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Urban

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Submitted By KristyKristy
Words 3428
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Question bank
1 mark questions
1. What is the valence shell electronic configuration of p-block elements ? Valence shell electronic configuration is ns2np1-6

2. What is Chile salt petre?/ Indian salt Petre ?
Chile saltpeter is NaNO3(sodium nitrate) and Indian salt petre is KNO3(postassium nitrate)

3. What is fluorapatite ?
The formula of fluorapatite is Ca9(PO4)6.CaF2. It is a mineral of phosphorous

4. What is the covalence of nitrogen in N2O5?
4

5. What is the basicity of H3PO2, H3PO3, H3PO4, H4P2O7
1,2,3,4,4
6. What is the maximum of covalency of N?
Ans. 4 8. What is the gas evolved when ammonium dichromate is heated? Ans. N2 9. What is the catalyst used in the conversion of ammonia to nitric oxide?
Ans Pt-Rh

10. Which is the stablest form of phosphorous?
Ans Black Phosphorous

11. What is the product formed by the hydrolysis of PCl3?
Ans. Phosphorous acid

12. What is the product formed by the hydrolysis of PCl5?
Ans. Phosphoric acid
13. in which chemical form PCl5 exists in solid state?
Ans. [(PCl4+)][(PCl6-)]

14. What are chalcogens?
Ans. Elements of group 16 are called chalcogens.

15. what is gypsum?
Ans. CaSO4.2H2O

16. what is Epsom salt?
Ans. MgSO4.7H2O
17. How many sulphur atoms in a sulphur molecule?
Ans. 8

18. Which is the radioactive element in halogens?
Ans. astatine

19. Which is the radioactive element in chalcogens?
Ans. Polonium.

20. Which is the radioactive element in noble gases?
Ans. Radon.

21. What happens when mercuric oxide is heated?
Ans. Mercury is formed.

22. What happens when silver oxide is heated?
Ans. silver is formed.

23. What is Oleum?
Ans. H2S2O7

24. Give an example of a nonmetal which exists as a liquid at room temperature.
Ans. Bromine

25. What is the formula of the first noble gas compound prepared?
Ans. XePtF6

2 mark questions
1. Bismuth exhibits +5 oxidation states rarely. Why?
The stability of +5 oxidation state decreases down the group due to inert pair effect. The only well characterized Bi (V) compound is BiF5.

2. Why are pentahalides of 15th group elements more covalent than trihalides?
In pentahalides, the central atom shows a higher positive oxidation state as a result of which the central atom has greater polarizing power. This results in increasing the covalent character.

3. Why is BiH3 the strongest reducing agent amongst all hydrides of Group 15 elements.
In group 15 elements, as we move down the group the atomic size of elements increases. As a result, the strength of element – H bond decreases. Accordingly the Bi-H bond is the weakest among the hydrides of elements of group 15 and hence BiH3 is the strongest reducing agent amongst all the hydrides of group 15.

4. Though nitrogen exhibits +5 oxidation state,it does not form pentahalide. Give reason.
Nitrogen with n = 2, has s and p orbitals only. It does not have d orbitals to expand its covalence beyond four. That is why it does not form pentahalide.

5. PH3 has lower boiling point than NH3 . Why?
Unlike NH3, PH3 molecules are not associated through hydrogen bonding in liquid state. That is why the boiling point of PH3 is lower than NH3.

6. Give two methods of preparation of N2 a)Dinitrogen is produced commercially by the liquefaction and fractional distillation of air.
b)Dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite. c) It can also be obtained by the thermal decomposition of ammonium dichromate. d) Very pure nitrogen can be obtained by the thermal decomposition of sodium or barium azide.

7. Give two uses of N2
The main use of dinitrogen is in the manufacture of ammonia and other industrial chemicals containing nitrogen, (e.g., calcium cyanamide).
It also finds use where an inert atmosphere is required (e.g., in iron and steel industry,inert diluent for reactive chemicals). Liquid dinitrogen is used as a refrigerant to preserve biological materials, food items and in cryosurgery.

