2nd PUC Chemistry Question Bank Chapter 7 The p-Block Elements

Students can Download 2nd PUC Chemistry Chapter 7 The p-Block Elements Questions and Answers, Notes Pdf, 2nd PUC Chemistry Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka 2nd PUC Chemistry Question Bank Chapter 7 The p-Block Elements

Question 1.
Give four anomalous properties of nitrogen.
Answer:
Nitrogen shows following anomalous properties.

  1. Smaller size.
  2. High electro negativity.
  3. High ionisation enthalpy.
  4. Non-availability of d-orbitals.

Question 2.
Nitrogen exhibits + 5 oxidation state, it does not form pentahalide. Give reason.
Answer:
Nitrogen does not have d-orbitals to expand the covalent beyond four. Hence it does not form pentahalide.

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 3.
PH3, has lower boiling point than NH3. Why?
Answer:
PH3 molecules does not associated through hydrogen bonding in liquid state.

Question 4.
Why pentahalides more covalent than trihalides?
Answer:
In pentahalides oxidation state of central atom is more (+5) than in trihalides (+3) i.e., central atom in pentahalides have larger polarising power and hence it gives more covalent character.

Question 5.
Why is BiH3 the strongest red using agent among all the hydrides of group 15 elements?
Answer:
In group 15 elements, BiH3 is least stable, because Bi has largest size. It easily loses hydrogen atom and acts as reducing agent.

Question 6.
Name the gas liberated when aqueous solution of ammonium chloride is mixed with sodium nitrite. Give equation.
Answer:
Nitrogen gas
NH4Cl + NaNO2 → N2 + 2H2O + NaCl

Question 7.
What happens when ammonium dichromate crystals are heated? Give equation.
Answer:
Ammonium dichromate crystals on heating gives nitrogen gas.
(NH4)2Cr2O7 → Cr2O3 + 4H2O + N2

Question 8.
How pure nitrogen gas is prepared? Give equation.
Answer:
Pure nitrogen gas is prepared by heating Barium azide.
Ba(N3)2 → Ba + 3N2

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 9.
How ammonia is manufactured by Habers process?
Answer:
Ammonia is manufactured by Haber’s process by the direct combination of nitrogen and hydrogen.
N2(g) + 3H2(g) → 2NH3(g)
According to Le-chatelier’s principle yield of ammonia is increased by

  • High pressure of 200 atm.
  • Optimum temperature of 700 K.
  • Use of catalyst like iron oxide with K2O and Al2O3

2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 1

Question 10.
Why does ammonia acts as a lewis base? Give an example.
Answer:
Nitrogen atom in ammonia has one lone pair of electrons which is available for donation.
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 2

Question 11.
Explain the manufacture of nitric acid by Ostwald’s process.
Answer:
Step – 1 : Catalytic oxidation of ammonia gives nitric oxide.
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 3

Step – 2: Nitric oxide thus formed combines with oxygen gives NO2
2NO + O2 ⇌ 2NO2

Step – 3: Nitrogen dioxide so formed dissolves in water to give
H2O + 3NO2 → 2HNO3 + NO

Question 12.
What is the action of Conc. HNO3 and dil. HNO3 on copper turnings.
Answer:
Conc. HNO3 oxidises copper to cupric nitrate and itself reduces to NO2
Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Dilute HNO3 oxidises copper to curpic nitrate and it reduces itself to NO
3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 13.
What is the action of Conc. HNO3 and dil HNO3 on zinc metal.
Answer:
Conc. HNO3 oxidises zinc to zinc nitrate and itself reduces to NO2.
Zn + 4HNO3 → Zn(NO3)2 + 2H2O + 2NO2
Dilute nitric acid Oxidises zinc to zinc nitrate and itself reduces to N2O.
4Zn + 10HNO3 → Zn(NO3)2 + 5H2O + N2O

Question 14.
Name the gas liberated when zinc reacts with dii HNO3
Answer:
N2O (Nitrous oxide).

Question 15.
Name the metals which do not react with Conc. HNO3. Give reason.
Answer:
Cr, Al metals do not react with Conc. HNO3, because of passive film of oxide on the surface.

Question 16.
Give two differences between white Phosphorus and red Phosphorus
Answer:

White Phosphorus Red Phosphorus
1. Consist of discrete tetrahedral units. 1. Consists of chains of a P4 tetrahedral units.
2. It glows in dark 2. It does not glows in dark
3. More reactive 3. Less reactive
4. Poisnous 4. Non-Poisnous

Question 17.
How is phosphine prepared from calcium phosphide
Answer:
Phosphine is Prepared by the action of water or dil HCl on calcium phosphide.
Ca3P2 + 6H2O → 3 Ca(OH)2 + 2PH3.
Ca3P2 + 6HCl → 3CaCl2 + 2PH3,

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 18.
How phosphine is prepared in the laboratory?
Answer:
Phosphine is prepared by heating white phosphorus with Cone. NaOH solution.
P4 + 3 NaOH + 3H2O → PH3 + 3NaH2PO2

Question 19.
Show that PH3 is basic in nature.
Answer:
PH3 reacts with acids like HI to form PH4I.
PH3 + Hl → PH4I

Question 20.
Give reason: PH3 has lower melting point than NH3.
Answer:
PH3 molecules does not form hydrogen bonds like NH3, hence it has low melting point.

