CBSE Class 12 Chemistry The p block elements Type MLL Questions Answers

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THE P-BLOCK ELEMENTS (TYPE MLL) - ONE MARK QUESTIONS

1. Why is H3PO3 diprotic? [1 Mark]

Answer: H₃PO₃ has two ionisable hydrogen atoms, the two P-OH hydrogens. The third hydrogen is bonded directly to phosphorus (P-H) and does not ionise.

Teacher's Note:
a) Only hydrogens attached to oxygen as O-H groups are acidic in oxyacids of phosphorus.

2. Nitrogen does not form pentahalide like phosphorous, why? [1 Mark]

Answer: Nitrogen has no vacant d-orbitals in its valence shell, so it cannot expand its octet.

Teacher's Note:
a) Phosphorus uses its empty 3d orbitals to accommodate five bonding pairs in PCl₅.

3. H2O is a liquid while H2s is a gas ,why? [1 Mark]

Answer: Water molecules are held together by strong intermolecular hydrogen bonding, whereas H₂S has only weak dipole-dipole forces.

Teacher's Note:
a) Oxygen is much more electronegative than sulphur, so the O-H...O hydrogen bonds are far stronger.

4. Arrange NH3 PH3 AsH3 BiH3 SbH3 in the increasing order of basic strength. [1 Mark]

Answer: BiH₃ < SbH₃ < AsH₃ < PH₃ < NH₃

Teacher's Note:
a) Basicity falls down the group as the size of the element increases and the lone pair becomes less available for donation.
b) NH₃ is the strongest base because nitrogen is the smallest and most electronegative atom in the group with a concentrated lone pair.

5. Oxygen is a gas while sulphur is a solid, why? [1 Mark]

Answer: Oxygen forms O₂ molecules held by weak van der Waals forces, while sulphur exists as large S₈ rings held by stronger van der Waals forces.

Teacher's Note:
a) Oxygen has strong pπ-pπ bonding, which keeps the molecule small. Sulphur does not form such bonds effectively, so it catenates into rings.
b) The larger, heavier S₈ molecules have much stronger intermolecular forces, which makes the solid stable at room temperature.

6. Write the formula of hyponitros acid. [1 Mark]

Answer: H₂N₂O₂

Teacher's Note:
a) The printed key gives HNO₂, which is nitrous acid and not hyponitrous acid. Hyponitrous acid has two nitrogen atoms.

7. Why does Al not react with conc. nitric acid? [1 Mark]

Answer: Aluminium forms a passive protective layer of oxide on its surface, which stops further reaction with the acid.

Teacher's Note:
a) This passivation also happens with chromium and iron in concentrated nitric acid.

8. Can PCl5 act as an oxidising agent and reducing agent? [1 Mark]

Answer: PCl₅ acts as an oxidising agent only, because phosphorus is in its highest oxidation state (+5) and cannot be oxidised further.

Teacher's Note:
a) Phosphorus can be reduced, for example PCl₅ → PCl₃ + Cl₂ shows the +5 state being reduced to +3.

9. Why does NO2 readily dimerise? [1 Mark]

Answer: NO₂ has an odd number of valence electrons. It dimerises to pair up the unpaired electron and form the more stable N₂O₄.

Teacher's Note:
a) The dimer N₂O₄ has all electrons paired, so it is diamagnetic and colourless.

10. Why is BiH3 the strongest reducing agent amongst all the hydrides of nitrogen family? [1 Mark]

Answer: The Bi-H bond has the lowest bond dissociation enthalpy in the group, so BiH₃ loses hydrogen most easily.

Teacher's Note:
a) Bond strength falls down the group as the atoms get larger, which increases the reducing power of the hydrides.

11. Write the chemical formula of peroxodisulphuric acid. [1 Mark]

Answer: H₂S₂O₈

Teacher's Note:
a) It contains a peroxide (O-O) linkage, which makes it a strong oxidising agent.

12. Why does NH3 act as a Lewis base /complexing agent? [1 Mark]

Answer: NH₃ has a lone pair of electrons on nitrogen which it can donate to a metal ion or an electron-deficient species.

Teacher's Note:
a) This is why ammonia forms complexes such as [Cu(NH₃)₄]²⁺ and [Ag(NH₃)₂]⁺.

13. What is the basicity of H3PO4? [1 Mark]

Answer: The basicity of H₃PO₄ is 3.

Teacher's Note:
a) Basicity is the number of ionisable O-H hydrogens, and all three hydrogens in H₃PO₄ are ionisable.

14. Why does R3P = O exist but R3N = O does not (R = alkyl group)? [1 Mark]

Answer: Phosphorus has vacant d-orbitals and can form a P=O bond, while nitrogen has no d-orbitals and cannot expand its octet to form the R₃N=O bond.

Teacher's Note:
a) This is the same reason that nitrogen does not form pentahalides.

15. Why does the reactivity of nitrogen differ from phosphorus? [1 Mark]

Answer: Nitrogen has no d-orbitals, shows a strong tendency to form pπ-pπ multiple bonds, and forms strong N≡N triple bonds, so it is unreactive.

Teacher's Note:
a) Phosphorus uses its single bonds and empty d-orbitals, so it is much more reactive than nitrogen.

16. Why is white phosphorous highly reactive? [1 Mark]

Answer: White phosphorus has a strained tetrahedral P₄ structure with P-P-P bond angles of only 60°, so it reacts readily.

Teacher's Note:
a) The strain stores extra energy in the molecule, which makes it easy to break the bonds and react.

17. Why group 16 members are called chalcogens? [1 Mark]

Answer: Group 16 elements are called chalcogens (ore-forming elements) because many metals occur in nature as their oxide or sulphide ores.