8. Why is N2 less reactive at room temperature?
Dinitrogen is rather inert at room temperature because of the high bond enthalpy of N N bond(945 kJ mol-1). Reactivity, however, increases rapidly with rise in temperature.

9. Why does ammonia act as a Lewis base?
The presence of a lone pair of electrons on the nitrogen atom of the ammonia molecule makes it a Lewis base.

10. How does ammonia react with a solution of Cu2+? 11. Draw the structures of N2O, NO, N2O3 , NO2, N2O4, N2 12. Why does NO2 dimerise?
NO2 contains odd number of valence electrons. It behaves as a typical odd molecule. On dimerisation, it is converted to stable N2O4 molecule with even number of electrons.

13. Mention three uses of ammonia.
Ammonia is used to produce various nitrogenous fertilisers(ammonium nitrate, urea, ammonium phosphate and ammonium sulphate)
It is used in the manufacture of some inorganic nitrogen compounds, like nitric acid. Liquid ammonia is also used as a refrigerant.

14. Give two examples for the oxidizing property of conc.Nitric acid.
Concentrated nitric acid is a strong oxidising agent and attacks most metals except noble metals such as gold and platinum. The products of oxidation depend upon the concentration of the acid, temperature and the nature of the material undergoing oxidation.

15. Mention three uses of nitric acid.
Nitric acid is used in the manufacture of ammonium nitrate for fertilisers and other nitrates for use in explosives and pyrotechnics.
It is also used for the preparation of nitroglycerin, trinitrotoluene and other organic nitro compounds.
It is used in the pickling of stainless steel,etching of metals and as an oxidiser in rocket fuels.

16. Why is white phosphorous more reactive?
White phosphorus is less stable and therefore, more reactive than the other solid phases under normal conditions because of angular strain in the P4 molecule where the angles are only 60°.

17. How is white phosphorous converted to red phosphorous?
Red phosphorus is obtained by heating white phosphorus at 573K in an inert atmosphere for several days
18. What are Holme’s signals?
Phosphine prepared is spontaneously inflammable due to the presence of phosphorus dihydride(P2H4). This principle is used in Holmes’s signal. A mixture of calcium carbide and calcium phosphide is placed in metallic containers. Two holes are made and the container is thrown into the sea. Water enters and produces acetylene and phospine respectively. The gaseous mixture catches fire spontaneously. Acetylene produces a bright luminous flame which serves as a signal to the approaching ship.

19. Bond angles in PH4+ is higher than that in PH3. Why?
20. P in PH3 is sp3 hybridized. It has three bond pairs and one lone pair around P. due to greater lone pair – bond pair repulsion than bond pair- bond pair repulsion, the tetrahedral angle decreases from 109028’ to 93.6 0. As a result PH3 is pyramidal. However when it reacts with proton, it forms PH4+ ion which has four bond pairs and no lone pair. Due to the absence of lone pair – bond pair repulsion and presence of four identical bond pair – bond pair interactions PH4+ assumes tetrahedral geometry with a bond angle of 109028’.

21. What is the structure of PCl3, PCl5 ? PCl3

22. Why does PCl3 fume in moisture?
PCl3 hydrolyses in the presence of moisture giving fumes of HCl.

23. Are all the five bonds in PCl5 equivalent? Justify your answer
24. PCl5 has a trigonal bipyramidal structure and the three equatorial P-Cl bonds are equivalent, while the two axial bonds are different and longer than equatorial bonds. This is due to the fact that the axial bond pairs suffer more repulsion as compared to equatorial bond pairs.

25. Draw the structures of H3PO2, H3PO3, H3PO4, H4P2O7, ( H PO3 )n
26. Draw the structures of H3PO2, H3PO3, H3PO4, H4P2O7, ( H PO3 )n

27. Elements of Group 16 generally show lower value of first ionization potential compared to the corresponding elements of group 15. Why?
Due to extra stable half-filled p orbital electronic configurations of Group 15 elements, larger amount of energy is required to remove electrons compared to Group 16 elements.

