2. Why does the reactivity of nitrogen differ from phosphorus?

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1 1. Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity. (i) Electronic configuration The valence shell electronic configuration of these elements is ns 2 np 3.The s orbital in these elements is completely filled and p orbitals are half filled, making their electronic configuration extra stable. (ii) Oxidation state The common oxidation state of the elements are -3, +3 and +5. The tendencies to exhibit -3 oxidation state decreases down the group due to increase in size and metallic group. In the last member of the group, bismuth hardly forms any compound in -3 oxidation state. The stability of +5 oxidation state decreases down the group. The stability of +5 oxidation state decreases and that of +3 state increases (due to inert pair effect) down the group. Nitrogen exhibits +1, +2, +4 oxidation states also when it reacts with oxygen. Phosphorus also shows +1 and +4 oxidation states in some oxo acids. (iii) Atomic size: Covalent and ionic (in a particular state) radii increase in size down the group. There is a considerable increase in size down the group. There is a considerable increase in covalent radius from N to P. However, from As to Bi only a small increase in covalent radius is observed. This is due to the presence of completely filled d and f orbitals in heavier members. (iv) Ionization enthalpy: Ionisation enthalpy decreases down the group due to gradual increase in atomic size. Because of the extra stable half filled p orbitals electronic configuration and smaller size, the ionization enthalpy of the group 15 elements is much greater than that of group 14 elements in the corresponding periods. The order of successive ionization ehthalpies are expected as H 1 < H 2 < H 3. (v) Electronegativity: The electronegativity value, in general, decreases down the group with increasing atomic size. However, amongst the heavier elements, the different is not that much pronounced. 2. Why does the reactivity of nitrogen differ from phosphorus? The reactivity of nitrogen is different from phosphorus because of the following reasons (i) Nitrogen has a small size, high electronegativity, and high ionization enthalpy as compared to phosphorus. 1

2 (ii) Nitrogen does not contain vacant d orbitals in its valence shell whereas phosphorus contains vacant d orbitals in its valence shell. (iii) Nitrogen has ability of form (N N) triple bond as a result of which its bond enthalpy (941.4 kj mol -1 ) is very high making it less reactive. 3. Discuss the trends in chemical reactivity of group 15 elements. The elements of Group 15 differ from one another appreciably in their chemical reactivity. Nitrogen has a very high dissociation energy (941.4 kj mol -1 ) and is practically inert, which is why it has accumulated in large amounts in the atmosphere. Phosphorus, in one of its allotrophic forms i.e. white phosphorus is extremely reactive. The strained structure of P 4 is responsible for high chemical activity. It catches fire when exposed to air forming P 4 O 10. The other allotrophe, red phosphorus is stable in air at room temperature, though it reacts on heating. The heavier elements, As, Sb and Bi are less reactive. Arsenic is stable in dry air. When heated in air it sublimes at 615 C forming As 4 O 6.Antimony is less reactive and stable towards water and air. On heating in air it forms Sb 4 O 6, Sb 4 O 8 or Sb 4 O 10. Bismuth is also known to form Bi 2 O 3 on heating. 4. Why does NH 3 form hydrogen bond but PH 3 does not? In ammonia, the nitrogen atom forms hydrogen bond because of the following reasons: (i) small size of nitrogen (ii) High electronegativity of nitrogen (3.0) Due to more difference of electronegativity between N and H atom then N-H bond is polar forming H bond. On the contrary, Phosphoru due to its larger size and lesser electronegativity (2.1) does not form hydrogen bond because P-H bond is almost purely covalent and no hydrogen bond is formed. 5. How is nitrogen prepared in the laboratory? Write the chemical equationsof the reactions involved. Dinitrogen is prepared in the laboratory by treating an aqueous solution of ammonium chloride with sodium nitrite. NH 4 Cl (aq) + NaNO 2 (aq) N 2 (g) + 2H 2 O (l) + NaCl (aq). The impurities of NO and HNO 3 obtained in small amounts is removed by passing the gas through aqueous sulphuric acid containing potassium dichromate. 6. How is ammonia manufactured industrially Haber Process for the Production of Ammonia In 1909 Fritz Haber established the conditions under which nitrogen, N 2 (g), and hydrogen, H 2 (g), would combine using medium temperature (~500 o C) very high pressure (~250 atmospheres, ~351kPa) a catalyst (a porous iron catalyst prepared by reducing magnetite, Fe 3 O 4 ). Osmium is a much better catalyst for the reaction but is very expensive. This process produces an ammonia, NH 3 (g), yield of approximately 10-20%. The Haber synthesis was developed into an industrial process by Carl Bosch. The reaction between nitrogen gas and hydrogen gas to produce ammonia gas is exothermic, releasing 92.4kJ/mol of energy at 298K (25 o C). 2