Teacher's Note:
a) The word chalcogen comes from the Greek word for ore, and oxygen and sulphur are the most common examples.

18. OF4 is not known but SF4 is known .Explain [1 Mark]

Answer: Oxygen has no vacant d-orbitals, so it cannot expand its octet to accommodate four fluorine atoms. Sulphur has vacant 3d orbitals, so SF₄ can exist.

Teacher's Note:
a) The printed key gives the same reason in short form, and the explanation is the absence of d-orbitals in oxygen.

19. Solid PCl5 exists as an ionic solid, Why? [1 Mark]

Answer: In the solid state PCl₅ exists as [PCl₄]⁺[PCl₆]^-, which is an ionic solid.

Teacher's Note:
a) The tetrahedral PCl₄⁺ cation and the octahedral PCl₆^- anion are formed by transfer of a chloride ion between molecules.

20. Bismuth is a strong oxidizing agent in pentavalentstate, why? [1 Mark]

Answer: Due to the inert pair effect, the 6s electrons of bismuth are held tightly, so the +5 state is unstable and Bi(V) is easily reduced to the more stable +3 state.

Teacher's Note:
a) Bi(V) gains two electrons to become Bi(III), which is why it acts as a strong oxidising agent.

THE P-BLOCK ELEMENTS (TYPE MLL) - TWO MARKS QUESTIONS

21. Draw the shapes of SF4, BrF3 on the basis of VSEPR theory. [2 Marks]

Answer:
1. SF₄ has five electron pairs with one lone pair, so its shape is see-saw (distorted trigonal bipyramid).
2. BrF₃ has five electron pairs with two lone pairs, so its shape is T-shaped.

Teacher's Note:
a) In SF₄ the lone pair sits in an equatorial position to reduce repulsion.
b) In BrF₃ both lone pairs occupy equatorial positions, leaving the three fluorine atoms in a T shape.

22. (a) Why is atomic radius of Argon more than that of Chlorine ?
(b) Why is ionization enthalpy of Nitrogen more than oxygen? [2 Marks]

Answer:
(a) Argon's radius is a van der Waals radius, which is measured between non-bonded atoms and is larger than a covalent radius. Chlorine's radius is a covalent radius.
(b) Nitrogen has a half-filled, stable 2p³ configuration, so more energy is needed to remove an electron from it than from oxygen.

Teacher's Note:
a) Comparing a van der Waals radius with a covalent radius is not a fair comparison, so the difference is mostly due to the type of radius used.
b) Oxygen has one paired electron in a p-orbital, and removing it is easier because of the electron repulsion in that orbital.

23. (a) Arrange F2,Cl2,Br2,I2 in the increasing order of bond dissociation energy.
(b) Arrange HOClO,HOClO2,HOClO3 in the increasing order of acidic strength. [2 Marks]

Answer:
(a) I₂ < F₂ < Br₂ < Cl₂
(b) HOClO < HOClO₂ < HOClO₃

Teacher's Note:
a) The printed key gives Cl₂ > Br₂ > F₂ > I₂, which is the decreasing order and does not match the question. The increasing order is I₂ < F₂ < Br₂ < Cl₂.
b) Acidity rises with the number of oxygen atoms on the central chlorine, because the extra oxygens withdraw electron density from the O-H bond.

24. Explain giving suitable reasons:
(i) NH3 has higher boiling point than PH3
(ii) SbF5 is known but BiF5 is unknown. [2 Marks]

Answer:
(i) NH₃ molecules are linked by intermolecular hydrogen bonding, while PH₃ has no hydrogen bonding, so more energy is needed to boil NH₃.
(ii) Bismuth shows the inert pair effect, so the +5 state is unstable and BiF₅ cannot form, while antimony in the +5 state is stable.

Teacher's Note:
a) Hydrogen bonding requires an electronegative atom such as nitrogen, which phosphorus does not provide.
b) The 6s pair of bismuth stays inert, so Bi(III) is the stable oxidation state.

25. Explain giving suitable reasons:
(i) SF6 is well known but SH6 is not known.
(ii) Proton affinity of NH3 is more than PH3. [2 Marks]

Answer:
(i) The six small, highly electronegative fluorine atoms surround and protect the sulphur atom sterically, and S-F bonds are strong. Hydrogen atoms are too small to shield sulphur, and S-H bonds are weak, so SH₆ does not exist.
(ii) Nitrogen is smaller and more electronegative than phosphorus, so its lone pair is more concentrated and more readily donated to H⁺.

Teacher's Note:
a) The printed key explains (i) by the bond enthalpy of H-H and F-F, which is not the real reason for SF₆ being stable.
b) Proton affinity is a measure of basicity in the gas phase, and NH₃ is the stronger base of the two.

26. Explain giving suitable reasons.
(i) Sulphur in vapour form is paramagnetic in nature.
(ii) Catenation properties of Phosphorous is more than Nitrogen. [2 Marks]

Answer:
(i) In the vapour state sulphur exists as S₂, which is like O₂. It has two unpaired electrons in antibonding π* orbitals, so it is paramagnetic.
(ii) The P-P single bond is stronger than the N-N single bond, so phosphorus forms long chains and rings much more readily than nitrogen.

Teacher's Note:
a) The printed key for (ii) gives the reason about pπ-pπ bonding, which explains multiple bonding and not catenation.
b) The N-N bond is weak because of repulsion between the lone pairs on the small nitrogen atoms.

