p-Block Elements

Group 15 (Nitrogen Family)

N [He]2s22p3 , Z=7P [Ne] 3s23p3, Z=15As [Ar]3d104s24p3, Z=33Sb [Kr] 4d10 5s25p3, Z=51Bi [Xe]4f14 5d10 6s2 6p3, Z=83

1. Why are atomic and ionic radii of group 15 elements smaller than the atomic radii of the corresponding group 14 elements?

2. Why does atomic radius increase on going down the group?

3. The first ionization enthalpy of group 15 elements are higher than than the corresponding members of the group 14 elements. Why?4. On going down the group, the ionization enthalpies decrease. Why?5. Why is electronegativity of group 15 elements more than group 14 elements?6. Metallic character is less than group 14 elements but it increases down the group. Why?7. 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. Why?8. The order of successive ionization enthalpies of group 15 elements is ∆iH1< ∆iH2< ∆iH3. Why?9. N,P : Non metals As, Sb : Metalloids Bi: Metal

10. A nitrogen atom posseses five valence electrons, but it does not form the compound of the type NCl5. Give reason. 11. Explain why phosphorus forms PF5, whereas nitrogen does not form NF5?12. Why is it that molecular nitrogen is not particularly reactive?13. Explain why nitrogen exists as a diatomic molecule, N2, whereas phosphorus exists as tetratomic molecule.14. The common oxidation states are -3,+3 and +5. The tendency to exhibit -3 oxidation state decreases down the group. Why?15. Bi hardly forms any compound in -3 oxidation state. Why?16. The stability of +5 oxidation state decreases down the group and +3 oxidation state increases. Why?17. The only well characterized Bi(V) compound is BiF5. Why?18. Bismuth is a strong oxidizing agent in pentavalent state.19. Nitrogen exhibits +1.+2,+4 oxidation states when it reacts with oxygen and phosphorus also shows +1 and +4 oxidation states in some oxoacids. Why?Give reasons:20. Nitrogen is restricted to a maximum covalence of 4 whereas heavier elements can expand their covalence as in PF6

-. 21. Why is it that molecular nitrogen is not particularly reactive?22. Why PH3 is weaker base than NH3?

23. Unlike phosphorus nitrogen shows little tendency for catenation. Explain 24. Molecular nitrogen exist in the gaseous state while other member of group 15 exist in solid state. 25. Nitrogen exists as a diatomic molecule with a triple bond and hence bond enthalpy is also high. 26. P, As,Sb forms single bonds 27. Single bond N-N bond is weaker than single P-P bond.

28. Nitrogen cannot form dπ-pπ bond as the heavier elements can e.g. R3P=O or R3P=CH2

29. P and As can form compounds (dπ-pπ) with transition elements like P(C2H5)3 and As(C6H5)3

30. On being slowly passed through water PH3 forms bubbles but NH3 dissolves. Why is it so? 31. Why PH3 is weaker base than NH3? 32. Why is BiH3 the strongest reducing agent amongst all the hydrides of group 15? 33. The stability of hydrides decreases from NH3 to BiH3 whereas reducing character increases. 34. Basic character decreases in the order NH3 > PH3 >AsH3 >SbH3 >BiH3

35. All these elements form two types of oxides E2O3 and E2O5. The acidic character of oxides decreases down the group. Or oxides of arsenic and antimony are amphoteric and those of Bismuth basic. 36. They react with metals exhibiting -3 oxidation state and form binary compounds, e.g. Ca3N2, Ca3P2, Na3As2, Zn3Sb2 and Mg3Bi2. 37. What is the covalence of nitrogen in N2O5.

The covalence of nitrogen in the above structure is four because it has four shared pair of electrons.

38. Draw the structure of all the oxides of nitrogen.

39. Why does NO2 dimerises? Explain. 40. Nitric oxide becomes brown when released in air. 41. Concentrated HNO3 turns yellow on exposure to sunlight. Why? Ans. On exposure to sunlight, nitric acid decomposes into NO2, O2 and H2O. The presence of NO2 in the partially decomposed nitric acid, to gives it yellow colour.