3 N 2 (g) nitrogen + 3H 2(g) hydrogen heat, pressure, catalyst 2NH 3 (g) ammonia H = kj mol -1 By Le Chetalier's Principle: increasing the pressure causes the equilibrium position to move to the right resulting in a higher yeild of ammonia since there are more gas molecules on the left hand side of the equation (4 in total) than there are on the right hand side of the equation (2). Increasing the pressure means the system adjusts to reduce the effect of the change, that is, to reduce the pressure by having fewer gas molecules. decreasing the temperature causes the equilibrium position to move to the right resulting in a higher yield of ammonia since the reaction is exothermic (releases heat). Reducing the temperature means the system will adjust to minimise the effect of the change, that is, it will produce more heat since energy is a product of the reaction, and will therefore produce more ammonia gas as well However, the rate of the reaction at lower temperatures is extremely slow, so a higher temperature must be used to speed up the reaction which results in a lower yield of ammonia. 7. Illustrate how copper metal can give different products on reaction with HNO 3. The reaction of copper with nitric acid produces different t products depending upon the concentration of nitric acid. (i) Reaction with cold dilute HNO 3 forms nitric oxide 2HNO 3 H 2 O + 2NO + 3(O) [Cu + [O] CuO] x 3 [CuO + 2HNO 3 Cu(NO 3 ) 2 + H 2 O] x 3 3Cu +8HNO 3 3Cu(NO 3 ) 2 + 2NO + 4H 2 O (ii) Reaction with hot conc. HNO 3 forms nitrogen dioxide 2HNO 3 H 2 O + 2NO 2 + (O) Cu + [O] CuO CuO + 2HNO 3 Cu(NO 3 ) 2 + H 2 O Cu +4HNO 3 Cu(NO 3 ) 2 + 2NO 2 + 2H 2 O 8. Give the resonating structures of NO 2 and N 2 O 5. The resonating structures of NO 2 and N 2 O 5 are as follows NO 2 3

4 N 2 O 5 : 9. The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be explained on the basis of sp 3 hybridisation in NH 3 and only s p bonding between hydrogen and other elements of the group]. The bond angle of the hydrides of group 15 are given below: Hydride NH 3 PH 3 AsH 3 SbH 3 Bond angle The HNH angle in ammonia is higher as compared to other hydrides of the group. It is because nitrogen in NH 3 is sp 3 hybridised. However due to lone pair of electrons the bond angle contracts from to The decreased bond angle in other hydrides is because of the fact that the sp 3 hybridisation becomes less and le ss distinct with increasing size of the central atom i.e. pure p- orbitals are utilized in M-H bonding or in simple words the s-orbital of H atom overlaps with orbital having almost pure p-character. 10. Why does R 3 P = O exist but R 3 N = O does not (R = alkyl group)? Compounds like R 3 N = O does not exist because of the absence of d-orbitals in the valence shell of nitrogen atom. As a result of this nitrogen cannot expand its covalency beyond 4 (lack of formation of d -p ). On the contrary R 3 P = O.exists because phosphorus has vacant d-orbitals in its valence shell and can expand it covalency beyond 4 by forming d -p. 11. Explain why NH 3 is basic while BiH 3 is only feebly basic. The hydrides of group 15 are basic in character because of the presence of lone pairs of electrons on the nitrogen atom (Lewis bases). Ammonia is strongly basic in character because it can easily donate its electron pair due to small size of nitrogen atom. As result of which electron density of lone pair is concentrated over a small region. BiH 3 on the contrary is feebly basic because of its larger size. Due to increase in size the electron density gets diffused over a larger region and hence the ability to donate the electron pair decreases. 4

