oxidation states; chlorides/bromides/iodides favour lower states.

Covalent halides are gases, liquids or low-melting solids; many hydrolyse to oxoacids: SiCl₄ + 4H₂O → Si(OH)₄ + 4HCl. Formation often by direct combination: C + 2Cl₂ → CCl₄; 2As + 3Cl₂ → 2AsCl₃.

Covalent Character of Halides Lower OS more ionic Higher OS more covalent PbCl₂ → PbCl₄ F stabilizes high OS · Cl/Br/I prefer lower OS
Higher oxidation state → more covalent halide. Classic: PbCl₂ vs PbCl₄.

Exam Connections and Chapter Summary

This is a trends chapter — exams test comparison questions: smaller atom, higher IE, more negative ΔegH, more acidic oxide/hydride, more covalent halide. Memorise exceptions: N>O for IE; Cl>F for electron gain enthalpy; inert pair for Tl⁺, Pb²⁺, Bi³⁺; period-2 CN limit of 4; pπ–pπ for C, N, O only.

Intext 18.1–18.3 practice pairs (F vs Cl size; Be vs B IE; F vs Cl ΔegH; SO₂ most acidic oxide; HI most acidic hydride; SnCl₄ to CCl₄ covalent order; SiCl₄ hydrolysis; NH₃→SbH₃ bond angles; BCl₃ more covalent than AlCl₃; PbCl₄ more covalent than PbCl₂).

L18 sets the language for L19 and L20 (detailed p-block compounds). Every later group discussion will reuse size, IE, EN, inert pair, and oxide/hydride/halide trends introduced here. Master the exceptions and you master half of p-block exam chemistry.

Link back to L17 s-block: vertical similarity is stronger in s-block; p-block mixes metals, metalloids and non-metals in one block, so horizontal change (metal → non-metal) is as important as vertical family trends. Link to L4 bonding: VSEPR angles in hydrides, H-bonding for N/O/F, and Fajans-type arguments for covalent character of high-OS halides all reappear here as periodic trends rather than isolated facts.

MCQ Quiz — L18 General Characteristics of p-Block Elements

0 / 10 correct

Flashcards — L18

1 / 18

Golden Rules — L18 General Characteristics of p-Block Elements

Most exam-important points from this chapter:

Config & size

p-Block = Groups 13–18, ns²np¹–⁶. Size ↓ across period, ↑ down group. Controls all other trends.

IE & Δ_egH

IE ↑ across (exception N>O, P>S half-filled). Δ_egH more −ve across; Cl more −ve than F (size/repulsion).

EN & metals

EN: F>O>N. Metallic character ↓ across, ↑ down. Non-metals dominate right side of p-block.

Anomalies & inert pair

First elements: pπ–pπ, H-bonds, CN≤4. Inert pair: Tl⁺, Pb²⁺, Bi³⁺ preferred for heavy 13–15.

Compounds

Hydrides covalent (VSEPR angles). Oxides: basic ↑ down, acidic ↑ OS. Halides: covalent ↑ OS; SiCl₄ hydrolyses.

Groups 13–18 · ns²np¹–⁶
Size ↓ across · ↑ down
IE ↑ across · ↓ down
EN: F > O > N
ΔegH: Cl more −ve than F
Metal ← left · Non-metal → right
Inert pair: Tl⁺ · Pb²⁺ · Bi³⁺
Hydrides: covalent · VSEPR
Oxides: acidity ↑ with OS

Section 1: Trends & Key Ideas

NIOS Chemistry 313, Module 6 — General Characteristics of the p-Block Elements (sections 18.1–18.10).

Electronic Configuration — ns² np¹–⁶

Groups 13–18 · Five periods of p-block (2p–6p filling)

Occurrence

O, N, C, S free and combined · Noble gases free only · Heavier group members less abundant · Abundant: O, Si, Al, N

Atomic Size Trends

Across period: decreases (same shell, ↑ nuclear charge) · B 88 → F 64 pm

Down group: increases (new shells) · B 88 → Tl 178 pm (Group 13)

Ionization Enthalpy & Electron Gain Enthalpy

IE generally ↑ left→right; ↓ down group · Exception: Group 15 > Group 16 (half-filled p³)

ΔegH more negative across period; less negative down group · Cl more negative than F (F small → e⁻–e⁻ repulsion)

Exam

Q: Why is IE of N > O?

