p-Block Elements II — Nitrogen & Oxygen Families (Groups 15–16)

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Groups 15 and 16 are the richest part of p-block chemistry for exams: dozens of compounds, but a handful of ideas. The first member (N, O) is small and forms multiple bonds; hydrides lose stability and gain acidity or reducing power down the group; and oxoacids are counted by their O–H groups.

What you will be able to do

• Explain trends in groups 15 and 16 and the anomalous behaviour of N and O
• Compare hydrides of groups 15 and 16: bond angle, stability, basicity, acidity and reducing character
• Write preparation and key reactions of N₂, NH₃, HNO₃, PH₃, PCl₃, PCl₅, O₃, SO₂ and H₂SO₄
• Describe oxides of nitrogen and oxoacids of phosphorus and sulphur, including their structures and basicity
• Predict products of metals with dilute and concentrated nitric and sulphuric acids

The big picture

Small first members, weaker bonds below.

In one line: N and O are small and electronegative, form pπ–pπ multiple bonds and hydrogen bonds, and cannot expand beyond four bonds; down each group hydrides become less stable and more acidic or reducing, and the lower oxidation state (+3 for Bi, +4 for Te) becomes favoured.

Property	Group 15 (N → Bi)	Group 16 (O → Po)
Valence configuration	ns² np³	ns² np⁴
Oxidation states	−3 to +5; +3 grows down (Bi³⁺)	−2, +2, +4, +6; +4 grows down
Character	N, P non-metals; As, Sb metalloids; Bi metal	O, S non-metals; Se, Te metalloids; Po metal
First member	N₂ gas, N≡N; max covalency 4	O₂ gas, O=O; mostly −2; H-bonding

Think of it like this: The first member of each group is like the youngest, smallest sibling who can only hold four friends’ hands at once — but holds two hands with the same friend (multiple bonds). The bigger siblings have more hands (d orbitals) and hold up to five or six partners, but each grip is weaker.

Group 15: trends

Data first.

Property	N	P	As	Sb	Bi
Covalent radius / pm	70	110	121	141	148
ΔiH₁ / kJ mol⁻¹	1402	1012	947	834	703
Electronegativity	3.0	2.1	2.0	1.9	1.9

• Ionisation enthalpy is higher than in group 14 — the half-filled p³ subshell is stable.
• Nitrogen is anomalous: small, highly electronegative, forms pπ–pπ multiple bonds (N≡N, N=O), has no d orbitals (max covalency 4), and its N–N single bond is weak (lone-pair repulsion).
• Inert pair effect: +5 stability falls down the group; Bi(V) is a strong oxidant (NaBiO₃) and BiF₅ is the only well-characterised Bi(V) compound.
• Nitrogen forms NCl₃ but no NCl₅; P forms both PCl₃ and PCl₅.

Hydrides EH₃

Hydride	Bond angle	Thermal stability	Basicity	Reducing power
NH₃	107.8°	Highest	Highest	Lowest
PH₃	93.6°	↓	↓	↑
AsH₃	91.8°	↓	↓	↑
SbH₃	91.3°	↓	↓	↑
BiH₃	90°	Lowest	Lowest	Highest

• Bond angle falls because the central atom grows and bonding pairs move away from it, so they repel each other less (and the larger atoms use almost pure p orbitals).
• Boiling points: PH₃ < AsH₃ < NH₃ < SbH₃ < BiH₃ — NH₃ is high because of hydrogen bonding.

Dinitrogen and ammonia

The two nitrogen compounds asked most.

Dinitrogen

• Lab: NH₄Cl(aq) + NaNO₂(aq) → N₂ + 2H₂O + NaCl (warm).
• From ammonium dichromate: (NH₄)₂Cr₂O₇ → N₂ + Cr₂O₃ + 4H₂O (green Cr₂O₃ — the “volcano” experiment).
• Very pure N₂: heat sodium or barium azide: Ba(N₃)₂ → Ba + 3N₂.
• Inert at room temperature (N≡N bond enthalpy 941.4 kJ mol⁻¹); reacts with Li at room temperature and with Mg, Ca at high temperature to give nitrides; with O₂ only at about 2000 K (lightning) → NO.