5 12. Nitrogen exists as diatomic molecule and phosphorus as P 4. Why? Nitrogen exist6s as diatomic molecule (N") because due to small size of nitrogen atom and absence of vacant d-orbitals in its valence shell it has strong ability to form multiple bonds (N N). The other members of group 15 i.e. phosphorus, arsenic and antimony all exists as discrete tetra atomic tetrahedral molecules viz, P4, As4 and Sb 4 because these are not capable of forming multiple bonds to their bigger size and repulsions between non bonded electrons of the inner core. Now, since the angles P-P-P is 60, the p -p bonding is therefore not possible. 13. Write main differences between the properties of white phosphorus and red phosphorus. Properties White Phosphorus Red Phosphorus 1.Colour White but turns yellow on exposure to air. Dark red and no effect of exposure to air. 2.State and melting point Waxy solids, low melting point Brittle solids, high melting point 3. Density 1.84 g/cm g/cm 3 4. Ignition temperature 307 K 543 K 5. Stability less stable at ordinary temperature More stable at ordinary temperature 6. Chemical reactivity very reactive Less reactive 7. Toxic nature highly toxic Non toxic 8. Structure The four P- atoms lie at the corners of a regular tetrahedron. Each is phosphorus is bonded to other three P-atoms, by covalent bonds, so that each P-atom completes its valence shell. P 4 molecules are linked by covalent bonds forming giant (P 4 ) n molecule. Structure of white and red phosphorus are shown below: 5

6 14. Why does nitrogen show catenation properties less than phosphorus? The tendency of catenation of nitrogen is less than phosphorus. It is because the single N-N bond is weaker than the single P-P bond due to high inter electronic repulsion of the non bonding electrons in the case of nitrogen atom (smaller size). 15. Give the disproportionation reaction of H 3 PO 3. The phosphorous acid (H3PO3) when heated at 473 K disproportionate to form Ortho phosporic acid and Phosphine 4H 3 PO 3 3H 3 PO 4 + PH Can PCl 3 act as an oxidising as well as a reducing agent? Justify. PCl 3 can act both as an oxidizing agent as well as reducing agent. (a) As a reducing agent: The following reactions support the reducing character PCl 3 + SO 2 Cl 2 PCl 5 + SO 2 PCl 3 + SO 3 POCl 3 + SO 2 (b) As an oxidizing agent: It oxidizes metals to their respective chlorides. 6

7 12Ag + 4PCl 3 6Na + PCl 3 12AgCl + P 3NaCl + Na 3 P 17. Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation. Electronic configuration:-all these elements have the same valence shell electronic configuration of ns 2 np 4.and hence must be placed in the same group. Oxidation states:-oxygen predominantly shows -2 oxidation state and sulphur exhibits -2 to some extent. Except oxygen all the elements exhibit maximum oxidation state of +6. They also exhibit +2 and+4 states also. Hydrides:- All elements of the group form oxides of formula E 2 O On the basis of above facts it is justified to place them in the same group. 18. Why is dioxygen a gas but sulphur a solid? Dioxygen is a gas because due to small size of oxygen atom and absence of d-orbitals in its valence shell. It has strong tendency to form p -p bonds multiple bonds (O=O) so as to complete its octet and hence can exists as discrete O 2 molecule. On the contrary sulphur is a solid and exists as staggered 8 atom ring. It is because the tendency of S=S bond formation is missing in sulphur atom due to its larger size and low S-S double energy. As a result sulphur atoms complete their octet through formation of S-S bonds in S 8 molecule. This leads to increase the forces of attraction and hence its physical state is solid. 19. Knowing the electron gain enthalpy values for O O and O O 2 as 141 and 702 kj mol 1 respectively, how can you account for the formation of a large number of oxides having O 2 species and not O?(Hint: Consider lattice energy factor in the formation of compounds). The second electron gain enthalpy (O - + e - O 2- ) is positive (702 kj mol-1) at a large number of oxides have O2- species. This is attributable to high lattice energy released during formation of such oxides. As a result off this the lattice energy realeased by the process compensates for the second electron gain enthalpy. 2M + + O 2- M + O Which aerosols deplete ozone? 7