27. Give chemical equations ,when :
(a). Ammonium dichromate is heated ?
(b). Sodium azide is heated? [2 Marks]

Answer:
(a) (NH₄)₂Cr₂O₇ → N₂ + Cr₂O₃ + 4H₂O
(b) 2NaN₃ → 2Na + 3N₂

Teacher's Note:
a) Ammonium dichromate decomposes to give nitrogen gas and green chromium(III) oxide.
b) Heating sodium azide to about 300 °C is a laboratory method for preparing pure nitrogen.

28. Complete the following equations:
(i) HgCl2 + PH3
(ii) P4 + NaOH + H2O [2 Marks]

Answer:
(i) 3HgCl₂ + 2PH₃ → Hg₃P₂ + 6HCl
(ii) P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂

Teacher's Note:
a) Mercury(II) chloride gives a black precipitate of mercury phosphide with phosphine.
b) In reaction (ii) phosphorus is both oxidised and reduced, which is a disproportionation reaction.

29. Write main differences between the properties of white phosphorus and red phosphorus. [2 Marks]

Answer:
1. White phosphorus is made of P₄ tetrahedra and is highly reactive, poisonous and glows in the dark, while red phosphorus is a polymeric solid that is less reactive, non-poisonous and does not glow.
2. White phosphorus is insoluble in water but soluble in carbon disulphide, while red phosphorus is insoluble in water and in carbon disulphide.

Teacher's Note:
a) The printed key says red phosphorus is soluble in water, which is wrong. Red phosphorus is insoluble in all common solvents.
b) White phosphorus must be stored under water because it catches fire in air.

30. Arrange H2O,H2S,H2Se, H2Te in the increasing order of (i) Acidic Character (ii) Thermal stability . [2 Marks]

Answer:
(i) Acidic character: H₂O < H₂S < H₂Se < H₂Te
(ii) Thermal stability: H₂Te < H₂Se < H₂S < H₂O

Teacher's Note:
a) Acidic character rises down the group because the H-E bond becomes weaker and the anion is larger and more stable.
b) Thermal stability falls down the group because the bonds get longer and weaker.

THE P-BLOCK ELEMENTS (TYPE MLL) - THREE MARKS QUESTIONS

31. (i) Why is ICl more reactive than I2?
(ii) Interhalogen compounds are strong Oxidizing agents ,why?
(iii) Bleaching of flowers by Cl2 is permanent while SO2 is temporary,why? [3 Marks]

Answer:
(i) The I-Cl bond is weaker and more polar than the I-I bond, so ICl breaks apart and reacts more easily.
(ii) Interhalogen compounds have weak, polar X-X' bonds that break easily to release halogen atoms, so they act as strong oxidising agents.
(iii) Cl₂ bleaches by oxidation, destroying the colouring matter permanently. SO₂ bleaches by reduction, forming a colourless compound that is slowly reoxidised by air, so the colour returns.

Teacher's Note:
a) A weaker bond means a lower bond dissociation enthalpy, which is the main reason for the higher reactivity.
b) Polarity adds to the reactivity, since the more electronegative halogen pulls electron density towards itself.
c) Permanent and temporary bleaching are due to the difference between oxidation and reduction.

32. Give Reasons:
(a) Iodine is more soluble in KI solution than in water.
(b) HF is stored in wax-coated bottles.
(c) HCl is not used to make the medium acidic in titrations involving KMnO4. [3 Marks]

Answer:
(a) Iodine reacts with iodide ions to form the soluble I₃^- complex, so its solubility increases.
(b) HF attacks glass, which is made of silica (SiO₂), so it is stored in wax-coated or plastic bottles.
(c) HCl is oxidised by KMnO₄ to Cl₂, so it consumes the oxidant and gives wrong titration results. Dilute H₂SO₄ is used instead.

Teacher's Note:
a) The reaction is I₂ + I^- ⇌ I₃^-.
b) The printed key writes HF reacts with silica, which is the correct reason in short form.
c) The printed key writes the oxidation of HCl to Cl₂, which is the same explanation.

33. Explain giving suitable reasons:
(i) SbF5 is known but BiF5 is unknown.
(ii) CN- ion is known but CP- is not .
(iii) Compounds of Noble gases are known with Xe and fluorine. [3 Marks]

Answer:
(i) Due to the inert pair effect, bismuth does not readily use its 6s electrons, so the +5 state of bismuth is too unstable to form BiF₅.
(ii) Carbon can form a strong pπ-pπ triple bond with nitrogen in CN^-. Phosphorus cannot form such a stable pπ-pπ bond, so CP^- is not stable.
(iii) Xenon has a low ionisation enthalpy, and fluorine is the most electronegative element and a small atom, so Xe-F compounds form readily.

Teacher's Note:
a) The inert pair effect is strongest in the heavy elements of the p-block.
b) Fluorine is able to stabilise xenon in higher oxidation states because its bonds are strong.
c) These points are often asked together, so learn each reason with its example.

34. Explain giving suitable reasons:
(i) PH3 has lower boiling point than NH3.
(ii) Nitric oxide becomes brown when released in air.
(iii) When HCl reacts with finely powered iron, it forms ferrous chloride and not ferric chloride. [3 Marks]

Answer:
(i) PH₃ has no intermolecular hydrogen bonding, unlike NH₃, so less energy is needed to boil it.
(ii) Nitric oxide is oxidised by atmospheric oxygen to nitrogen dioxide, NO₂, which is brown: 2NO + O₂ → 2NO₂.
(iii) HCl reacts with iron to give Fe²⁺ and hydrogen gas. The hydrogen produced prevents further oxidation of Fe²⁺ to Fe³⁺, so FeCl₃ does not form: Fe + 2HCl → FeCl₂ + H₂.