28. Electron gain enthalpy of oxygen is less negative than sulphur. Why? Because of the compact nature of oxygen atom (1s2 2s2 2p4), it has less negative electron gain enthalpy than sulphur

29. H2S is less acidic than H2Te . Why?
Due to the decrease in bond (E–H) dissociation enthalpy down the group, acidic character increases. H2Te bonds are easily broken and provide H+ ions easily hence it is more acidic.

30. Te usually shows +4 oxidation state than +6. Why?
The stability of + 6 oxidation state of group 16 elements decreases down the group and stability of + 4 oxidation state increases due to inert pair effect

31. Give two methods of preparation of O2
1. By heating oxygen containing salts such as chlorates, nitrates and permanganates. 2. By the thermal decomposition of the oxides of metals low in the electrochemical series 3. Electrolysis of water leads to the release of hydrogen at the cathode and oxygen at the anode.
4. Industrially, dioxygen is obtained from air by first removing carbon dioxide and water vapour and then, the remaining gases are liquefied and fractionally distilled to give dinitrogen and dioxygen.

32. Give two uses of O2
It is used in normal respiration and combustion processes.
Oxygen is used in oxyacetylene welding.
In the manufacture of many metals, particularly steel.
Oxygen cylinders are widely used in hospitals, high altitude flying and in mountaineering.
The combustion of fuels, e.g., hydrazines in liquid oxygen, provides tremendous thrust in rockets.

33. What are amphoteric oxides? .Give examples
Some metallic oxides show characteristics of both acidic as well as basic oxides. Such oxides are known as amphoteric oxides. They react with acids as well as alkalies.Eg. Al2O3

34. Why does O3 act as a powerful oxidizing agent?
Ozone is a powerful oxidizing agent because it easily decomposes to give nascent oxygen. It oxidizes lead sulphide to lead sulphate and iodide ions to iodine.

35. How is O3 estimated quantitatively?
Ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (pH 9.2) to liberate iodine. It is titrated against a standard solution of sodium thiosulphate using starch as indicator. From the volume of thiosulphate concentration of iodine and thereby amount of ozone can be determined.
O3 + 2KI + H2O  I2 + O2 + 2KOH 2Na2S2O3 + I2  Na2S2O6 + 2 NaI

36. Why is O3 explosive at higher concentration?
Ozone is thermodynamically unstable with respect to oxygen since its decomposition into oxygen results in the liberation of heat (ΔH is negative) and an increase in entropy (ΔS is positive).
O3 (g)  O2 (g) + O ; ΔH = -142KJ mol-1
These two effects reinforce each other, resulting in large negative Gibbs energy change (ΔG) for its conversion into oxygen. Hence it is explosive
[ ΔG =ΔH – TΔS ( when ΔH is negative and ΔS is positive ΔG is highly negative)]

37. Why is supersonic jet aeroplanes a threat to ozone layer?
Nitrogen monoxide decomposes ozone easily. Exhaust systems of supersonic jet aero planes emit nitrogen oxides leading to depletion ozone layer in the upper atmosphere. Hence these planes are a threat to ozone layer.

38. Give two uses of O3
It is used as a germicide, disinfectant and for sterilising water.
It is used for purification of air in hospitals, railway station tunnels etc.
It acts as an oxidising agent in the manufacture of potassium permanganate
39. Which form of sulphur shows paramagnetic behavior? Why?
In vapour state, sulphur partly exists as S2 molecule which has two unpaired electrons in the antibonding π* orbitals like O2 and, hence, exhibits paramagnetism.

40. How is monoclinic sulphur prepared from rhombic sulphur? What is its transition temperature?
Monoclinic sulphur is prepared by melting rhombic sulphur in a dish and cooling, till crust is formed. Two holes are made in the crust and the remaining liquid poured out. On removing the crust, colourless needle shaped crystals of monoclinic sulphur are formed.
Transition temperature:
Monoclinic sulphur is stable above 369 K and transforms into α-sulphur below it. Conversely, Rhombic sulphur is stable below 369 K and transforms into monoclinic sulphur above this. At 369 K both the forms are stable. Hence it is the temperature at which both rhombic and monoclic sulphur are stable

41. Draw the resonating structures of SO2. 42. Give two uses of SO2
Sulphur dioxide is used
(i) in refining petroleum and sugar (ii) in bleaching wool and silk
(iii) as an anti-chlor for removing chlorine from a fabric after bleaching.
(iv) Liquid SO2 is used as a solvent to dissolve a number of organic and inorganic chemicals.