8 The aerosols responsible largely for the depletion of ozone layer are (i) nitric oxide emitted from the exhaust systems of supersonic jet aeroplanes. NO (g) + O 3 NO 2 (g) + O 2 (g) (ii) Freons (Chlorofluoro hydrocarbons) which are used in aerosol sprays and refrigerants. CCl 2 F 2 CF 2 Cl + Cl Cl + O 3 ClO + O Describe the manufacture of H 2 SO 4 by contact process? The Contact Process makes sulphur dioxide; convers the sulphur dioxide into sulphur trioxide (the reversible reaction at the heart of the process); converts the sulphur trioxide into concentrated sulphuric acid. Making the sulphur dioxide This can either be made by burning sulphur in an excess of air:... or by heating sulphide ores like pyrite in an excess of air: In either case, an excess of air is used so that the sulphur dioxide produced is already mixed with oxygen for the next stage.converting the sulphur dioxide into sulphur trioxidethis is a reversible reaction, and the formation of the sulphur trioxide is exothermic. 8

9 The reasons for all these conditions will be explored in detail further down the page.converting the sulphur trioxide into sulphuric acid This can't be done by simply adding water to the sulphur trioxide - the reaction is so uncontrollable that it creates a fog of sulphuric acid. Instead, the sulphur trioxide is first dissolved in concentrated sulphuric acid: The product is known as fuming sulphuric acid or oleum.this can then be reacted safely with water to produce concentrated sulphuric acid - twice as much as you originally used to make the fuming sulphuric acid. 22. How is SO 2 an air pollutant? Sulphur dioxide and sulphur trioxide are harmful gaseous pollutants.so 2 is released into atmosphere through volcanic eruptions and accounts for 67% of total SO 2 present in atmosphere. The remaining 33% comes from human activity.the SO 2 present in atmosphere undergoes photolytic and catalytic oxidation to SO 3 which gets converted to H 2 SO 4 in the presence of moisture. This comes down in the form of acid rain.so 2 is strongly irritating to respiratory of humans and animals. It causesacute irritation to eyes. It also produces leaf injuries (called necrotic blotching). 23. Why are halogens strong oxidising agents? Halogens act as strong oxidizing agents because of their high electronegativity values and high electron affinity values. The oxidizing power of the halogens is comparable in terms of their reduction potential values given below: Element: F Cl Br I Red. Potential (E 0 Volt) Oxidizing nature decreases As the reduction potential values decrease from fluorine to iodine, the oxidizing power also decreases. 24. Explain why fluorine forms only one oxoacid, HOF. Flourine is known to form only one oxoacid, HOF which is highly unstable. Other halogens form oxoacids of the type HOX, HXO 2, HXO 3, and HXO 4 (X=Cl,Br,I). Flourine due to its small size and high electro negativity cannot act as central atom in higher oxoacids and hence do not form higher oxoacids. 25. Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not. Oxygen atom can form H bonds whereas chlorine does not. The tendency for H bonding depends upon small size and high electronegativity value. Although the electreonegativity of O and Cl is nearly same yet chlorine does not form H bond due to its larger size (99pm) as compared to oxygen (66pm). 9

10 26. Write two uses of ClO 2. Chlorine dioxide is used as (i) oxidizing agent (ii) bleaching agent. 27. Why are halogens coloured? All the halogens are coloured. The colour of halogens deepens with raise in atomic number from fluorine to iodine. Elements F Cl Br I Colour Light yellow Yellow green Reddish Deep violet The colour is due to absorption of energy from visible light by their molecules for excitation of outer electrons of higher energy levels (the gap of energy between valence shell of halogens and higher energy shells is less). Flourine absorbs violet portion of the light and appears yellow while iodine absorbs yellow and green portions of the light and thus appears violet.change in colour on moving from F to I is calkled blue shift or bathochromic shift. 28. Write the reactions of F 2 and Cl 2 with water. (i) Flourine is highly reactive and decomposes water very readily even at low temperature and in dark forming a mixture of O2 And O3. 3F 2 + 3H 2 O 6HF + O 3 2F 2 + 2H 2 O 4HF + O 2 5F 2 + 5H 2 O 10HF + O 3 + O 2 (ii) Chlorinbe decomposes water in the presence of sunlight forming halogen acid and oxoacid. Cl 2 + H 2 O HCl + HClO 29. How can you prepare Cl 2 from HCl and HCl from Cl 2? Write reactions only. (i) Chlorine can be obtained from HCl by heating with oxidizing agents like MnO 2, KMnO 4, K 2 Cr 2 O 7 etc. MnO 2 + 4HCl MnCl 2 + 2H 2 O + Cl 2 2KMnO HCl 2KCl + 2MnCl 2 + 8H 2 O + 5 Cl 2 (ii) HCl can be obtained from chlorine by burning it in excess of hydrogen H 2 + Cl 2 2HCl 10