Teacher's Note:
a) The printed key says the hydrogen hinders the formation of FeCl₃, which is correct.
b) Nitrogen dioxide is the brown gas, while nitric oxide itself is colourless.
c) The hydrogen evolved is a reducing gas, which keeps iron in the +2 state.

35. (i) Which Xe compound has distorted octahedral shape?
(ii) How does Chlorine react with hot and concentrated NaOH ?
(iii) Write the chemical reaction involved in the ring test. [3 Marks]

Answer:
(i) XeF₆ has a distorted octahedral shape because of its lone pair.
(ii) Chlorine disproportionates: 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O.
(iii) NO₃^- + 3Fe²⁺ + 4H⁺ → 3Fe³⁺ + NO + 2H₂O, followed by [Fe(H₂O)₆]²⁺ + NO → [Fe(H₂O)₅NO]²⁺ + H₂O, which gives the brown ring.

Teacher's Note:
a) The printed key for (iii) is unbalanced, so the balanced equations are given above.
b) XeOF₄ is square pyramidal, while XeO₃ is pyramidal, so XeF₆ is the only one that matches the question.
c) Hot concentrated alkali gives chlorate, while cold dilute alkali gives hypochlorite.

36. (a) Are all the bonds in PCl5 equivalent in length?
(b) On the basis of structure show that H3PO2 is a good reducing agent.
(c) How many P-OH bonds are present in Pyrophosphoric acid? [3 Marks]

Answer:
(a) No. The two axial P-Cl bonds are slightly longer (about 2.19 Å) than the three equatorial P-Cl bonds (about 2.04 Å).
(b) H₃PO₂ has two P-H bonds and one P-OH bond. The P-H hydrogens are not ionisable, but they can be donated in reduction reactions, which makes H₃PO₂ a good reducing agent.
(c) Pyrophosphoric acid, H₄P₂O₇, has four P-OH bonds.

Teacher's Note:
a) The printed key says H₃PO₂ has one P-H bond, which is wrong. Its structure has two P-H bonds, which is why it reduces silver salts and other oxidising agents.
b) The axial bonds are longer because they experience greater bond pair repulsion.

37. Complete the following equations:
P4 + SOCl2
NH3 + CuSO4(aq)
XeF4 + H2O [3 Marks]

Answer:
1. P₄ + 8SOCl₂ → 4PCl₃ + 4SO₂ + 2S₂Cl₂
2. 4NH₃ + CuSO₄(aq) → [Cu(NH₃)₄]SO₄
3. 6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂

Teacher's Note:
a) The printed key for XeF₄ + H₂O is not balanced for oxygen, so the balanced equation above is the correct one.
b) In the ammonia reaction, the deep blue tetraammine copper(II) complex is formed.
c) Xenon(VI) oxide, XeO₃, is the product that forms by hydrolysis of XeF₄.

38. (i) Arrange M-F,M-Cl,M-Br,M-I in the increasing order of ionic character.
(ii) Arrange HF,HCl,HBr,HI in the increasing order of reducing behavior.
(iii) Arrange F2,Cl2,Br2,I2 in the increasing order of bond dissociation energy. [3 Marks]

Answer:
(i) M-I < M-Br < M-Cl < M-F
(ii) HF < HCl < HBr < HI
(iii) I₂ < F₂ < Br₂ < Cl₂

Teacher's Note:
a) Ionic character falls as the halide becomes less electronegative and larger.
b) Reducing power rises down the group because the H-X bond becomes weaker.
c) Fluorine has a low bond dissociation enthalpy because of the repulsion between its lone pairs, so it is lower than chlorine.

39. Starting from sulphur ,how would you manufacture H2SO4 by contact process. [3 Marks]

Answer:
1. S + O₂ → SO₂
2. 2SO₂ + O₂ ⇌ 2SO₃ (in the presence of V₂O₅ catalyst)
3. SO₃ + H₂SO₄ → H₂S₂O₇ (oleum), which is then diluted with water: H₂S₂O₇ + H₂O → 2H₂SO₄

Teacher's Note:
a) SO₃ is not dissolved directly in water because the reaction is very exothermic and produces a fine mist.
b) Conditions used are about 2 bar pressure and 720 K, which give a good yield of SO₃.
c) The catalyst is vanadium(V) oxide.

40. Write the reaction involved in formation of ammonia by Habers process?State the favorable conditions for good yield of ammonia . [3 Marks]

Answer:
1. N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH is negative.
2. Favourable conditions: low temperature (about 700 K as a compromise with rate), high pressure (about 200 atm), and a catalyst of finely divided iron with molybdenum as a promoter.

Teacher's Note:
a) Low temperature favours the exothermic forward reaction, but too low a temperature makes the reaction very slow.
b) High pressure favours the side with fewer gas molecules, which is ammonia.
c) The catalyst increases the rate but does not change the equilibrium position.

THE P-BLOCK ELEMENTS (TYPE MLL) - FIVE MARKS QUESTIONS

41. A gas "X" is soluble in water . Its aq. Solution turns red litmus blue with excess of aq. CuSO4 solution it gives deep blue colour and with FeCl3 solution a brownish ppt. soluble in HNO3 is obtained. Identify gas "X" and write reactions for changes observed . [5 Marks]

Answer:
1. Gas X is ammonia, NH₃. Its aqueous solution is basic, so it turns red litmus blue.
2. With excess aqueous CuSO₄: 4NH₃ + CuSO₄(aq) → [Cu(NH₃)₄]SO₄, a deep blue complex.
3. With aqueous FeCl₃: 3NH₃ + 3H₂O + FeCl₃ → Fe(OH)₃ + 3NH₄Cl, giving a brown precipitate.
4. The brown precipitate of Fe(OH)₃ dissolves in dilute nitric acid: Fe(OH)₃ + 3HNO₃ → Fe(NO₃)₃ + 3H₂O.