43. How is the presence of SO2 detected?
(i) SO2 is a pungent smelling gas
(ii) It discharges the color of potassium permanganate solution (pink to colorless) (iii) it turns orange coloured acidified potassium dichromate solution to green K2Cr2O7 + 3SO2 + H2SO4  K2SO4 + Cr2(SO4)3 + H2O

44. SO2 is used as an antichlor . Why?
Sulphur dioxide is a reducing agent. Hence it reduces chlorine to sulphuryl chloride. Thereby the excess chlorine from a bleached fabric can be removed.

45. Draw the structures of H2SO3, H2S2O7, H2SO4, H2S2O8,.

7. What happens when SO2 is passed through an aqueous solution of Fe(III) salt? Moist sulphur dioxide behaves as a reducing agent hence it converts iron(III) ions to iron(II) ions. Colour of the solution changes from reddish brown to pale green
8. What happens when SO2 is passed through an aqueous solution of KMnO4 ?
It discharges the color of potassium permanganate solution (pink to colorless)
46. Why is Ka2 HBrO > HIO
It is because greater is the size of the halogen atom lesser will be its electronegativity and hence lesser shall be the acidic character.
70.Write the composition of bleaching powder. How is it prepared?
When chlorine gas is passed through dry slaked lime bleaching powder[Ca(OCl)2] is formed. The actual composition of bleaching powder is Ca(OCl)2.CaCl2.Ca(OH)2.2H2O

3 MARK
1. Write a note on anomalous properties of Nitrogen.
Nitrogen has small size, high electronegativity, high ionisation enthalpy and non-availability of d orbitals.
Nitrogen has unique ability to formp∏- p∏ multiple bonds with itself and with other elements having small size and high electronegativity (e.g., C, O) nitrogen exists as a diatomic molecule with a triple bond its bond enthalpy(941.4 kJ mol–1) is very high.
Nitrogen has less capacity for catenation
Nitrogen does not form d∏- p∏ multiple bond

2. Describe briefly the manufacture of ammonia by Haber process.
Ammonia is manufactured by Habers process. It takes place according to the following equation According to Le Chatelier principle the following conditions are required for better yield.
a) high pressure 200 atm.
b) moderate temperature 773K
c) finely divided iron as catalyst, molybdenum as a promoter.
Step 1: Compression
A mixture of pure and dry nitrogen and hydrogen in the ratio 1:3 by volume is compressed to 200 atm pressure
Step 2: conversion
The compressed gases are passed through a converter containing catalyst and promoter kept at 773K. nitrogen combines with hydrogen to form ammonia.
The reaction is exothermic. About 20% of the mixture is converted to product
Step 3: Cooling
The mixture of gases is passed through cooling pipes. Ammonia gas condenses into liquid ammonia below 240K
Step 4: recycling
The unreacted gases are recirculated by pumping back and mixing with fresh reacting gases.

3. Describe briefly the manufacture of nitric acid by Ostwald’s process.
Step 1
Catalytic oxidation of NH3 by atmospheric oxygen. Step 2
Nitric oxide thus formed combines with oxygen giving NO2. Step 3
Nitrogen dioxide so formed, dissolves in water to give HNO3. NO thus formed is recycled and the aqueous HNO3 can be concentrated by distillation upto 68% by mass. Further concentration to 98% can be achieved by dehydration with concentrated H2SO4.