Ammonia

• Haber process: N₂ + 3H₂ ⇌ 2NH₃, ΔH = −46.1 kJ per mole of NH₃; about 200 atm, 700 K, iron oxide catalyst with K₂O and Al₂O₃ promoters.
• Trigonal pyramidal (sp³, one lone pair); highly soluble in water; weak base: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻.
• Lewis base: with Cu²⁺ it gives deep blue [Cu(NH₃)₄]²⁺; it dissolves AgCl as [Ag(NH₃)₂]⁺.
• Precipitates metal hydroxides: FeCl₃ + NH₄OH → Fe(OH)₃ (brown); ZnSO₄ + NH₄OH → Zn(OH)₂ (white).

Oxides of nitrogen and nitric acid

Five oxides, one important acid.

Oxide	Oxidation state of N	Nature	Appearance	Structure / note
N₂O	+1	Neutral	Colourless gas (laughing gas)	Linear N–N–O; from heating NH₄NO₃
NO	+2	Neutral	Colourless gas	Odd electron, paramagnetic; Cu + dil. HNO₃
N₂O₃	+3	Acidic	Blue solid	Anhydride of HNO₂
NO₂	+4	Acidic	Brown gas	Odd electron, paramagnetic; dimerises to colourless N₂O₄
N₂O₅	+5	Acidic	Colourless solid	Anhydride of HNO₃

Nitric acid — Ostwald process

1. 4NH₃ + 5O₂ → 4NO + 6H₂O (Pt/Rh gauze, about 500 K, 9 bar)
2. 2NO + O₂ ⇌ 2NO₂
3. 3NO₂ + H₂O → 2HNO₃ + NO (NO is recycled); concentrated to about 68% by distillation, 98% with conc. H₂SO₄

Metal + HNO₃	Main products
Cu + dilute (8HNO₃)	3Cu(NO₃)₂ + 2NO + 4H₂O
Cu + concentrated (4HNO₃)	Cu(NO₃)₂ + 2NO₂ + 2H₂O
Zn + dilute (10HNO₃)	4Zn(NO₃)₂ + N₂O + 5H₂O
Zn + concentrated (4HNO₃)	Zn(NO₃)₂ + 2NO₂ + 2H₂O
Cr, Al, Fe + concentrated	Passive (oxide film)

• Non-metals are oxidised to their oxoacids: I₂ + 10HNO₃ → 2HIO₃ + 10NO₂ + 4H₂O; C → CO₂; S₈ → H₂SO₄; P₄ → H₃PO₄.
• Brown ring test for NO₃⁻: add FeSO₄ and run conc. H₂SO₄ down the side; NO₃⁻ is reduced to NO, which forms brown [Fe(H₂O)₅(NO)]²⁺ at the junction.
• In HNO₃ the N is planar (sp²); one N–O is an ordinary single bond to OH and the other two share a double bond by resonance.

Solved example 01 — Predicting HNO₃ products

Write equations for copper with (a) dilute and (b) concentrated nitric acid, and name the gases.

1. (a) 3Cu + 8HNO₃(dil.) → 3Cu(NO₃)₂ + 2NO + 4H₂O.
2. (b) Cu + 4HNO₃(conc.) → Cu(NO₃)₂ + 2NO₂ + 2H₂O.

Answer: Dilute gives colourless NO (which turns brown in air); concentrated gives brown NO₂.

Check yourself

• C1. Which oxides of nitrogen are paramagnetic?
• C2. Why does NO₂ dimerise?

Answers at the end.

Phosphorus and its compounds

Allotropes, phosphine, halides and oxoacids.

Allotrope	Structure	Properties
White	Discrete P₄ tetrahedra (angle 60°, strained)	Waxy, poisonous, glows in the dark (chemiluminescence), catches fire in air (stored under water), soluble in CS₂
Red	Polymeric chains of P₄ units	Made by heating white P at 573 K in an inert atmosphere; less reactive, does not glow, insoluble in CS₂
Black	Layered (α and β forms)	Made by heating red P at high temperature or under pressure; least reactive

Phosphine, PH₃

• From calcium phosphide: Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PH₃.
• From white P: P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂ (inert atmosphere) — a disproportionation.
• Pure PH₃ does not catch fire; impure PH₃ (with P₂H₄ or P₄ vapour) does — used in Holme’s signals with acetylene.
• Weakly basic: PH₃ + HI → PH₄I. Forms black Cu₃P₂ with CuSO₄ solution.

