The p-Block Elements Class 12 Notes - CBSE Chemistry Chapter 7

Chapter summary

The p-Block Elements (Groups 15 to 18) covers the trends, oxidation states and key compounds of the nitrogen family, the oxygen family, the halogens and the noble gases. It includes the preparation and properties of ammonia, nitric acid, sulphuric acid, ozone, interhalogen compounds and xenon fluorides. This is a high-yield NEET chapter where questions test group trends, anomalous behaviour of the first element, industrial processes, and the structures of compounds like the xenon fluorides.

Chapter notes
🃏 Flash Cards: The p-Block Elements (Groups 15-18)

Class 12 Chemistry – swipe through all 9 cards to understand the whole chapter.

🧭Start here1/9

The p-Block at a Glance

Groups 15-18 fill the np orbitals, and each group’s top element behaves oddly because period-2 atoms have no d-orbitals.

Gp15 ns2np3 · Gp16 ns2np4 · Gp17 ns2np5 · Gp18 ns2np6

Down a group: metallic character rises, non-metallic falls.

  • First member (N, O, F, Ne) is anomalous: small size, high EN, no d-orbitals
  • Inert pair effect makes lower oxidation states more stable down a group
  • Maximum covalency of period-2 elements is limited (N=4, O=2, F=1)
🏛️Group 152/9

Nitrogen Family Trends

The nitrogen family (N, P, As, Sb, Bi) shows oxidation states −3, +3, +5 with predictable stability shifts down the group.

States: −3, +3, +5 · +5 stability ↓, +3 stability ↑ (Bi3⁺ most stable)

N≡N is a gas; P, As exist as tetrahedral P4 units.

  • Hydride bond angle: NH3 (107°) > PH3 (~93°) > AsH3 > SbH3
  • Hydride basicity & thermal stability ↓, reducing power ↑ down the group
  • N forms pπ–pπ multiple bonds; no NCl5 (no d-orbitals), but PCl5 exists
⚗️Group 153/9

NH3, HNO3 & N-Oxides

Ammonia and nitric acid are the key industrial Group-15 compounds, with nitrogen oxides spanning +1 to +5.

N2 + 3H2 ⇌ 2NH3 (Haber, Fe, ~200 atm, ~700 K)

HNO3 (Ostwald, Pt/Rh): conc.→NO2, dilute→NO, never H2.

  • NH3 is a pyramidal Lewis base; gives deep blue [Cu(NH3)4]2
  • Oxides: N2O (+1), NO (+2, paramagnetic), NO2 (+4, brown), N2O5 (+5)
  • Aqua regia = 3 HCl : 1 HNO3 dissolves Au and Pt
🔥Group 154/9

Phosphorus & Its Oxoacids

Phosphorus has reactive white and stable red allotropes, and its oxoacids differ by the number of P–OH groups.

Basicity = no. of P–OH · H3PO2 (mono) · H3PO3 (di) · H3PO4 (tri)

P–H bonds give reducing power; H3PO4 (0 P–H) is not reducing.

  • White P4 is reactive, glows, stored under water; red P is polymeric and stable
  • H3PO2 (+1, 2 P–H) is a strong reducing agent, H3PO3 (+3, 1 P–H)
  • PH3 from P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2; weaker base than NH3
💨Group 165/9

Oxygen Family (Chalcogens)

The chalcogens (O, S, Se, Te, Po) sit two electrons short of an octet and favour the −2 state.

States: −2, +2, +4, +6 · O is gas (O2, pπ–pπ); S is solid (S8 crown ring)

O exceptions: +2 in OF2, −1 in peroxides; max covalency of O = 2.

  • Hydride acidity ↑ down: H2O < H2S < H2Se < H2Te
  • H2O is liquid (H-bonding); bond angle H2O (104.5°) > H2S > H2Se
  • Ozone O3 is angular (~117°), a powerful oxidiser: 2O3 → 3O2
🧪Group 166/9

SO2 & Sulphuric Acid

Sulphuric acid is made by the Contact process and acts as acid, dehydrating agent and oxidiser.