Teacher's Note:
a) Ammonia is the only common gas that gives a deep blue complex with copper(II) sulphate solution.
b) The brown precipitate is iron(III) hydroxide, which is insoluble in water but soluble in acids.
c) The question asks for reactions, so each change is written as a separate equation.
d) Excess ammonia is needed to form the complex.

42. Write the reaction involved in formation of Nitric acid by Osrwald,s process? State the favorable conditions for good yield of Nitric oxide. [5 Marks]

Answer:
1. 4NH₃ + 5O₂ → 4NO + 6H₂O (Pt-Rh catalyst)
2. 2NO + O₂ → 2NO₂
3. 3NO₂ + H₂O → 2HNO₃ + NO
4. Favourable conditions for nitric oxide: high temperature (about 800 to 900 °C), low pressure (about 4 to 6 atm), and a platinum-rhodium catalyst with short contact time.

Teacher's Note:
a) The printed key gives only the three equations and no conditions for nitric oxide, so the conditions above are added.
b) The first step is strongly exothermic, so the temperature must be controlled to keep the yield of NO high.
c) The NO formed in step 3 is recycled back to the oxidation step.
d) Low pressure favours the formation of NO because the number of gas molecules increases.

43. Complete the following equations:
(I) XeF4 + H2O
(ii) XeF6 + PF5
(iii) Cl2 + F2 (excess)
(iv) HgCl2 + PH3
(v) SO3 + H2SO4 [5 Marks]

Answer:
(i) 6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂
(ii) XeF₆ + PF₅ → [XeF₅]⁺[PF₆]^-
(iii) Cl₂ + 3F₂ (excess) → 2ClF₃
(iv) 3HgCl₂ + 2PH₃ → Hg₃P₂ + 6HCl
(v) SO₃ + H₂SO₄ → H₂S₂O₇

Teacher's Note:
a) In (i) the oxygen must balance, which the printed key does not do, so the balanced equation is given.
b) In (ii) XeF₆ acts as a fluoride ion donor to PF₅, forming an ionic salt.
c) In (iii) excess fluorine converts chlorine to chlorine trifluoride.
d) In (v) sulphur trioxide dissolves in sulphuric acid to form oleum.

44. A translucent white waxy solid 'A' on heating in an inert atmosphere is converted in to its allotropic form (B). Allotrope 'A' on reaction with very dilute aqueous KOH liberates a highly poisonous gas 'C' having rotten fish smell. With excess of chlorine 'A' forms 'D' which hydrolysis to compound 'E'. Identify compounds 'A' to 'E' . [5 Marks]

Answer:
A = White phosphorus (P₄)
B = Red phosphorus
C = Phosphine, PH₃
D = Phosphorus pentachloride, PCl₅
E = Phosphoric acid, H₃PO₄

Teacher's Note:
a) The reaction with KOH is P₄ + 3KOH + 3H₂O → PH₃ + 3KH₂PO₂.
b) The reaction with excess chlorine is P₄ + 10Cl₂ → 4PCl₅.
c) Hydrolysis of PCl₅ gives PCl₅ + 4H₂O → H₃PO₄ + 5HCl.
d) The printed key gives the same identities, and the reasoning above links each step.

45. What happens when Concentrated H2SO4 is added to/ Give the reactions of H2SO4 with(i) calcium fluoride (ii) KCl, (iii) Sugar (iv) Cu turnings.(v) Sulphur [5 Marks]

Answer:
(i) CaF₂ + H₂SO₄(conc.) → CaSO₄ + 2HF
(ii) KCl + H₂SO₄(conc.) → KHSO₄ + HCl (at room temperature); 2KCl + H₂SO₄ → 2HCl + K₂SO₄ (on strong heating)
(iii) C₁₂H₂₂O₁₁ + conc. H₂SO₄ → 12C + 11H₂O (sugar is charred)
(iv) Cu + 2H₂SO₄(conc.) → CuSO₄ + SO₂ + 2H₂O
(v) 3S + 2H₂SO₄(conc.) → 3SO₂ + 2H₂O

Teacher's Note:
a) Calcium fluoride gives hydrogen fluoride, which is a gas that attacks glass and must be handled with care.
b) With potassium chloride, the product depends on the temperature, as shown above.
c) Concentrated sulphuric acid removes water from sugar, which leaves black carbon behind.
d) Copper is oxidised by hot concentrated acid, and the acid is reduced to sulphur dioxide.

THE P-BLOCK ELEMENTS - ADDITIONAL QUESTIONS

1. Why does NO₂ dimerise?
2. Give chemical reactions involved in brown ring test to confirm nitrates.
3. Give the structure of nitric acid.
4. Give equations in each step of Ostwald's process.
5. Give flow chart for preparation of ammonia by Haber's process.
6. Explain preparation of nitrogen.
7. Why do chromium and aluminium not react with the most oxidizing agent?
8. Name the oxides of nitrogen and give oxidation number of each oxide.
9. Give conditions which favors formation of ammonia as it is a reversible reaction.
10. Draw the structures of white phosphorus and red phosphorus. Which one of these two types of phosphorus is more reactive and why?
11. Which oxoacids of phosphorus are reducing in nature?
12. Why is phosphorus acid diprotic and phosphoric acid triprotic in spite of 3 hydrogens in both?
13. Give the structure of oxy acids of phosphorus and list the anions formed.
14. Why does PCl₃ fume in moisture?
15. Show that PH₃ is basic in nature.
16. Give reason that NCl₅ is not formed but PCl₅ is formed.
17. Show that hydrogen peroxide behaves both as an oxidizing and reducing agent.
18. What is oleum? Draw its structure.
19. What happens when sulphur is passed through conc. H₂SO₄ solution and SO₂ is passed through an aqueous solution of Fe(III) salt?
20. Why are halogens coloured?