Phosphorus halides

	PCl₃	PCl₅
Preparation	P₄ + 6Cl₂ → 4PCl₃; or P₄ + 8SOCl₂ → 4PCl₃ + 4SO₂ + 2S₂Cl₂	P₄ + 10Cl₂ → 4PCl₅; or P₄ + 10SO₂Cl₂ → 4PCl₅ + 10SO₂
Shape	Pyramidal (sp³)	Trigonal bipyramidal (sp³d); axial bonds (240 pm) longer than equatorial (202 pm)
Hydrolysis	PCl₃ + 3H₂O → H₃PO₃ + 3HCl	PCl₅ + H₂O → POCl₃ + 2HCl; then POCl₃ + 3H₂O → H₃PO₄ + 3HCl
With alcohols / acids	3ROH → 3RCl + H₃PO₃	ROH → RCl + POCl₃ + HCl
Solid state	Molecular	Ionic: [PCl₄]⁺[PCl₆]⁻

Oxoacids of phosphorus

Acid	Formula	Oxidation state of P	P–OH groups (basicity)	P–H bonds	Note
Hypophosphorous	H₃PO₂	+1	1 (monobasic)	2	Strong reducing agent
Phosphorous	H₃PO₃	+3	2 (dibasic)	1	Reducing; 4H₃PO₃ → 3H₃PO₄ + PH₃ on heating
Orthophosphoric	H₃PO₄	+5	3 (tribasic)	0	Not reducing
Pyrophosphoric	H₄P₂O₇	+5	4	0	P–O–P bridge
Metaphosphoric	(HPO₃)ₙ	+5	1 per P	0	Cyclic or chain

Key idea: Basicity = number of ionisable P–OH groups, not the number of H atoms. H atoms bonded directly to P are not acidic but give reducing power: acids with P–H bonds (H₃PO₂, H₃PO₃) reduce AgNO₃ to Ag.

Solved example 02 — Basicity of H₃PO₃

Why is H₃PO₃ dibasic although it has three hydrogen atoms?

1. Structure: one P=O, two P–OH and one P–H bond around a tetrahedral P.
2. Only H attached to O can ionise; the P–H hydrogen is not acidic.

Answer: Two ionisable P–OH groups → dibasic; it forms NaH₂PO₃ and Na₂HPO₃ only.

Check yourself

• C3. Why is white phosphorus more reactive than red?
• C4. How many P–H bonds are there in hypophosphorous acid?

Answers at the end.

Group 16: trends

Oxygen stands apart again.

Property	O	S	Se	Te	Po
ΔiH₁ / kJ mol⁻¹	1314	1000	941	869	813
Electron gain enthalpy / kJ mol⁻¹	−141	−200	−195	−190	−174
Electronegativity	3.50	2.44	2.48	2.01	1.76

• Electron gain enthalpy of O is less negative than S — the small O atom repels the incoming electron (compare F and Cl).
• Ionisation enthalpy of group 16 is lower than group 15 — removing an electron from p⁴ relieves pairing (N > O, P > S).
• Oxygen is anomalous: small, highly electronegative, H-bonding (water is liquid, H₂S a gas), max covalency 2 to 4, shows −2 almost always (+2 only in OF₂, +1 in O₂F₂).
• Down the group, +4 becomes more stable than +6 (inert pair). Polonium is radioactive.

Hydrides H₂E

Hydride	Bond angle	b.p. / K	Thermal stability	Acid strength
H₂O	104.5°	373	Highest	Lowest
H₂S	92°	213	↓	↑
H₂Se	91°	232	↓	↑
H₂Te	90°	269	Lowest	Highest

Acid strength rises down the group because the H–E bond gets longer and weaker; reducing power rises for the same reason (H₂Te is the strongest reducing agent).

Dioxygen and ozone

O₂ and its reactive allotrope.

• Lab O₂: 2KClO₃ → 2KCl + 3O₂ (MnO₂ catalyst, heat); also 2H₂O₂ → 2H₂O + O₂ and 2HgO → 2Hg + O₂.
• O₂ is paramagnetic (two unpaired π∗ electrons). It combines with most elements, often only on heating.
• Oxides: acidic (SO₂, CO₂, Cl₂O₇, Mn₂O₇), basic (Na₂O, CaO), amphoteric (Al₂O₃, ZnO), neutral (CO, NO, N₂O).