4HNO3 (l) → 4NO2 (g) + O2 (g) + 2H2O (l) 42. Ammonia is a good complexing agent. Explain. Ans. Ammonia is a good complexing agent because of the presence of lone pair of electrons on nitrogen. This lone pair can easily be donated to electron deficient compounds forming complexes. For example, it reacts with Cu2+ ion to form a deep blue complex.

Cu2+ (aq) + 4NH3 (aq) → [Cu(NH3)2]2+

Deep blue complex

43. Why does iron become passive when dipped in conc. HNO3? Iron becomes passive when dipped in conc. HNO3 due to the formation of a thin protective layer of the metal oxide on its surface. This protective layer corresponds to ferrosoferric oxide, FeO.Fe2O3 and prevents further action of the metal.

44. In the ring test of nitrates what chemical compound is formed?

45. Give one reaction in which ammonia acts as a reducing agent. When ammonia is passed over heated cupric oxide, copper is formed

3CuO + 2NH3 → 3Cu + N2 + 3H2O 46. N2O supports combustion more vigorously than air. Explain. Ans. N2O decomposes to give O2 which is about 1/3 of the volume of gases produced (2N2 + O2). On the other hand, air contains 1/5 th part of O2 of its volume. Due to larger content of O2, N2O supports combustion more vigorously than air.

47. Mention the conditions required to maximize the yield of ammonia. Ans.

N2 + 3H2 2NH3 ΔH0 = - 46.1 kJ mol-1

(i) Low temperature of the order of about 700 K.(ii) High pressure of 200 x 105 Pa(iii) Presence of catalyst such as iron oxide with small amount of K2O

and Al2O3. 48. Write the short note on allotropic forms of phosphorus,. Phosphorus is found in many allotropic foms, the important ones being white, red and black.WHITE PHOSPHORUS : - It is a translucent white waxy solid. It is poisonous, insoluble in water but soluble in carbon disulphide and glows in

dark (chemiluminescence). It dissolves in boiling NaOH solution in an inert atmosphere giving PH3.

P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2 (sodium hypophosphite)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 600. It readily catches fire in air to give dense white fumes of P4O10.

P4 + 5O2 → P4O10

It consists of discrete tetrahedral P4 molecule.RED PHOSPHORUS:- It is obtained by heating white phosphorus at 573 K in an inert atmosphere for several days.When red phosphorus is heater under high pressure, a series of phases of black phosphorus are formed.

Red phosphorus possesses iron grey lustre. It is odourless, non toxic and insoluble in water as well as in carbon disulphide. Chemically, red phosphorus is much less reactive than white phosphorus. It does not glow in the dark. It is polymeric consisting of chains of P4 tetrahedra linked together in the manner.BLACK PHOSPHORUS : - It has two forms α – black phosphorus and β – black phosphorus. α – blackphosphorusis formed when red phosphorus is heated in a sealed tube at 803 K. It can be sublimed in airand has opaque monoclinic or rhombohedral crystals. It does not oxidize in air. β – black phosphorus isprepared by heating white phosphorus at 473 K under high pressure. It does not burn in air upto 673 K.

49. PCl5 is ionic in solid state. Ans.PCl5 is ionic in the solid state because it exists as [PCl4]+[PCl6]- in which the cation is tetrahedral and anion is octahedral.

50. Bond angle in PH4+ is higher than that in PH3. Why?

51. Write a balanced equation for the hydrolytic reaction of PCl5 with heavy water. Ans. PCl5 + D2O → POCl3 + 2DCl

52. What happens when PCl5 is heated?Ans. On heating, PCl5 first sublimes and then decomposes on strong heating.

PCl5 → PCl3 + Cl2

53. Why does PCl3 fume in moisture?Ans. PCl3 gets hydrolysed in the presence of moisture and gives fumes of HCl.

PCl3 + 3H2O → H3PO3 + 3HCl 54. In what way it can be proved that PH3 is basic in nature.

Ans. PH3 reacts with acids like HI to form phosphonium iodide, PH4I.PH3 + HI → PH4I

This shows that PH3 is basic in nature. This basic nature of PH3 is due to the presence of lone pair on phosphorus atom and therefore, it acts as a Lewis base.