11 30. What inspired N. Bartlett for carrying out reaction between Xe and PtF 6? N. Bartlett in 1962, prepared a compound for reacting oxygen with PtF 6, a powerful oxidizing agent. The X-Ray examination of solid compound, O 2 PtF 6, showed that it consisted of O 2 + and PtF 6 - ions. Bartlett thought that a similar could be prepared with xenon because the ionization enthalpy of xenon is quite close to the ionization enthalpy of oxygen (1314 kj mol -1 ). Accord8ingly he reacted xenon with PtF6 and got a red solid compound Xe + PtF 6 XePtF 6 Red solid 31. What are the oxidation states of phosphorus in the following: (i) H 3 PO 3 (ii) PCl 3 (iii) Ca 3 P 2 (iv) Na 3 PO 4 (v) POF 3? Compound Oxidation state of phosphorus H 3 PO 3 +3 PCl 3 +3 Ca 3 P 2-3 Na 3 PO 4 +5 POF Write balanced equations for the following: (i) NaCl is heated with sulphuric acid in the presence of MnO 2. (ii) Chlorine gas is passed into a solution of NaI in water. (i) Chlorine gas is produced. 2NaCl + MnO 2 + 2H 2 SO 4 Na 2 SO 4 + MnSO 4 + 2H 2 O + Cl 2 (ii) Iodine is liberated 2NaI + Cl 2 2NaCl + I How are xenon fluorides XeF 2, XeF 4 and XeF 6 obtained? All three binary fluorides of Xenon are formed by direct union of elements under approximate experimental conditions. XeF2 can also be prepared by irradiating a mixture of Xenon and fluorine with sunlight or light from a pressure-mercury arc lamp. Xe (g) + F 2 (g) XeF 2 (g) 1 vol 2 vol sealed Ni vessel Xenon diflouride Xe (g) + F 2 (g) 1 vol 2 vol Xe (g) + F 2 (g) 1 vol 20 vol sealed Ni vessel sealed Ni vessel XeF 4 (g) XeF 6 (g)

12 34. With what neutral molecule is ClO isoelectronic? Is that molecule a Lewis base? ClO - species has 26 electrons. The neutral molecule which is isoelectronic with it is ClF Since it can form ClF 3 and ClF 5, it is a Lewis base. 35. How are XeO 3 and XeOF 4 prepared? 6XeF H 2 O 4Xe + 2XeO HF + 3O 2. XeF 6 + 3H 2 O XeF 6 + H 2 O XeO 3 + 6HF XeOF 4 + 2HF 36. Arrange the following in the order of property indicated for each set: (i) F 2, Cl 2, Br 2, I 2 - increasing bond dissociation enthalpy. (ii) HF, HCl, HBr, HI - increasing acid strength. (iii) NH 3, PH 3, AsH 3, SbH 3, BiH 3 increasing base strength. (i) Cl-Cl > Br-Br > F-F > I-I - Bond energy is highest for chlorine molecule. (ii) HI > HBr > HCl > HF - HI is the strongest acid among the halogen acids. (iii) NH 3 > PH 3 > AsH 3 > SbH 3 > BiH 3 Ammonia is the most basic hydride. 37. Which one of the following does not exist? (i) XeOF 4 (ii) NeF 2 (iii) XeF 2 (iv) XeF 6 The sum of the first two ionization energies of Ne is too high to be compensated. 38. Give the formula and describe the structure of a noble gas species which is isostructural with: (i) ICl 4 - (ii) IBr 2 - (iii) BrO 3 - (i) XeF 4 (ii) XeF 2 (iii) XeO Why do noble gases have comparatively large atomic sizes? 12