2SO2 + O2 →(V2O5) 2SO3 · SO3 + H2SO4 → oleum (H2S2O7)

In H2SO4, S is sp3 and +6; SO2 bleaching is temporary (by reduction).

  • Contact process: burn S → SO2, oxidise to SO3 over V2O5 catalyst
  • SO2 is acidic and reducing; its bleaching reverses on exposure to air
  • Oleum is diluted with water to give concentrated H2SO4
Group 177/9

Halogens & Fluorine Anomaly

The halogens (F, Cl, Br, I, At) are the most electronegative group and the strongest oxidisers.

Oxidising power: F2 > Cl2 > Br2 > I2 · Bond enthalpy: Cl2 > Br2 > F2 > I2

Electron gain enthalpy: Cl > F (F small → strong e⁻–e⁻ repulsion).

  • F shows only −1 (no d-orbitals); Cl, Br, I reach +1, +3, +5, +7
  • HF is liquid (H-bonding) and a weak acid; acidity HF < HCl < HBr < HI
  • HCl prep: NaCl + H2SO4 → NaHSO4 + HCl
🔗Group 178/9

Interhalogens & Oxoacids

Halogens combine with each other and with oxygen to give reactive interhalogens and a ladder of oxoacids.

Acid strength: HClO < HClO2 < HClO3 < HClO4 (oxidising power reverse)

Interhalogens (XX’ₙ) are more reactive than the parent halogens.

  • ClF3 is T-shaped, IF5 square pyramidal, IF7 pentagonal bipyramidal
  • Acidity and oxidation state of Cl oxoacids rise from HOCl to HClO4
  • HOCl is the strongest oxidiser, HClO4 the strongest acid
🎈Group 189/9

Noble Gases & Xenon Compounds

Noble gases have a completely filled shell making them inert, yet xenon’s low ionisation enthalpy lets it react with F and O.

ns2np6 (He = 1s2) · XeF2 linear · XeF4 square planar · XeF6 distorted octahedral

Use VSEPR (count lone pairs) to get xenon-compound geometry.