55. All the five bonds in PCl5 are not equivalent. Justify. 56. What happens when white phosphorus is heated with

concentrated NaOH solution in an inert atmosphere of CO2. Ans. Phosphine is formed. P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2 (sodium hypophosphite)

57. Give the structure of all the oxoacids of phosphorus. What are there basicity?

58. What is the difference between the nature of pi – bonds present in H3PO3 and HNO3 molecules?Ans.In HNO3 there is p π- pπ bonding and in H3PO3 there is p π -d π bonding. 59. How is the nature of pi bonding in H3PO4 different from that in HNO3?In HNO3 p π - p π forms multiple bonding whereas in H3PO4 there is no p π - p π multiple bonding.

60.Which of the following is a tri – basic acid? H3PO3 and H3PO4

61. Why H3PO3 is diprotic?This is due to because out of the three hydrogens, one is directly attached to the phosphorus, which cannot be ionized.Only two hydrogens attached to the oxygens, give two H+ ions hence it is diprotic.

62. Explain why nitrogen forms a large no. of oxides than phosphorus.

63. Give the disproportionation products of H3PO3.On heating H3PO3 disproportionates to give phosphoric acid and phosphine.

4 H3PO3 → 3 H3PO4 + PH3

64. How do you account for the reducing behaviour of H3PO2 on the basis of its structure. 65. What happens when orthophosphorous acid is heated? Ans. On heating, H3PO3 disproportionates to give phosphoric acid and phosphine.

4H3PO3 → 3 H3PO4 + PH3

66. Phosphoric acid has high viscosity and high melting point. Why? Ans. Phosphoric acid has a tendency to form hydrogen bonding in concentrated solutions. Therefore, it has high viscosity and is a syrupy liquid and has high boiling point.

HNO3

Group 16 Elements (Chalcogens)

What do you mean by inert pair effect? Ans. When a pair of s – electrons of outermost orbit has a tendency of not participating in chemical bond formation, it is known as inert pair effect.

1. Why does atomic and ionic radii decrease down the group?2. Why does ionization enthalpy decrease down the group?3. Ionisation enthalpy of group 16 elements is less than group 15

Elements. Why?4. Electron gain enthalpy of oxygen atom is less than sulphur. Why?5. Electronegativity decreases from O to Po.6. O,S: Non-metals, Se, Te : Metalloids, Po : Metal (Radioactive).7. Melting point…………………8. Oxygen exists as a diatomic molecule whereas sulphur exists as Polyatomic molecule (S8).9. Elements of group 16 show -2, +2,+4,+6 oxidation states but +4 +6 are more common.10. S, Se, Te show + 4 oxidation state with oxygen but +6 with Fluorine.11. The stability of +6 oxidation state decreases down the group But + 4 oxidation state increases.12. Oxygen behaves anomalously.13. Oxygen limits its covalency till four.14. Acidic character increases from H2O to H2Te.15. Thermal stability of hydrides decreases from H2O to H2Po.16. All the hydrides of this group except H2O are reducing agents.17. H2O is a liquid and H2S is a gas at room temperature.18. Reducing property of dioxides decreases from SO2 to TeO2 or TeO2 is oxidizing agent/SO2 is reducing agent.19. EO2 and EO3 both the type of oxides are acidic in nature. 20. Explain the difference in structures of SO2 and SeO2.

Ans.Due to small size of sulphur, p – orbital of sulphur can overlap with p – orbital of oxygen to form strong pπ – pπ bond, so SO2 molecules exist as discrete molecules. Due to larger size of the p – orbital of selenium, 3p – orbital of oxygen cannot effective overlap 4p – orbital of selenium. So, SeO2 exists as

one giant molecule 21. Explain the hybridization in SF6 molecule. What is the shape of this molecule? 22. Give the structure of SF4 and SF6

23. Why is OF6 compound not known? 24. Explain why SF6 is known but SH6 is not known. 25. Why should binary compounds of oxygen and fluorine be called oxygen fluorides and not fluorine oxides. What kind of bond is expected between oxygen and fluorine in oxygen fluoride?