P-BLOCK - ANSWERS

1. NO₂ contains an odd number of valence electrons. It behaves as an odd electron molecule and therefore undergoes dimerisation to form the stable N₂O₄ molecule with an even number of electrons.

2. The brown ring test depends on the ability of ferrous ion to reduce nitrates to nitric oxide, which reacts with ferrous ion to form a brown coloured complex.
NO₃^- + 3Fe²⁺ + 4H⁺ → NO + 3Fe³⁺ + 2H₂O
[Fe(H₂O)₆]²⁺ + NO → [Fe(H₂O)₅(NO)]²⁺ + H₂O

3. Structure of nitric acid: (a) planar molecule with angles about 105° (H-O-N) and 120° (O-N-O); (b) two resonance structures.

4. Nitric acid is produced industrially by the Ostwald process, which involves three steps:
Step 1: Catalytic oxidation of ammonia: 4NH₃ + 5O₂ → 4NO + 6H₂O
Step 2: Oxidation of nitric oxide: 2NO + O₂ → 2NO₂
Step 3: Reaction with water and oxygen to form nitric acid: 4NO₂ + 2H₂O + O₂ → 4HNO₃

5. nitrogen + hydrogen → ammonia (+ heat). The process operates continuously and the recycling of the unreacted nitrogen and hydrogen gives an overall conversion of about 98%.

6. Air is liquefied, and the oxygen, which is about 20.9%, gets boiled off at -183°C, leaving liquid nitrogen behind, which boils at -196°C. This process is known as fractional distillation. Nitrogen can also be made by heating NaN₃ to 300 degrees C. Annual worldwide production is around 44,000,000 tons.

7. These elements form a passive layer of oxide on the surface and prevent the metal from reacting with nitric acid.

8. The common oxides of nitrogen include examples of nitrogen with every oxidation number from +1 to +5:
N as +1: N₂O
N as +2: NO
N as +3: N₂O₃
N as +4: NO₂
N as +5: N₂O₅

9. The reaction is reversible. Only about 15-20% of the reactants are converted into products. The forward reaction is exothermic. The amount of product or yield from a reversible reaction depends on temperature, pressure and catalyst. Decreasing the temperature favors exothermic reactions. Increasing the pressure favors smaller volume. Using a catalyst gives the equilibrium conditions more quickly.

10. White phosphorus is less stable and therefore more reactive than red phosphorus under normal conditions because of angular strain in the P₄ molecule, where the angles are 60° only.

The more reactive form is white phosphorus, because of the angular strain in the P₄ molecule (angles of 60° only).

11. All acids which have a P-H bond are reducing in nature. Examples: phosphorous acid, H₃PO₃ (one P-H bond, O=P(OH)₂H) and hypophosphorous acid, H₃PO₂ (two P-H bonds, O=P(OH)H₂).

12. Phosphoric acid, H₃PO₄, has 3 ionisable hydrogens (three P-OH groups). In phosphorous acid, H₃PO₃, there are 2 ionisable hydrogens (two P-OH groups); the third one is bonded directly to phosphorus (P-H).

13. Oxyacids of phosphorus and the oxyanions formed:

14. It hydrolyses in moisture giving fumes of HCl. PCl₃ + 3H₂O → H₃PO₃ + 3HCl

15. PH₃ has a lone pair of electrons and readily reacts with acids like HI and forms PH₄I: PH₃ + HI → PH₄I.

16. Nitrogen does not have usable d orbitals and cannot expand its octet. Phosphorus can expand its valence shell to hold more than eight electrons, but nitrogen cannot.

17. It produces oxygen and acts as an oxidizing agent in both acid and basic medium: Mn²⁺ + H₂O₂ → Mn⁴⁺ + 2OH^-
As a reducing agent: HOCl + H₂O₂ → H₂O + Cl^- + O₂ (with H⁺)

18. Oleum is an oxoacid of sulphur and is pyrosulphuric acid, H₂S₂O₇.

19. When sulphur is passed through conc. H₂SO₄ solution it forms SO₂: 3S + 2H₂SO₄(conc.) → 3SO₂ + 2H₂O.
When SO₂ is passed through an aqueous solution of Fe(III) salt, it converts Fe(III) ions to Fe(II): 2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^2- + 4H⁺

20. Absorption of radiations in the visible region by halogen atoms results in the excitation of outer electrons to higher energy levels. By absorbing radiation of different wavelengths, they display different colours. For example, F₂ is yellow, Cl₂ is greenish yellow, Br₂ is red and I₂ is violet.