Question 21.
Nitrogen is less reactive at room temperature.
Answer:
High bond dissociation enthalpy makes nitrogen less reactive.

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 22.
Which allotropic form of phosphorus has discrete tetrahedral P4 molecules?
Answer:
White phosphorus.

Question 23.
Bond angle in PH+4 is higher than that in PH3. Why?
Answer:
In both PH+4 and PH3, atom phosphorus is in sp3 hybridised state. PH+4 is tetrahedral in Shape and bond angle is 109° 281. In PH3 due to the presence of one lone pair of electrons bond angle decreases from tetrahedral angle.

Question 24.
Name the acid obtained when PCI undergoes hydrolysis. Give equation.
Answer:?
Orthophosphoric acid
PCl5 + H2O → POCl3 + 2HCl
POCl3 + 3H2O → H3PO4 + 3HCl

Question 25.
White phosphorus is heated with excess of dry chlorine to get X. X on hydrolysis finally forms an oxoacids of phosphorus Y. What are X and Y? What is the basicity of Y?
Answer:
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 4

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 26.
Are all the five bonds in PCl5 molecule equivalent? Justify your answer.
Answer:
No. PCl5 has a trigonal bipyramidal structure. Three equitorial P-Cl bonds are equivalent, while the two axial bond are different and longer than equitorial bonds.

Question 27.
Write the structure for the following oxoacids of phosphorus.
1. Hypophosphorous acid
2. Orthophosphorous acid
3. Orthophosphoric acid
4. Pyrophosphoric acid.
Answer:
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 5

Question 28.
How do you account for the reducing behaviour of H3PO2 ?
Answer:
The acids which contain P-H bond have strong reducing properties. In H3PO2 two H atoms are bonded directly to P atom which imparts reducing character to the acid.

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 29.
Mention any two reasons for the anomalous behaviour of oxygen.
Answer:
Oxygen shows following anomalous properties.

  1. Smaller size
  2. High electronegativity
  3. Non availability of d-orbitals.

Question 30.
H2S is less acidic than H2Tc. Give reason.
Answer:
Due to the decrease in bond dissociation enthalpy down the group, acidic character increases.

Question 31.
Among the following which one is more acidic? Give reason.
H2O, H2S, H2Sc, H2Te
Answer:
Due to the decrease in bond dissociation enthalpy down the group, acidic character increases.

Question 32.
Why is H2O a liquid and H2S a gas?
Answer:
Due lo hydrogen bonding.

Question 33.
Give an example for amphoteric oxide.
Answer:
Al2O3.

Question 34.
Describe the preparation of ozonised oxygen with an equation.
Answer:
When slow dry steam of oxygen is passed through a silent electrical discharge, conversion of oxygen to ozone takes place. The product is known as ozonised oxygen.
3O2 → 2O3, ΔH = +142 kJ mol-1

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 35.
Give two examples to show that ozone is an oxidising agent.
Answer:
1. Ozone oxidises lead sulphide to lead sulphate.
PbS + 4O3 → PbSO4 + 4O2

2. Ozone oxidises iodide ions to iodine.
2I + H2O + O3 → I2 + O2 + 2OH

Question 36.
Write the equation for the action of ozone with lead sulphide.
Answer:
1. Ozone oxidises lead sulphide to lead sulphate.
PbS + 4O3 → PbSO4 + 4O2

2. Ozone oxidises iodide ions to iodine.
2I + H2O + O3 → I2 + O2 + 2OH

Question 37.
Mention the allotropic form of sulphur which is more stable above 369 K and below 369 K.
Answer:
Stable above 369 K is β – Sulphur (Monoclinic sulphur).
Stable below 369 K is α – Sulphur (Rhombic sulphur).

Question 38.
Complete the following equation NO + O3
Answer:
NO + O3 → NO2 + O2

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 39.
Give the conversion of
1. SO2 to SO2Cl2
2. SO2 to SO3
Answer:
1. Sulphur dioxide reacts with chlorine in presence of charcoal. Sulphuryl chloride is obtained.
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 6

2. Sulphur dioxide is oxidised to sulphur trioxide in presence of V2O5 catalyst
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 7

Question 40.
Give two examples to show that moist SO2 is a reducing agent.
Answer:
1. Moist sulphur dioxide reduces Fe3+ [Iron (III)] to Fe2+ [lron (II)]
2Fe3+ + SO2 + 2H2O → SO2-4 + 2Fe2+ + 4H+

2. Moist sulphur dioxide decolourises potassium permanganate.
2 MnO4 + 5SO2 + 2H2O → 5SO2-4 + 4H+ + 2Mn2+

Question 41.
Complete the following equation.
2Fe3+ + SO2 + 2H2O →
Answer:
1. Moist sulphur dioxide reduces Fe3+ [Iron (III)] to Fe2+ [lron (II)]
2Fe3+ + SO2 + 2H2O → SO2-4 + 2Fe2+ + 4H+