26. Why is SF6 much less reactive than SF4? 27. SF6 is not easily hydrolysed. Explain. 28. SF4 is a gas, SeF4 is a liquid, TeF4 is a solid. 29. What kind of bond is expected between oxygen and fluorine in oxygen fluoride? Ans. Covalent type of bond, because there is very small difference in the electronegativity. 30. SO3 has zero dipole moment. Why? 31. Why does O3 act as a powerful oxidizing agent? 32. How is ozone estimated quantitatively? Ans. Ozone oxidizes potassium iodide to iodine as: 2KI + O3 + H2O → 2KOH + O2 + I2 The liberated iodine may be titrated against a standard solution of sodium thiosulphate. I2 + 2Na2S2O3 → Na2S4O6 + 2NaI. Thus, to estimate O3 quantitatively, ozone is allowed to react with known amount of excess potassium iodide solution buffered with a borate buffer (pH = 9.2). The liberated I2 is titrated against Na2S2O3 solution using starch as an indicator. 33. Sulphur in vapour state exhibits paramagnetic behaviour. Why? Ans. Sulphur exists as S2 molecule in vapour state which has two unpaired electrons in the antibonding orbital like O2 molecule. Because of this reason sulphur exhibits paramagnetism.

34. Give the structure of oxo-acids of sulphur.

35. Mention three areas in which H2SO4 plays an important Role. 36. Write the steps involved in the production of H2SO4 by contact process. Ans. 1. Burning of sulphur or sulphide ores in air to generate SO2. 2. Conversion of SO2 to SO3 by the reaction with oxygen in The presence of a catalyst (V2O5) 3. Absorption of SO3 in H2SO4 to give oleum (H2S2O7) 4. Dilution of oleum with water gives H2SO4 (96-98% pure) The key step in the manufacture of H2SO4 is the catalytic Oxidation of SO2 with O2 to give SO3 in the presence of V2O5 (Catalyst). 2SO2(g) + O2(g) 2SO3(g) ∆rH = -196.6 KJmol-1. The reaction is exothermic, reversible and the forward reaction leads to a decrease in volume. 37. Write the conditions to maximize the yield of H2SO4 by Contact process. Ans. Low temperature, 720 K High Pressure, 2 bar The temperature should not be very low otherwise rate of Reaction will become slow. 38. Why is ka2<<Ka1 for H2SO4 in water? 39. What are the characteristics of sulphuric acid?

a) Low volatility, b.p. is 611 K, freezing point 283K b) Strong acidic character c) Strong affinity for water d) Ability to act as an oxidizing agent. 40. Write the structure of O3 and SO2 41. What are the allotropic forms of sulphur? How are they obtained? 42. Comment on the nature of two S – O bonds formed in SO2 molecule. Are the two S – O bonds in this molecule equal? 43. How is the presence of SO2 dectected? Ans. Sulphur dioxide can be detected by the following tests: (i) It has a pungent characteristic smell. (ii) It decolourises acidified potassium permanganate solution. (iii) It turns acidified potassium dichromate solution green. (iv) It turns blue litmus red.

Group 17 (Halogens)

1. Halogens have smallest atomic radii in their repective periods.2. Atomic radii increase from fluorine to iodine.3. Halogens have high ionization enthalpy.4. Halogens have maximum negative electron gain enthalpy in the corresponding periods.5. Electron gain enthalpy of the elements of the group becomes less negative down the group. However, the negative electron gain enthalpy of fluorine is less than chlorine.6. Halogens have very high electronegativity.

7. Melting and boiling point increase with atomic number.8. All halogens are coloured, F2 is yellow, Cl2 is greenish yellow, Br2 is Red and I2 is violet colour.9. Fluorine and chlorine are gases, bromine is a liquid and iodine is a solid.10. Enthalpy of dissociation follows the following order: F-F < Cl-Cl > Br-Br > I-I 11. Fluorine and chlorine react with water.12. Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a sronger oxidizing agent than Chlorine. Why?13. All halogens exhibit -1 oxidation state. However, chlorine, bromine Iodine exhibit +1,+3,+5 and +7 oxidation states.14. All the halogens are very reactive. They react with metals and non-metals to form halides.15. The reactivity decreases down the group.16. Despite its lower electron affinity fluorine is a stronger oxidizing agent than chlorine. Explain.Ans. The reduction potential of fluorine is higher than the other halogens, therefore despite having lower value of electron affinity, fluorine is a stronger oxidizing agent. 17. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidizing power of F2 and Cl2.