4. Write main differences between properties of white P and red and black P
Property White P Red P Black P
Color White but turns yellow on exposure to air Dark red Black
State Waxy solid, can be cut with knife Brittle Crystalline with greasy touch
Smell Garlic Odourless -
Density 1.84 2.1 2.69
Ignition temperature 307 K 533 K 673 K
Melting point 317 K Does not melt 860 K
Toxic nature Highly toxic Non toxic
Chemical reactivity Highly reactive Less reactive Least reactive

5. How is phosphine prepared? How is it purified?
(a) Phosphine is prepared by the reaction of calcium phosphide with water or dilute HCl. (b) it is prepared by heating white phosphorus with concentrated NaOH solution in an inert atmosphere of CO2. It is purified by absorbing in HI to form phosphonium iodide which on treatment with KOH give pure phosphine. [PH4I + KOH  KI + H2O + PH3]

6. Write a note on Anomalous properties of oxygen.
Oxygen is a gas while other elements are solids
Oxygen can form strong hydrogen bonds while other elements do not
Oxygen can show a maximum covalency of four while other elements can show higher covalency
All the hydrides except water possess reducing property and this character increases from H2S to H2Te.

7. How is ozone prepared?
When a slow dry stream of oxygen is passed through a silent electrical discharge, conversion of oxygen to ozone (10%) occurs [Silent electrical discharge:(spark less electric discharge). A discharge which does not produce audible or visible effects. If an electromotive force (emf) is applied to a pair of metallic conductors that are separated by a small gap, the flow of current is essentially zero. Now, if we increase EMF progressively to very high levels say thousands of volts and the gap is viewed in the dark, a faint glow, associated with a tiny current flow, becomes visible. This is the region of the silent electric discharge. This glow becomes much stronger when EMF is increased further. Eventually, at some critical higher EMF region, the system enters the highly visible arc or spark region. This means system is not in Silent electric discharge anymore. ]

8. How is sulphuric acid manufactured?
Sulphuric acid is manufactured by the Contact Process. it involves three steps:
(i) burning of sulphur or sulphide ores in air to generate SO2.
(ii) conversion of SO2 to SO3 by the reaction with oxygen in the presence of a catalyst (V2O5)
(iii) absorption of SO3 in H2SO4 to give Oleum (H2S2O7).
Oleum is diluted to get sulphuric acid of desired concentration.
Chemical reactions
Step 1: S + O2  SO2
Step 2: According to Le Chatelier’s principle the following conditions are needed for better yield
a) optimum temperature of 720 K
b) optimum pressure of 2 bar
c) vanadium pentoxide catalyst
d) to prevent poisoning of catalyst, the gases are purified in several stages.
9. Purification stages:
Dust precipitator: a chamber in which dust in the gases are settled down
Scrubber: soluble impurities are washed away by water
Drying tower: gases are dried by the spray of concentrated sulphuric acid
Arsenic purifier: As2O3 impurity is removed by treating with gelatinous ferric hydroxide
Testing box: purity of gases are checked by the principle of scattering effect.
The pure gases are passed through the catalytic chamber and converted to SO3.
Step 3:
A tall tower is filled with quartz crystals. Sulphur trioxide gas is passed into it from the bottom. Con. Sulphuric acid is sprayed from the top. The gas is absorbed into sulphuric acid to form oleum. It is diluted to get sulphuric acid. H2S2O7 + H2O  H2SO4
10. FLOW CHART

11. Write a note on anomalous properties of fluorine.
Most of the reactions of fluorine are exothermic (due to the small and strong bond formed by it with other elements).
It forms only one oxo acid while other halogens form a number of oxoacids.
Hydrogen fluoride is a liquid (b.p. 293 K) due to strong hydrogen bonding. Other hydrogen halides are gases. Fluorine exhibits only –1 oxidation state whereas other halogens exhibit variable oxidation states also

12.
What is aqua regia? Explain its action with equations.
When three parts of concentrated HCl and one part of concentrated HNO3 are mixed, aqua regia is formed which is used for dissolving noble metals, e.g., gold, platinum.

13. Who prepared the first compound of noble gases and how?
Neil Bartlett,
When Xe and PtF6 are mixed, a rapid reaction occurs and a red solid with the formula Xe+[PtF6]- is formed.
Xe + PtF6  Xe+[PtF6]-…...

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