N has half-filled 2p³; removing e⁻ from stable configuration costs more energy

Answer: Half-filled stability

Electronegativity & Metallic Character

EN ↑ across period, ↓ down group · Order: F > O > N

Metallic character ↓ across period, ↑ down group · Non-metallic character opposite

Anomalous First Members & Inert Pair Effect

First row (C, N, O, F): small size · compact 2p · strong H-bonds · pπ–pπ multiple bonds · max CN = 4 (no d-orbitals)

Inert pair: ns² reluctant to bond · lower OS more stable down group · Tl⁺ stable · Pb²⁺ preferred · Bi³⁺ preferred

Causes of inert pair

High promotion energy s²pⁿ → valence state · Poor orbital overlap → weaker bonds for heavy atoms

Hydrides, Oxides, Halides

Hydrides: covalent · CH₄ 109.5° · NH₃ 107° · H₂O 104° · acid strength ↑ left→right and top→bottom (HI most acidic among hydrogen halides)

Oxides: basic character ↑ down group (same OS) · acidity ↑ with oxidation state of element

Halides: mostly covalent · covalent character ↑ with higher OS (PbCl₄ > PbCl₂) · SiCl₄ + 4H₂O → Si(OH)₄ + 4HCl

Section 2: Detailed Definitions

Ionization enthalpy: Energy to remove outermost e⁻ from gaseous atom (kJ mol⁻¹).

Electron gain enthalpy: Energy change when e⁻ added to gaseous atom; usually negative (energy released).

Electronegativity: Ability of atom in a bond to attract shared pair toward itself.

Inert pair effect: Reluctance of ns² electrons to participate in bonding; lower oxidation states stabilize down Groups 13–15.

pπ–pπ bonding: Multiple bonds via p-orbital overlap — strong for C, N, O; weak for Si, P, S (larger 3p orbitals).

Section 3: Diagrams & Visuals

p-Block Trends Map — L18 Size ↓ IE ↑ EN ↑ across → Size ↑ IE ↓ metal ↑ down ↓ Anomalous top: C=C, N≡N, O=O · CN max 4 · H-bonds N/O/F Inert pair: Tl(+1) · Pb(+2) · Bi(+3) more stable than higher OS Hydrides acid ↑ · Oxides basic ↓ group · Halides covalent ↑ OS

Periodic trends + anomalies + compound chemistry of p-block

COORDINATION NUMBER LIMIT ═══════════════════════════════════════ Period 2: max CN = 4 (BH₄⁻, BF₄⁻, CF₄, NH₄⁺) Heavier: CN 6 possible ([AlF₆]³⁻, [SiF₆]²⁻, [PCl₆]⁻) Reason: d-orbitals available for n ≥ 3 ═══════════════════════════════════════

Section 5: Comprehensive Q&A (12 Questions)

Q1: Outer electronic configuration of p-block elements?

ns² np¹–⁶ for Groups 13–18.

Q2: Why does atomic size decrease across a period?

Electrons added to same shell; nuclear charge increases → stronger pull → smaller radius.

Q3: Why is first IE of N greater than O?

Half-filled 2p³ in N is extra stable; removing an electron from O (2p⁴) is easier.

Q4: Why is ΔegH of Cl more negative than F?

Very small F atom: added electron faces strong interelectronic repulsion. Cl is larger → less repulsion, more favourable.

Q5: Most electronegative element?

Fluorine, then oxygen, then nitrogen.

Q6: Why does O₂ exist as gas but S as solid?

Oxygen forms strong pπ–pπ double bonds (O=O) as discrete molecules. Sulphur prefers single-bonded catenated S₈ rings — solid molecular solid.

Q7: Two reasons for anomalous first p-block elements?

Small size / compact 2p orbitals; absence of d-orbitals (max CN 4); ability to form pπ–pπ bonds; high EN and H-bonding (N, O, F).

Q8: What is inert pair effect? Consequence for Tl and Pb?

ns² electrons reluctant to bond. Tl prefers +1; Pb prefers +2 over higher oxidation states.

Q9: Most acidic among H₂Se, H₂O, HCl, HI?

HI — acid strength of hydrides increases down Group 17 and is high for hydrogen halides.

Q10: Which is more covalent — PbCl₂ or PbCl₄?

PbCl₄ — higher oxidation state increases covalent character (Fajans / polarizing power).

Q11: Product when SiCl₄ reacts with water?

SiCl₄ + 4H₂O → Si(OH)₄ + 4HCl (hydrolysis to oxoacid/silicic acid).

Q12: Bond angle trend NH₃, PH₃, AsH₃, SbH₃?

Decreases from ~107° toward ~90° down the group (less hybridization, larger central atom).

Section 6: Tips, Tricks & Exam Hacks

Memory Aids

  • IE exception: "N before O, P before S"
  • Δ_egH: "Cl beats F" (more negative)
  • EN: "FON — Fluorine, Oxygen, Nitrogen"
  • Inert pair: "Heavy likes low OS — Tl⁺, Pb²⁺"
  • CN: "Period 2 max four"

Exam Tips

  • Always mention half-filled/full-filled exceptions for IE
  • Inert pair: Groups 13, 14, 15 only
  • Covalent character: higher OS → more covalent
  • Oxide acidity ↑ with oxidation number of central atom
  • First element of each group is always special

Section 7: Connections & Relationships

Builds on: Periodic table, VSEPR (L4), s-block trends (L17).
Leads to: L19–L20 detailed p-block compounds (Groups 13–18 chemistry).
Related: Fajans rules for covalent character; H-bonding (L4, L7, L17).