Ozone, O₃

• Made by silent electric discharge through dry O₂: 3O₂ ⇌ 2O₃, ΔH = +142 kJ mol⁻¹ (endothermic, so the discharge avoids heat that would decompose it).
• Structure: bent (about 117°), both O–O bonds equal (128 pm) by resonance; diamagnetic.
• Strong oxidant (it gives nascent O): PbS + 4O₃ → PbSO₄ + 4O₂; 2KI + H₂O + O₃ → 2KOH + I₂ + O₂ — the liberated I₂ is titrated to estimate ozone.
• The stratospheric ozone layer absorbs UV. NO from supersonic jets and Cl atoms from CFCs destroy it: NO + O₃ → NO₂ + O₂.

Sulphur, SO₂ and sulphuric acid

The industrial heart of group 16.

Allotrope	Form	Stable	Note
Rhombic (α)	Yellow crystals; S₈ crown rings	Below 369 K	Insoluble in water, soluble in CS₂
Monoclinic (β)	Needle-shaped; S₈ rings	Above 369 K	Made by melting and cooling rhombic S
S₂ vapour	Diatomic	At high temperature	Paramagnetic, like O₂

The transition temperature between rhombic and monoclinic sulphur is 369 K.

Sulphur dioxide

• From burning sulphur, or Na₂SO₃ + 2HCl → 2NaCl + H₂O + SO₂; industrially by roasting sulphide ores.
• Acidic: SO₂ + H₂O → H₂SO₃; with NaOH → Na₂SO₃.
• Reducing agent (in the presence of water): turns acidified K₂Cr₂O₇ from orange to green and decolourises acidified KMnO₄ — tests for SO₂.
• Bleaches by reduction (temporary — the colour returns in air); chlorine bleaches by oxidation (permanent).
• Bent molecule (sp², about 119°) with two equal S–O bonds by resonance.

Sulphuric acid — contact process

1. Burn sulphur or roast sulphide ores → SO₂.
2. 2SO₂ + O₂ ⇌ 2SO₃, ΔH = −196.6 kJ; V₂O₅ catalyst, about 720 K, 2 bar.
3. Absorb SO₃ in conc. H₂SO₄ → H₂S₂O₇ (oleum); dilute oleum with water → H₂SO₄ (96–98%). SO₃ is not absorbed directly in water because it forms an acid mist.

Property	Example
Strong dibasic acid	Ka₁ very large; Ka₂ = 1.2 × 10⁻²
Dehydrating agent	Chars sugar: C₁₂H₂₂O₁₁ → 12C + 11H₂O; dries gases (not NH₃ or H₂S)
Oxidising agent (hot, conc.)	Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O; C + 2H₂SO₄ → CO₂ + 2SO₂ + 2H₂O
Low volatility	Displaces volatile acids: NaCl + H₂SO₄ → NaHSO₄ + HCl

Oxoacids of sulphur

Acid	Formula	Feature
Sulphurous	H₂SO₃	S +4; exists in solution only
Sulphuric	H₂SO₄	S +6; tetrahedral
Thiosulphuric	H₂S₂O₃	One O replaced by S
Pyrosulphuric (oleum)	H₂S₂O₇	S–O–S bridge
Peroxomonosulphuric (Caro’s acid)	H₂SO₅	One O–O bond
Peroxodisulphuric (Marshall’s acid)	H₂S₂O₈	S–O–O–S peroxo bridge

Solved example 03 — SO₂ versus Cl₂ as bleaches

Why is the bleaching by SO₂ temporary but by Cl₂ permanent?

1. SO₂ + 2H₂O → H₂SO₄ + 2[H]; nascent hydrogen reduces the dye to a colourless form.
2. In air the reduced dye is slowly re-oxidised, and the colour returns.
3. Cl₂ + H₂O → HCl + HOCl → nascent [O] oxidises the dye irreversibly.

Answer: SO₂ bleaches by reduction (reversible); Cl₂ by oxidation (permanent).

Check yourself

• C5. What is the catalyst in the contact process?
• C6. What is the transition temperature between rhombic and monoclinic sulphur?

Answers at the end.

Advanced problems

Each needs two ideas from the chapter.

Solved example 04 — NH₃ vs PH₃

Explain why NH₃ is more basic than PH₃ and has a larger bond angle.