13 The atomic size in the case of noble gases is expressed in terms Vander waal s radii whereas the atomic size of other members of the period is either metallic radii or covalent radii. As the Vander waal s radii is larger than both metallic as well as covalent radii, ther3efore the atomic size of the noble gas is quite large. Among the noble gases, the atomic size increases down the group due to addition of new electronic shells. 40. List the uses of neon and argon gases. Uses of neon: It is used in neon discharge lamps and signs which are used for advertising purposes.it is used in safety devices for protecting electrical instruments because it has a property of carrying exceedingly high currents under high voltage. Uses of Argon: a. It is widely used for filling in candescent metal filament electric bulbs. b. It is used for filling radio-valves, rectifiers and fluorescent tubes. c. It is used in producing inert atmosphere during welding and extraction of various metals. 41. Why are pentahalides more covalent than trihalides? Higher the positive oxidation state of central atom, more will be its polarising power, which, in turn, increases the covalent character of bond, formed between the central atom and the other atom. 42. Why is BiH 3 the strongest reducing agent amongst all the hydrides of Group 15 elements? Because BiH 3 is the least stable among the hydrides of Group Why is N 2 less reactive at room temperature? Because of strong pπ pπ overlap resulting into the triple bond, N N. 44. Mention the conditions required to maximise the yield of ammonia. The conditions required to maximize the yield of ammonia: 1. An optimum temperature of about 750 K 2. A pressure of about 200 atmospheres 3. Using finely divided iron as catalyst and molybdenum as promoter. 45. How does ammonia react with a solution of Cu 2+? It gives a deep blue solution with a solution of copper (II) sulphate due to the formation of [Cu(NH 3 ) 4 ]SO 4. 13

14 46. What is the covalence of nitrogen in N 2 O 5? From the structure of N 2 O 5 it is evident that covalence of nitrogen is four 47. Bond angle in PH 4 + is higher than that in PH 3. Why? Both are sp 3 hybridised. In PH 4 + all the four orbitals are bonded whereas in PH 3 there is a lone pair of electrons on P, which is responsible for lone pair-bond pair repulsion in PH3 reducing the bond angle to less than What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO 2? When white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO 2, Phosphine gas is produced. P 4 + 3NaOH + 3H 2 O 3NaH 2 PO 2 + PH What happens when PCl 5 is heated? On heating phosphorus pentachloride sublimes at 160 C. Whenheated it dissociates into PCl 3 and Cl 2. PCl 5 PCl 3 + Cl Write a balanced equation for the hydrolytic reaction of PCl5 in heavy water. PCl 5 + D 2 O POCl 3 + 2DCl 51. What is the basicity of H 3 PO 4? Three P OH groups are present in the molecule of H 3 PO 4. Therefore, itsbasicity is three. 52. What happens when H 3 PO 3 is heated? When H 3 PO 3 is heated, it undergoes auto oxidation and reduction to form phosphoric acid and phosphine. 4H 3 PO 3 H 3 PO 4 + PH List the important sources of sulphur. Sulphur occurs mainly as the sulphides and sulphates in the combined state. For example: lead sulphide galena, copper pyrites, copper sulphide copper glance, calcium 14

15 sulphate gypsum etc. Traces of sulphur also occur as hydrogen sulphide gas coming out of the volcanoes. Organic substances such as eggs, proteins, garlic, onion, hair and wool, etc., also contain a small percentage of sulphur Write the order of thermal stability of the hydrides of Group 16 elements. The members of Group 16 family forms volatile hydrides such as H 2 O,H 2 S,H 2 Se,H 2 Te and H 2 Po.The energy needed to dissociate these molecules decreases down the group. Hence the thermal stability of these hydrides follows the order: H 2 O > H 2 S > H 2 Se > H 2 Te > H 2 Po. 55. Why is H 2 O a liquid and H 2 S a gas? Because of small size and high electronegativity of oxygen, molecules of water are highly associated through hydrogen bonding resulting in its liquid state. 56. Which of the following does not react with oxygen directly?zn, Ti, Pt, Fe Platinum does not react with oxygen directly. 57. Complete the following reactions: (i) C 2 H 4 + O 2 (ii) 4Al + 3 O 2 (i) C 2 H 4 + O 2 2CO 2 + 2H 2 O. (ii) 4Al + 3 O 2 2Al 2 O Why does O 3 act as a powerful oxidising agent? Ozone is a powerful oxidizing agent because on decomposition it forms atomic oxygen. O 3 O 2 + O. 59. How is O 3 estimated quantitatively? When ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (ph 9.2), iodine is liberated which can be titrated against a standard solution of sodium thiosulphate. This is a quantitative method for estimating O 3 gas. 15