Section 8: Complete Quick Reference

• Config: ns²np¹–⁶ · Groups 13–18

• Size ↓ across, ↑ down · IE ↑ across (N>O exception), ↓ down

• ΔegH: Cl > F (magnitude negative) · EN: F > O > N

• Metal ↓ across, ↑ down · First elements anomalous

• Inert pair: lower OS stable for heavy 13–15

• Hydrides covalent · Oxides basic ↑ down · Halides covalent ↑ OS

Remember: ✓ Period 2 CN≤4 ✓ pπ–pπ for C,N,O ✓ PbCl₄ more covalent than PbCl₂ ✓ HI very acidic

PYQ — Previous Year Questions

Extracted from NIOS Chemistry (313) board exam papers in your PDF. Chapter L18 — General Characteristics of p-Block Elements only. Use Model Answer for marking points; Explanation for concept clarity.

L18 — General Characteristics of p-Block Elements

1 question(s) · Sources: 313/MAY/205C

Section B — Short / Long answer (from papers)

PYQ1. Aluminium forms [AlF6]3– ion but boron does not form [BF6]3– ion. Explain. Eobw{‘{Z¶‘ [AlF6]3– Am¶Z ~ZmVm h¡, O~{H$ ~moam°Z [BF6]3– Am¶Z Zht ~ZmVm h¡& ì¶m»¶m

2 marks · Q34 · 313/MAY/205C

Model Answer

Give the chemical reason linked to structure/bonding/equilibrium. Start with the principle, then apply to the species named in the question.

Explanation

Reasoning marks require principle + application. Cite electron effects, stability, or Le Chatelier as relevant.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205C · Q34 · 2 mark(s) · L18.

Problem Solving — L18 General Characteristics of p-Block Elements

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). If the question says draw, a labelled pencil sketch is provided. Explanations open by default.

Question 1 of 6p-block

Draw a sketch of s and p blocks. Which groups form the p-block?

ns² np¹–⁶

Pencil sketch (labelled)

s / p / d blocks (idea) s d-block p group → · period ↓
Pencil sketch: block layout of periodic table

Solution — step by step with formulas

  1. Groups 13–18; valence electrons in np orbitals.

Final answer: Groups 13–18

Formulas used in this problem

ns² np¹–⁶

Textbook formal language

p-block shows metals, metalloids and non-metals.

Working formulas: ns² np¹–⁶. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Right-hand side of the long form table after d-block.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — p-block location

Noble gases complete the block.

Linked to chapter notes (L18). Remember: ns² np¹–⁶. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write ns² np¹–⁶ before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 2 of 6Inert pair

What is the inert pair effect?

Solution — step by step with formulas

  1. Reluctance of ns² electrons of heavier p-block elements to participate in bonding; lower OS more stable.

Final answer: ns² pair less reactive down group

Textbook formal language

Explains Tl⁺, Pb²⁺, Bi³⁺ relative stability.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Heavier atoms keep their s electrons more ‘idle’.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Inert pair effect

Important for group 13–15 heavier members.

Linked to chapter notes (L18). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 3 of 6Catenation

Which element shows maximum catenation?

Solution — step by step with formulas

  1. Carbon (strong C–C bonds, tetravalency).

Final answer: Carbon

Textbook formal language

Catenation: self-linking of atoms.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Carbon forms long chains and rings—basis of organic chemistry.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Catenation

Si catenates much less.

Linked to chapter notes (L18). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 4 of 6Allotropes

Name three allotropes of carbon.

Solution — step by step with formulas

  1. Diamond, graphite, fullerenes.

Final answer: Diamond, graphite, fullerenes

Textbook formal language

Same element, different structures and properties.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Hard 3D network vs layered graphite vs ball cages.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Carbon allotropes

Bonding explains hardness/conductivity.

Linked to chapter notes (L18). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 5 of 6Oxides

How does acidic character of oxides generally change down a p-block group?

Solution — step by step with formulas

  1. Often decreases (more metallic/basic character down the group).

Final answer: Acidic character decreases down group (general)

Textbook formal language

Non-metal oxides acidic; metallic oxides basic/amphoteric.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Going down, oxides become less acidic/more basic.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Oxide character

Check specific group examples in notes.

Linked to chapter notes (L18). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 6 of 6Anomaly

Give one reason the first member of a p-block group is anomalous.

Solution — step by step with formulas

  1. Small size, high electronegativity, absence of d-orbitals (period 2).

Final answer: Small size / no d-orbitals

Textbook formal language

Period-2 elements differ strongly from heavier congeners.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Tiny atoms behave differently from big ones in the same group.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — First member anomaly

E.g. N₂ vs P₄ structural difference.

Linked to chapter notes (L18). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.