QUESTIONS (1x20)

1. Which of NH₃ and H₃O⁺ has higher bond angle and why?
2. Which of PH₄I and PH₄Cl is more stable and why?
3. What is the basicity of H₃PO₃ and H₃PO₄?
4. NH₃ is more easily liquefiable than PH₃?
5. Which of NH₃ and PH₃ is stronger Lewis base and why?
6. Why does NO₂ dimerise?
7. N₂O₄ is colourless but NO₂ is brown in colour?
8. Write the products of hydrolysis of ClF₃.
9. Why is S₂ paramagnetic?
10. Why does not SO₃ disproportionate?
11. Why Cl₂ bleaches permanently but SO₂ temporarily?
12. Why is He used in observation balloons?
13. Why is SF₆ resistant to hydrolysis?
14. What is the geometry and shape of ClF₅?
15. Why is ICl more reactive than I₂?
16. Arrange following in increasing order of their reactivity: IF, F₂ and I₂
17. Xe is more reactive than He. Why?
18. Does the hydrolysis of XeF₆ lead to a redox reaction? Why?
19. H₂S is less soluble in water than H₂Se. Why?
20. Trimethylamine is more basic than trisilylamine. Why?

Answer: (20x1)

1. H₃O⁺ because the central atom has higher electronegativity, thus it pulls bond pairs of electrons towards itself and bp-bp repulsion increases.
2. PH₄I is more stable. PH₄⁺ is a bigger cation and I^- is a bigger anion, so it offers effective crystal packing and has larger lattice enthalpy.
3. 2 and 3
4. Due to intermolecular H-bonding in NH₃.
5. NH₃, because of its smaller size it has greater charge density on nitrogen.
6. In NO₂, there is an odd electron.
7. In N₂O₄ there is no unpaired electron.
8. ClF₃ + 2H₂O → 3HF + HClO₂
9. Due to presence of unpaired electrons on anti-bonding orbitals.
10. Sulphur is in its highest oxidation state.
11. Cl₂ bleaches by oxidation but SO₂ bleaches by reduction.
12. He is very light.
13. SF₆ is sterically protected.
14. Geometry: octahedral and shape: square pyramidal.
15. ICl has low bond dissociation enthalpy.
16. I₂ < IF < F₂ (increasing reactivity).
17. Less I.E.
18. No, because the oxidation state of Xe does not change.
19. H₂Se has stronger van der Waals forces with water.
20. Due to presence of vacant d-orbitals in Si, the pair of electrons lying on N disperses via dπ-pπ back bonding.

QUESTIONS (10x2)

1. Give reason for the following: (A) Phosphorus is reactive but Nitrogen is much stable. (B) Nitrogen is linear but Phosphorus is tetrahedral.
2. Explain why? (A) Nitrogen has stronger tendency of multiple bonding than that of Phosphorus. (B) NCl₅ does not exist but PCl₅ exists.
3. Explain the chemistry of ring test of nitrate ion.
4. Draw the structures of following using VSEPR Model: XeO₂F₂ and XeO₃
5. Complete the following reactions: (A) Ca₃P₂ + H₂O → (B) Cu + Conc. HNO₃ →
6. Give reason for the following: (a) NO₂ has net dipole moment but N₂O₄ does not have? (b) Phosphorus has greater catenation tendency than Nitrogen?
7. Explain following: (1) Interhalogens are covalent, diamagnetic. (2) I₂ is soluble in aqueous KI.
8. Arrange following according to the property shown against each: (a) HClO, HClO₃, HClO₄, HClO₂ - Increasing order of acidity; (b) PH₃, NH₃, SbH₃, AsH₃, BiH₃ - Decreasing order of basicity
9. Arrange following according to the property shown against each: (a) I₂, F₂, Cl₂, Br₂ - Increasing order of ease of liquefaction; (b) ClO, Cl₂O₃, Cl₂O₅, Cl₂O₇ - Increasing acidity strength
10. Complete the following reactions: (1) P₄ + NaOH + H₂O → (2) NaCl + MnO₂ + H₂SO₄ →

Answer (10x2)

1. (A) N₂ has much higher bond enthalpy. (B) In N₂, N is sp-hybridized but in P₄, P is sp³-hybridized.
2. (A) N has three unpaired electrons in p-orbitals, has high effective nuclear charge and small atomic size to undergo greater overlapping of the orbitals. (B) N does not have d-orbitals.
3. 3Fe²⁺ + NO₃^- + 4H⁺ → 3Fe³⁺ + NO + 2H₂O
   [Fe(H₂O)₆]²⁺ + NO → [Fe(H₂O)₅(NO)]²⁺ + H₂O (brown ring)
4. Each correct structure one mark. XeO₂F₂: see-saw (trigonal bipyramidal arrangement with one lone pair); XeO₃: pyramidal.
5. (A) Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PH₃ (B) Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O
6. (a) In NO₂ there is an odd electron on nitrogen. (b) The P-P bond is stronger than the N-N bond.
7. (1) Because they have high electronegativity and paired electrons. (2) It forms KI₃.
8. (a) HClO < HClO₂ < HClO₃ < HClO₄ (b) NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃
9. (a) F₂, Cl₂, Br₂, I₂ (b) ClO, Cl₂O₃, Cl₂O₅, Cl₂O₇
10. (1) P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂ (2) 4NaCl + MnO₂ + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + Cl₂ + 2H₂O

QUESTIONS (10x3)