16 60. What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt? Sulphur dioxide acts as a fairly strong reducing agent in the presence of moisture. Hence it reduces ferric sulphate to ferrous sulphate. Fe 2 (SO 4 ) 3 + SO 2 + 2H 2 O 2FeSO 4 + 2H 2 SO Comment on the nature of two S O bonds formed in SO 2 molecule. Are the two S O bonds in this molecule equal? Both the S O bonds are covalent and have equal strength due to resonatingstructures. 62. How is the presence of SO 2 detected? SO 2 gas when passed through acidified potassium permanganate decolourises it due to its reducing property. 2MnO SO 2 +2H 2 O 2Mn 2+ +5SO H Mention three areas in which H 2 SO 4 plays an important role. 1. It is used in the manufacture of fertilizers. 2. It is used in storage batteries 3. IT is used in petroleum refining and making of paints, detergents etc. 64. Write the conditions to maximise the yield of H 2 SO 4 by Contact process. The conditions to maximise the yield of H 2 SO 4 by Contact process are: 1. an optimum temperature of K. 2. an optimum pressure of atm. 3. using platinised asbestos or divanadium pentoxide as catalyst. 4. using oxygen is excess. 5. preventing poisoning of catalyst, by removing impurities like arsenic oxide, dust particles and moisture. 65. Why is K a2 << K a1 for H2SO 4 in water? 16

17 H 2 SO 4 is a very strong acid in water largely because of its first ionisation to H 3 O + and HSO 4. The ionisation of HSO 4 to H 3 O + and SO 4 2 is very very small. That is why K a2 << K a1. HSO 4 - does not have the tendency to lose the proton readily. 66. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F 2 and Cl 2. diss H of fluorine (79.1 kj) is smaller than that of chlorine (122kJ). Electron gain enthalpy of chlorine (-384 kj) is higher than that of fluorine (-333kJ). The value of hyd H of fluoride ion (-460 kj) is much higher than that of chloride ion (-383 kj). It is due to small size and high charge density of fluoride ion. Thus net energy released for ½ F 2 (g) F - (aq) is -752 kj whereas that for ½ Cl 2 (g) Cl - (aq) is kj, which accounts for higher oxidizing power of F Give two examples to show the anomalous behaviour of fluorine. Ionisation enthalpy, electro negativity, enthalpy of bond dissociation and electrode potentials are all higher for fluorine than expected from the real trends set by other halogens. Also, ionic and covalent radii, melting point and boiling point and electron gain enthalpy are quite lower than expected. Fluorine can exhibit only -1 oxidation state and cannot show positive oxidation states like other halogens. Fluorine can form hydrogen bonding and so HF is much less acidic than other hydrogen halides. 68. Sea is the greatest source of some halogens. Comment. Sea water contains some of the halogens like chlorine, bromine and traces of fluorine. It is evident from the electrolysis of sea water which yields chlorine and bromine. Sea water contains chlorides, bromides and iodides of sodium, magnesium,calcium. Sea weeds contain iodide. 69. Give the reason for bleaching action of Cl 2. In the presence of moisture, chlorine acts as an oxidizing and a bleaching agent. Chlorine reacts with water forming HCl and HClO (hypochlorous acid). HClO is not so stable and decomposes giving nascent oxygen, which is responsible for oxidizing and bleaching properties of chlorine. Cl2 + H2O HCl + HClO 17

18 HClO HCl + O Cl2 + H2O 2HCl + O Coloured matter + Nascent oxygen colourless matter 70. Name two poisonous gases which can be prepared from chlorine gas.. Phosgene (COCl 2 ) and tear gas (CCl 3.NO 2 ) are the poisonous gases that can be prepared from chlorine gas. 71. Why is ICl more reactive than I 2? In general, interhalogen compounds are more reactive than halogens due to weaker X X bonding than X X bond. Thus, ICl is more reactive than I Why is helium used in diving apparatus? Helium oxygen mixture is used by deep-sea divers in preference to nitrogen oxygen mixtures. It is much less soluble in blood than nitrogen. This prevents bends which is the pain caused by formation of nitrogen bubbles in blood veins when a diver comes to the surface. 73. Balance the following equation: XeF 6 + H 2 O XeO 2 F 2 + HF XeF 6 + 2H 2 O XeO 2 F 2 + 4HF 74. Why has it been difficult to study the chemistry of radon? Radon is radioactive with very short half-life which makes the study of chemistry of radon difficult. 18