Ans. F2, is stronger oxidizing agent than Cl2. This can be explained on the basis of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy. The process of oxidizing behaviour may be expressed as: ½ X2 (g) → X (g) ΔHdiss. X (g) → X- (g) ΔHeg X- (g) → X- (aq) ΔHhyd The overall tendency for the change (i.e., oxidizing behaviour) depends upon the net effect of

three steps. As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy change for this step is positive. On the other hand, energy is released in step (II) and (III), therefore, enthalpy change for these steps is negative. Now although fluorine has less negative electron gain enthalpy, yet it is stronger oxidizing agent because of low enthalpy of dissociation and very high enthalpy of hydration. In other words, large amount of energy required in step (I) overweigh the smaller energy released in step (II) for fluorine. As a result, the ΔH overall is more negative for F2 than for Cl2. Hence, F2 is stronger oxidizing agent than Cl2. 18. Fluorine behaves anomalously. 19. Most of the reactions of fluorine are exothermic. 20. HF is a liquid. 21. Acidic strength follows the order : HI> HBr>HCl>HF 22. Bond dissociation follows the order: H-F>H-Cl>H-Br>H-I 23. Ionic character of the metal halides decreases in the order: MF> MCl>MBr>MI. 24. SnCl4, PbCl4,SbCl5,UF6 are more covalent than SnCl2, PbCl2, SbCl3 and UF4 respectively. 25. Considering the parameter such as bond dissociation enthalpy, electron gain enthalpy & hydration enthalpy, compare the oxidizing power of F2 and Cl2. 26. What are interhalogen compounds? 27. Interhalogen compounds are more reactive than halogens. 28. How is chlorine a powerful bleaching agent? Ans. Bleaching action is due to oxidation. Cl2 + H2O 2HCl + O Coloured substance + O Colourless substance It bleaches vegetable or organic matter in presence of

Moisture. Bleaching effect of chlorine is permanent. 29. Name poisonous gases which can be prepared from chlorine gas. Ans. Phosgene (COCl2), tear gas (CCl3NO2), Mustard gas (ClCH2CH2SCH2CH2Cl) 30. Chlorine water on standing loses its yellow colour. Ans. Due to the formation of HCl and HOCl (hypochlorous acid) which gives nascent oxygen which is responsible for oxidizing and bleaching properties of chlorine. 31. HCl when reacts with finely divided iron forms ferrous chloride and not ferric chloride. Why? Ans. HCl reacts with finely divided iron and produces H2 gas. Fe + 2HCl → FeCl2 + H2. Liberation of hydrogen prevents the formation of ferric chloride. 32. Draw the structures of chloric (I) acid, chloric (III) acid, chloric (V) acid, chloric (VII) acid.

33. Fluorine forms only fluoric (I) acid,HOF.Ans. Due to high electronegativity and small size of fluorine.34. Arrange the following oxoacids in increasing order of acid strength and decreasing order of oxidizing power. HClO4, HClO3, HClO2, HClO.

Ans. HClO < HClO2 < HClO3 < HClO4 (increasing order of strength) HClO > HClO2 > HClO3 > HClO4 (decreasing order of oxidizing power) 35. Draw the structures of interhalogen compounds according to VSEPR theory: XX’3; XX’5; XX’7 BrF5, ICl5, BrF3, ICl3 36. ClF3 exists but FCl3 does not. Explain.

Ans. Chlorine has vacant d – orbitals, thus one p – electron on promotion to d – orbital gives 3s2 3p4 3d1 (three unpaired electrons) which forms ClF3. Fluorine cannot form FCl3 due to the absence of d – orbitals.

37. NCl3 gets readily hydrolysed while NF3 does not. Why? Ans. In NCl3, Cl has vacant d – orbitals to accept lone pair of electrons present on oxygen atom of water molecule

38. With the increase in oxidation no. of a particular halogen atom, the acidic character of corresponding oxoacid Increases. Explain. (HClO < HClO2 < HClO3 < HClO4) Ans. Here oxygen is more electronegative than chlorine. The more the oxygen atom bonded to the chlorine atom more the electrons will be pulled away from the OH bond, and the more the bond will be weakened. Thus HClO4 requires the least energy to break the O — H bond and form H+. Hence HClO4 is the strongest acid. The strength of acid is also explained on the basis of stability of ion formed. Thus the stability will be ClO4- > ClO3- > ClO2- > ClO– Hence the acidic strength will also decreases from HClO4 to HClO.