1. Give reason for the following: i) NO₂ is acidic oxide but NO is neutral? ii) NH₃ has greater tendency of complex formation than PH₃? iii) PH₃ dissolves in HI. Why?
2. Explain for the following: a) H₃PO₃ shows disproportionation reactions? b) PCl₅ in solid state exists as an ionic compound? c) BiCl₅ is a strong oxidizing agent?
3. Explain the structure of the following compounds using VSEPR theory: a) PCl₆^- b) SF₄ c) ICl₂^-
4. Give the comparative account of the chemistry of carbon and silicon with regard to their: [i] property of catenation [ii] stability of hydrides and oxides
5. Account for the following: [i] Ammonia has higher boiling point than phosphine [ii] Trimethyl amine is pyramidal and trisilyl amine is planar. [iii] Ammonia is stronger base than phosphine.
6. Describe the following about halogen family: [i] Oxidising power [ii] Relative acidic strength of their hydrides [iii] Relative reducing strength of their hydrides.
7. Give reason for the following observations. [i] Noble gases are mostly chemically inert [ii] Nitrogen does not form pentahalide [iii] Bismuth is a strong oxidising agent in pentavalent state
8. Arrange following according to the property shown against each: a) NO₂, P₂O₃, N₂O₅, P₂O₅, As₂O₃, Bi₂O₃ - Increasing acidity strength; b) PH₃, NH₃, SbH₃, AsH₃, BiH₃ - Decreasing order of thermal stability; c) Xe, He, Ne, Kr, Ar - Increasing order of ease of liquefaction
9. Complete following reactions: 1) NH₃ + AgCl → 2) XeF₂ + H₂O → 3) NaNO₂ + NH₄Cl →
10. How does ozone react with the following: i) NO ii) PbS iii) Aq. KI

Answer (10x3)

1. i) More oxygen content in NO₂ than NO. ii) NH₃ is a stronger Lewis base. iii) PH₃ is a base and HI is an acid, so it gives PH₄I.
2. a) The O.S. of P is +3, so it can undergo oxidation as well as reduction. b) It exists as [PCl₄]⁺[PCl₆]^-. c) Bi is in +5 oxidation state but its stable O.S. is +3.
3. Correct structures: PCl₆^- is octahedral; SF₄ is see-saw; ICl₂^- is linear.
4. [i] P has greater catenation tendency than N. [ii] The oxides and hydrides of N are more stable than those of P.
5. [i] Due to intermolecular H-bonding. [ii] Presence of d-orbitals in Si can allow dπ-pπ back bonding, thereby dispersing the lone pair. [iii] The lone pair lying on N in NH₃ can be easily donated due to greater charge density on nitrogen.
6. [i] F₂ > Cl₂ > Br₂ > I₂ [ii] HI > HBr > HCl > HF [iii] HI > HBr > HCl > HF
7. [i] They have very high I.E. and completed octet. [ii] Absence of d-orbitals. [iii] Inert pair effect.
8. a) Bi₂O₃ < As₂O₃ < P₂O₃ < P₂O₅ < NO₂ < N₂O₅ b) NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃ c) He, Ne, Ar, Kr, Xe
9. 1) AgCl + 2NH₃ → [Ag(NH₃)₂]Cl 2) 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂ 3) NaNO₂ + NH₄Cl → NaCl + N₂ + 2H₂O
10. i) NO + O₃ → NO₂ + O₂ ii) PbS + 4O₃ → PbSO₄ + 4O₂ iii) 2KI + O₃ + H₂O → 2KOH + I₂ + O₂

QUESTIONS (5x5)

1. A white coloured salt (A) on treatment with conc. H₂SO₄ gives a pungent smelling gas (B) which turns moist blue litmus to red. The gas (B) oxidizes in presence of MnO₂ to yield a greenish yellow gas (C). The gas (C) is used in disinfecting drinking water and decolourising the wood pulp in paper industries. Identify A, B and C and write necessary equations.
2. Give reason for the following observations. a) HF is weakest acid and HI is strongest. b) Fluorides of Xe undergo hydrolysis readily. c) Oxygen is diatomic but S is octatomic? d) Reaction of NaBr and H₂SO₄ does not form HBr but it forms Br₂ gas. e) HF is liquid but HCl is a gas.
3. Arrange following according to the property shown against each: a) HClO, HClO₃, HClO₄, HClO₂ - Increasing order of acidity; b) HClO, HBrO, HIO, HFO - Increasing order of acidity; c) F₂, O₂, Cl₂, Br₂ - Increasing order of oxidizing tendency; d) PH₃, NH₃, SbH₃, AsH₃, BiH₃ - Decreasing order of bond angle; e) I₂, F₂, Cl₂, Br₂ - Increasing order of b.p.
4. An element (X) on heating with conc. NaOH yields a poisonous gas (Y) and spontaneously catches fire. The gas (Y) reacts with aq. HgCl₂ to form precipitate (Z). Identify X, Y and Z and write necessary equations.
5. Draw the structure of the following compounds using VSEPR theory: XeF₆, XeOF₄, XeO₃, PCl₃, White Phosphorus (P₄)

Answer: 5 Marks

1. (A) = NaCl, (B) = HCl, (C) = Cl₂
   NaCl (A) + H₂SO₄ → NaHSO₄ + HCl (B)
   4HCl + MnO₂ → MnCl₂ + 2H₂O + Cl₂ (C)
2. a) Bond dissociation enthalpy of HF is more than HI. b) Presence of vacant d-orbitals in Xe. c) O=O is stronger than S=S. d) Because H₂SO₄ oxidises HBr to Br₂. e) Presence of intermolecular H-bonding in HF.
3. a) HClO < HClO₂ < HClO₃ < HClO₄ b) HIO < HBrO < HClO < HFO c) Cl₂, Br₂, O₂, F₂ (as given in the answer key) d) NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃ e) F₂ < Cl₂ < Br₂ < I₂
4. X = P₄, Y = PH₃, Z = Hg₃P₂
   P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂
   2PH₃ + 3HgCl₂ → Hg₃P₂ + 6HCl
5. Structures: P₄ (tetrahedral, 60° angles); PCl₃ (pyramidal with one lone pair on P); XeF₆ (distorted octahedral, one lone pair); XeOF₄ (square pyramidal, one lone pair); XeO₃ (pyramidal, one lone pair).