1. N is smaller and more electronegative; its lone pair is in a compact sp³ orbital and is readily donated.
2. P’s lone pair is in a larger, more diffuse orbital (largely s character), and is less available.
3. In NH₃ the bonding pairs are close to N and repel strongly (107.8°); in PH₃ they are farther away and P uses almost pure p orbitals (93.6°).

Answer: Higher electron density and a more available lone pair make NH₃ more basic; stronger bond-pair repulsion gives a larger angle.

Solved example 05 — Volume of oxygen

What volume of O₂ at STP is produced by heating 12.25 g of KClO₃ completely? (KClO₃ = 122.5 g mol⁻¹)

1. 2KClO₃ → 2KCl + 3O₂: 2 mol KClO₃ → 3 mol O₂.
2. 12.25 g = 0.100 mol → 0.150 mol O₂.
3. 0.150 × 22.4 L.

Answer: 3.36 L of O₂.

Solved example 06 — Hydrolysis products

Compare the products of complete hydrolysis of PCl₃ and PCl₅, and give the basicity of each acid formed.

1. PCl₃ + 3H₂O → H₃PO₃ + 3HCl — phosphorous acid, dibasic.
2. PCl₅ + 4H₂O → H₃PO₄ + 5HCl — orthophosphoric acid, tribasic.

Answer: H₃PO₃ (dibasic, P +3) and H₃PO₄ (tribasic, P +5).

Solved example 07 — Why N₂ is so unreactive

Explain why dinitrogen is almost inert at room temperature while white phosphorus catches fire in air.

1. N₂ has a triple bond (one σ, two π) with a very high bond enthalpy, 941.4 kJ mol⁻¹.
2. P₄ has six single P–P bonds in a strained tetrahedron with 60° angles, which break easily.

Answer: The N≡N triple bond is very strong; the strained P–P bonds of P₄ are weak and reactive.

Solved example 08 — Ozone estimation

O₃ is passed through excess moist KI, and the iodine liberated needs 20.0 mL of 0.100 M Na₂S₂O₃. Find the moles of O₃.

1. O₃ + 2KI + H₂O → 2KOH + I₂ + O₂: 1 mol O₃ → 1 mol I₂.
2. I₂ + 2Na₂S₂O₃ → 2NaI + Na₂S₄O₆: 1 mol I₂ ≡ 2 mol thiosulphate.
3. Thiosulphate = 2.00 × 10⁻³ mol → I₂ = 1.00 × 10⁻³ mol.

Answer: 1.00 × 10⁻³ mol of O₃.

Solved example 09 — Concentrated H₂SO₄ as an oxidant

Write the reactions of hot concentrated H₂SO₄ with copper, carbon and sulphur.

1. Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O.
2. C + 2H₂SO₄ → CO₂ + 2SO₂ + 2H₂O.
3. S + 2H₂SO₄ → 3SO₂ + 2H₂O.

Answer: In each case H₂SO₄ is reduced to SO₂.

Solved example 10 — Acid strength of H₂E

Why is H₂Te a stronger acid than H₂O, although O is more electronegative?

1. Acid strength of binary hydrides down a group is decided by H–E bond strength.
2. The H–Te bond is long and weak; the H–O bond is short and strong.

Answer: Weaker H–Te bond releases H⁺ more easily: H₂O < H₂S < H₂Se < H₂Te.

Exam trap gallery

Where marks are lost in this chapter.

#	What students write	What is actually correct	Why
1	NCl₅ exists like PCl₅.	No — N has no d orbitals; max covalency 4.	Only PCl₃-type NCl₃ forms.
2	PH₃ has a larger bond angle than NH₃.	NH₃ 107.8° > PH₃ 93.6°.	Bond pairs are closer to the small N atom.
3	NH₃ has the lowest b.p. of the group 15 hydrides.	PH₃ has the lowest; NH₃ is raised by H bonding.	H bonding only in NH₃.
4	H₃PO₃ is tribasic.	Dibasic (two P–OH).	The P–H hydrogen does not ionise.
5	Cu + dilute HNO₃ gives NO₂.	Dilute gives NO; concentrated gives NO₂.	Concentration changes the reduction product.
6	Oxygen has a more negative electron gain enthalpy than sulphur.	S (−200) is more negative than O (−141).	Compact 2p subshell of O.
7	H₂O is the strongest acid among group 16 hydrides.	H₂Te is the strongest; H₂O the weakest.	H–E bond weakens down the group.
8	SO₂ bleaches by oxidation.	By reduction (temporary).	Cl₂ bleaches by oxidation.
9	SO₃ is dissolved directly in water in the contact process.	It is absorbed in H₂SO₄ to form oleum.	Direct absorption forms an acid mist.
10	Ozone is paramagnetic like O₂.	O₃ is diamagnetic.	All electrons are paired.
11	White phosphorus is stored in kerosene.	White P is stored under water (Na is stored in kerosene).	White P catches fire in air.
12	Concentrated H₂SO₄ can dry NH₃.	It reacts with NH₃ (and oxidises H₂S).	Use CaO for NH₃.