39. Preparation of interhalogen compounds.

40. Hydrolysis of interhalogen compounds give halide ion derived from the smaller halogen and a hypohalite (when XX’), halite (when XX’3), halate (when XX’5) and perhalate (when XX’7) anion from the larger halogen

XX’ + H2O HX’ + HOX XX’3 + 2H2O 3HX’ + HOXO

Group 18 (Noble Gases)

1. Why are the elements of Group 18 known as noble gases?2. Which noble gas is not found in atmosphere?Ans. Radon, it is obtained as a radioactive decay product of 226Ra. 226Ra 222

86Rn + 42He

3. Name the rarest elements of this group.Ans. Xenon and Radon4. What is the commercial source of Helium?Ans. Natural Gas5. Name the noble gases whose minerals are of radioactive origin.Ans. Helium and Neon6. Noble gases exhibit very high ionization enthalpy.7. Ionization enthalpy decreases with increase in atomic size.8. They have large positive values of electron gain enthalpy.

9. Noble gases are monoatomic.10. They are colourless, odourless and tasteless.11. They are sparingly soluble in water.12. They have very low melting and boiling point. 13. Noble gases are least reactive.14. What was the reasoning applied by Neil Bartlett for carrying out reaction of Xe with PtF6? Ans. In 1962, Neil Barlett noticed that platinum hexafluoride is a powerful oxidizing agent. He observed that an ionic compound is formed when PtF6 reacts with O2 i.e., PtF6 is capable to oxidize O2 into O2+. O2 + PtF6 → O2+ [PtF6]- He also noticed that the oxygen and xenon have some markable similarities, like nearly same ionization energies and both have same molecular

diameters. He thought that because of above similarities O2 could be replaced by Xe. He did the same and obtained an orange yellow solid Xe[PtF6], after the reaction of PtF6 and Xe (in excess). Xe + PtF6 → Xe+[PtF6]-

15. Which compound led to the discovery of compounds of noble gas? Ans. The discovery of compound O2+[PtF6]- led the discovery of compounds of noble gas.

16. Most of the known noble gas compounds are those of xenon. Explain.Ans. Because the ionization energy of Xe is lower than that of He, Ne, Ar thus it can form compounds easily.

17. Why do noble gases form compounds with fluorine and oxygen only? Ans. The first ionization potential of fluorine and oxygen is comparable to that of noble gases specially xenon.

18. Write the name of the compound which is formed by krypton. Ans. Krypton is known to form kryptondifluoride (KrF2).19. Neon is generally used in warning signal illumination. Why? Ans. Because neon light is visible through mist and fog and that too from long distance.

20. For protecting electrical instruments, neon is generally used in safety devices. Why? Ans. It is because of the property of neon of carrying exceedingly high current under high pressure it is used in safety devices for protecting electrical instruments.

21. Write one use of neon. Ans. Neon is used in fluorescent bulbs used for advertisement display.

22. Write two reactions to show that xenon fluorides act as fluoride ion donors.Ans. i) XeF2 + PF5 [XeF]+[PF6]-

ii) XeF4 + SbF5 [XeF3]+[SbF6]-

iii) XeF6 + MF M+[XeF7]- (M=Na,K,Rb or Cs) iv) Hydrolysis of XeF2, XeF4 and XeF6

23. Draw and name the structures of XeF2,XeF4, XeF6,XeOF4,XeO3.24. Does the hydrolysis of XeF6 lead to a redox reaction? Explain.25. Why is Helium used in diving apparatus?26. Why has it been difficult to study the chemistry of radon?27. Write the chemical equations for the following: a) Preparation of XeF2, XeF4, XeF6

b) XeF4 + O2F2 at 143K c) Hydrolysis of XeF2

d) Hydrolysis of XeF4, XeF6

e) Partial hydrolysis of XeF6.28. Write the uses of Helium, Neon and Argon.