  • Monatomic, largest (van der Waals) radii in their period, low boiling points
  • Xe reacts because it has the lowest ionisation enthalpy among stable noble gases
  • XeO3 pyramidal, XeOF4 square pyramidal, XeO4 tetrahedral
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📝 Practice The p-Block Elements (Groups 15-18) — 10 NEET PYQs
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Q1NEET 2021
Noble gases are named because of their inertness towards reactivity. Identify the INCORRECT statement about them:
Correct answer: B. Noble gases are held only by weak London dispersion forces, so they have very LOW melting and boiling points — statement (b) is incorrect. The other statements are correct.
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Q2NEET 2020
Which one of the following oxoacids of sulphur has an –O–O– (peroxide) linkage?
Correct answer: B. Peroxodisulphuric acid, H₂S₂O₈ (Marshall’s acid), contains a peroxide –O–O– linkage joining the two sulphur centres. The others have no peroxy bridge.
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Q3NEET 2019
A compound ‘X’ on reaction with H₂O gives a colourless gas ‘Y’ with a rotten-fish smell; ‘Y’ is absorbed in a CuSO₄ solution to give Cu₃P₂ as a product. The compound ‘X’ is:
Correct answer: A. The rotten-fish-smelling colourless gas is phosphine, PH₃. Calcium phosphide reacts with water: Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PH₃. The PH₃ then reacts with CuSO₄ to give Cu₃P₂. Hence X is Ca₃P₂.
🔎 See the full step-by-step solution in the app →
Q4NEET 2019
Which of the following oxoacids of phosphorus has the strongest reducing property?
Correct answer: C. Reducing power of phosphorus oxoacids is proportional to the number of P–H bonds. H₃PO₂ (hypophosphorous acid) has two P–H bonds — the maximum among these — so it is the strongest reducing agent. H₃PO₃ has one P–H, and H₃PO₄ has none.
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Q5NEET 2019
Which is the correct thermal stability order for H₂E (E = O, S, Se, Te and Po)?
Correct answer: B. Thermal stability of group-16 hydrides decreases down the group as the M–H bond weakens with increasing central-atom size. So stability order is H₂Po < H₂Te < H₂Se < H₂S < H₂O.
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Q6NEET 2018
The correct order of N-compounds in their decreasing order of oxidation states of nitrogen is:
Correct answer: C. Oxidation states of N: HNO₃ = +5, NO = +2, N₂ = 0, NH₄Cl = −3. Decreasing order is therefore HNO₃ (+5) > NO (+2) > N₂ (0) > NH₄Cl (−3).
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Q7NEET 2018
In the structure of ClF₃, the number of lone pairs of electrons on the central Cl atom is:
Correct answer: B. Cl has 7 valence electrons; three are used in bonding to three F atoms, leaving 4 electrons = 2 lone pairs. With 3 bond pairs + 2 lone pairs (sp³d) the molecule is T-shaped.
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Q8NEET 2017
The species having bond angles of 120° is:
Correct answer: D. BCl₃ is sp²-hybridised with no lone pair on the central B atom, giving a trigonal-planar shape with 120° bond angles. PH₃ (~93.5°) and NCl₃ (~107°) are pyramidal; ClF₃ is T-shaped (~90°).
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Q9NEET 2016
Which one of the following is the correct order of acidity of chlorine oxoacids?
Correct answer: A. As the oxidation state of Cl increases (HClO +1, HClO₂ +3, HClO₃ +5, HClO₄ +7), more terminal O atoms stabilise the conjugate base, so acidity increases: HClO < HClO₂ < HClO₃ < HClO₄.
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Q10NEET 2016
Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules?
Correct answer: A. Bond dissociation enthalpy normally falls down the group, but F₂ is anomalously low because of lone-pair repulsion in the small F–F bond. The correct order is therefore Cl₂ > Br₂ > F₂ > I₂.
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Frequently Asked Questions

What is the inert pair effect and how does it affect Group 15?

The inert pair effect is the reluctance of the outermost s-electrons to take part in bonding as we move down a group. In Group 15 it makes the +5 oxidation state less stable and the +3 state more stable down the group, so bismuth is most stable in the +3 state.

How are ammonia and nitric acid manufactured?

Ammonia is made by Haber’s process, the reaction of nitrogen with hydrogen over an iron catalyst at about 200 atmospheres and 700 K. Nitric acid is made by Ostwald’s process, the catalytic oxidation of ammonia over a platinum-rhodium catalyst followed by further oxidation and absorption in water.

Why does fluorine show only the minus one oxidation state?

Fluorine is the most electronegative element and has no available d-orbitals in its valence shell, so it cannot expand its octet. As a result it shows only the minus one oxidation state, whereas the other halogens (chlorine, bromine, iodine) can show positive states such as +1, +3, +5 and +7.

What are the shapes of XeF2, XeF4 and XeF6?

Xenon difluoride (XeF2) is linear, xenon tetrafluoride (XeF4) is square planar, and xenon hexafluoride (XeF6) is a distorted octahedron. Noble gases react only with the most electronegative elements fluorine and oxygen, and xenon does so because it has the lowest ionisation energy among the stable noble gases.

Why is the oxidising power order of halogens F2 greater than Cl2 greater than Br2 greater than I2?

Oxidising power depends on the ease of accepting an electron and the overall energetics of reduction. Fluorine is the strongest oxidiser because of its low bond dissociation enthalpy and the high hydration enthalpy of the fluoride ion, even though its electron gain enthalpy is anomalously lower than chlorine. The power then decreases down the group to iodine.

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