Memory hook: “N small: no NCl₅, high b.p. NH₃.” · “Basicity counts OH, not H.” · “Dilute copper gives NO, concentrated gives NO₂.” · “SO₂ reduces, Cl₂ oxidises.”

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. Laughing gas is: (Easy)

• (a) NO
• (b) N₂O
• (c) NO₂
• (d) N₂O₃

Q02. The brown gas formed when Cu reacts with conc. HNO₃: (Easy)

• (a) NO
• (b) N₂O
• (c) NO₂
• (d) N₂

Q03. Most basic hydride: (Easy)

• (a) NH₃
• (b) PH₃
• (c) AsH₃
• (d) BiH₃

Q04. Catalyst in the Ostwald process: (Easy)

• (a) Fe
• (b) V₂O₅
• (c) Pt/Rh
• (d) Ni

Q05. Ozone is prepared by: (Easy)

• (a) heating O₂
• (b) silent electric discharge
• (c) electrolysis of water
• (d) burning sulphur

Q06. Stable form of sulphur at room temperature: (Easy)

• (a) monoclinic
• (b) rhombic
• (c) plastic
• (d) S₂

Q07. Basicity of H₃PO₂: (Easy)

• (a) 1
• (b) 2
• (c) 3
• (d) 0

Q08. Which cannot form a pentahalide? (Moderate)

• (a) P
• (b) As
• (c) Sb
• (d) N

Q09. Bond angle order of group 15 hydrides: (Moderate)

• (a) NH₃ < PH₃
• (b) PH₃ > AsH₃ > NH₃
• (c) NH₃ > PH₃ > AsH₃
• (d) all 109.5°

Q10. Lowest boiling point: (Moderate)

• (a) NH₃
• (b) PH₃
• (c) AsH₃
• (d) SbH₃

Q11. Paramagnetic oxide of nitrogen: (Moderate)

• (a) N₂O
• (b) NO
• (c) N₂O₅
• (d) N₂O₃

Q12. Zn + dilute HNO₃ gives mainly: (Moderate)

• (a) NO₂
• (b) N₂O
• (c) H₂
• (d) NO

Q13. Structure of PCl₅ in the solid state: (Moderate)

• (a) PCl₅ molecules
• (b) [PCl₄]⁺[PCl₆]⁻
• (c) [PCl₆]⁺[PCl₄]⁻
• (d) P₂Cl₁₀ dimer

Q14. Strongest acid: (Moderate)

• (a) H₂O
• (b) H₂S
• (c) H₂Se
• (d) H₂Te

Q15. Brown ring in the nitrate test is due to: (Moderate)

• (a) NO₂
• (b) [Fe(H₂O)₅(NO)]²⁺
• (c) FeSO₄
• (d) Fe₂O₃

Q16. SO₂ turns acidified K₂Cr₂O₇: (Moderate)

• (a) yellow
• (b) green
• (c) colourless
• (d) blue

Q17. Oleum is: (Moderate)

• (a) H₂SO₄
• (b) H₂S₂O₇
• (c) H₂S₂O₈
• (d) H₂SO₅

Q18. PH₃ is obtained by heating white P with: (Moderate)

• (a) HCl
• (b) NaOH solution
• (c) H₂SO₄
• (d) water only

Q19. Number of P–OH bonds in pyrophosphoric acid: (Tough)

• (a) 2
• (b) 3
• (c) 4
• (d) 5

Q20. In PCl₅, the axial bonds are: (Tough)

• (a) shorter than equatorial
• (b) longer than equatorial
• (c) equal
• (d) absent

Q21. Marshall’s acid is: (Tough)

• (a) H₂SO₅
• (b) H₂S₂O₈
• (c) H₂S₂O₇
• (d) H₂S₂O₃

Q22. Oxidation state of P in H₃PO₃: (Tough)

• (a) +1
• (b) +3
• (c) +5
• (d) −3

Q23. Volume of O₂ at STP from 12.25 g KClO₃: (Tough)

• (a) 1.12 L
• (b) 2.24 L
• (c) 3.36 L
• (d) 6.72 L

Q24. Which acid reduces AgNO₃ to Ag? (Tough)

• (a) H₃PO₄
• (b) H₃PO₃
• (c) H₄P₂O₇
• (d) HNO₃

Q25. O₃ with moist KI gives: (Tough)

• (a) KIO₃
• (b) I₂
• (c) KI₃ only
• (d) HI

Answer key with explanations

1. (b) N₂O — Nitrous oxide.
2. (c) NO₂ — Concentrated acid.
3. (a) NH₃ — Most available lone pair.
4. (c) Pt/Rh — Pt/Rh gauze.
5. (b) silent electric discharge — Avoids decomposition.
6. (b) rhombic — Below 369 K.
7. (a) 1 — One P–OH.
8. (d) N — No d orbitals.
9. (c) NH₃ > PH₃ > AsH₃ — Decreases down.
10. (b) PH₃ — No H bonding, smallest.
11. (b) NO — Odd electron.
12. (b) N₂O — More reactive metal, dilute acid.
13. (b) [PCl₄]⁺[PCl₆]⁻ — Ionic solid.
14. (d) H₂Te — Weakest H–E bond.
15. (b) [Fe(H₂O)₅(NO)]²⁺ — Nitrosyl complex.
16. (b) green — Cr(VI) → Cr(III).
17. (b) H₂S₂O₇ — Pyrosulphuric acid.
18. (b) NaOH solution — Disproportionation.
19. (c) 4 — Tetrabasic.
20. (b) longer than equatorial — More repulsion at 90°.
21. (b) H₂S₂O₈ — Peroxodisulphuric acid.
22. (b) +3 — 3(+1) + x + 3(−2) = 0.
23. (c) 3.36 L — 0.15 mol O₂.
24. (b) H₃PO₃ — P–H bond.
25. (b) I₂ — Used to estimate ozone.

Check-yourself answers

• C1. NO and NO₂ (odd-electron molecules).
• C2. It has an unpaired electron; pairing two NO₂ molecules gives N₂O₄ with all electrons paired.
• C3. White P exists as strained P₄ tetrahedra (60° angles); red P is polymeric and more stable.
• C4. Two.
• C5. V₂O₅.
• C6. 369 K.

One-page revision

Read this page the night before the exam.

Group 15

• ns²np³; −3 to +5; Bi³⁺ stable
• N: no d, max covalency 4, N≡N
• NH₃ > PH₃ > … basicity and bond angle
• b.p. PH₃ lowest; NH₃ H-bonded

N₂, NH₃, HNO₃

• NH₄Cl + NaNO₂ → N₂; Ba(N₃)₂ pure N₂
• Haber: Fe, 200 atm, 700 K
• Ostwald: Pt/Rh; 3NO₂ + H₂O → 2HNO₃ + NO
• Cu dil → NO; conc → NO₂; Zn dil → N₂O

Phosphorus

• White P₄ (water), red polymeric
• P₄ + NaOH → PH₃ + NaH₂PO₂
• PCl₅ TBP; solid [PCl₄]⁺[PCl₆]⁻
• H₃PO₂ 1, H₃PO₃ 2, H₃PO₄ 3 (P–OH)

Group 16

• O: −2; H-bonding; ΔegH O < S
• H₂O → H₂Te: acidity, reducing ↑
• O₃: silent discharge; bent 117°; oxidant
• Oxides: acidic, basic, amphoteric, neutral

Sulphur and SO₂

• Rhombic ⇌ monoclinic at 369 K
• SO₂: reducing; K₂Cr₂O₇ green
• Bleach: SO₂ reduces (temporary)
• Na₂SO₃ + HCl → SO₂

H₂SO₄

• Contact: V₂O₅, 720 K, 2 bar
• SO₃ → oleum H₂S₂O₇
• Dehydrating, oxidising (Cu → SO₂)
• Caro H₂SO₅; Marshall H₂S₂O₈