2. Moist sulphur dioxide decolourises potassium permanganate.
2MnO4 + 5SO2 + 2H2O → 5SO2-4 + 4H+ + 2Mn2+

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 42.
Give the structure for
(a) Sulphurous acid
(b) Sulphuric acid
(c) Peroxosulphuric acid
(d) Pyrosulphuric acid (Oleum).
Answer:
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 8

Question 43.
Show that Cone. H2SO4 is a dehydrating agent.
Answer:
Sulphuric acid is manufactured by the contact process which involves following Steps.
Step -1: Burning of sulphur or sulphide ores in air to generate SO2
S + O2 → SO2

Step – II: Catalytic oxidation of SO2 with O2 in presence of V2O5 catalyst gives SO3
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 9

Step – III: SO3 gas is absorbed in Cone. H2SO4 to produce oleum. Dilution of oleum with water gives H2SO4 of desired concentration.
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 10

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 44.
Complete the following equation.
H2SO4 + SO3 → ?
Answer:
H2SO4+ SO3 → H2S2O7

Question 45.
Show that Conc. H2SO4 is a dehydrating agent.
Answer:
Conc. H2SO4 removes elements of water from the organic compounds. Example: Sugar undergoes charring when it is warmed with Conc. H2SO4
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 11

Question 46.
Complete the following equation.
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 12
Answer:
Cone. H2SO4 removes elements of water from the organic compounds. Example: Sugar undergoes charring when it is warmed with Conc. H2SO4
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 11

Question 47.
Complete the following equations and balance.
1. CU + H2SO4
2. S + H2SO4
3. C + H2SO4
4. 2NaCl + H2SO4 → _____ + Na2SO
Answer:
1. CU + 2H2SO4 → CuSO4 + 2H2O + SO2
2. S + 2H2SO4 → 2H2O + 3SO2
3. C + 2H2SO4 → 2H2O + 2SO2 + CO2
4. 2NaCl + H2SO4 → 2HCl + Na2SO4

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 48.
Give two anomalous properties of fluorine.
Answer:
The anomalous behavior of fluorine is due to its

  1. Smaller size.
  2. Highest electronegativity.
  3. Low F – F bond dissociation enthalpy.
  4. Non-availability of d-orbitals in valence shell.

Question 49.
Fluorine exhibits only-1 oxidation state, whereas other halogens exhibit +1, +3, +5 and +7 oxidation states. Explain.
Answer:
Fluorine is the most electronegative element and cannot exhibit any positive oxidation state.
Other halogens have d-orbitals and therefore can expand their octacts and show +1, +3, +5 and +7 oxidation states also.

Question 50.
Complete the following equations.
2F2 + 2H2O →
Answer:
2F2 + 2H2O → 4HF + O2

Question 51.
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 13
Answer:
2PbO2(s) → 2PbO(s) + O2

Question 52.
Give three methods of preparation of chlorine gas.
Answer:
1. Chlorine gas is prepared by heating manganese dioxide with Cone. HCl.
MNO2 + 4HCl → Mncl2 + 2H2O + Cl2

2. Chlorine gas is prepared when mixture of NaCl and MnO2 is treated with Conc. H2SO4
4NaCl + MnO2 + 4H2SO4 → 4NaHSO4 + MnCl2 + 2H2O + Cl2

3. Chlorine gas is prepared by treating Cone. HCl on KMnO4
2KMnO4 + 16 HCl → 2KCl + 2MnCl2 + 8H2O + 5Cl2

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 53.
Name the gas liberated when concentrated HCl is heated with MnO2. Give the equation for the reaction. Name the reagent used to obtain bleaching powder from chlorine.
Answer:

  • Chlorine gas
  • MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
  • Slaked lime.

Question 54.
How is chlorine prepared in the laboratory using KMnO4?
Answer:
Chlorine gas is prepared by treating Cone. HCl on KMnO4
2KMnO4 + 16 HCl → 2KCl + 2MnCl2 + 8H2O + 5Cl2

Question 55.
How chlorine gas is manufactured by Deacon’s process?
Answer:
Oxidation of HCl gas by atmosphere oxygen in the presence of CuCl2 at 723K.
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 14

Question 56.
Give three examples to show that chlorine has affinity towards hydrogen.
Answer:
It reacts with the compounds containing hydrogen lo form HCl.
H2 + Cl2 → 2HCl
H2S + Cl2 → 2HCl + S
C10H16 + 8Cl2 → 16 HCl + 10C

Question 57.
How excess of ammonia reacts with chlorine.
Answer:
Excess of ammonia reacts with chlorine forming ammonium chloride with N2 gas.
8NH3 + 3Cl2 → 6NH4Cl + N2

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 58.
How excess of chlorine reacts with ammonia.
Answer:
Ammonia reacts with excess of chlorine forming nitrogen trichloride.
NH3 + 3Cl2 → NCl3 + 3HCl

Question 59.
How cold and dilute sodium hydroxide (alkali) reacts with chlorine.
Answer:
Cold and dilute sodium hydroxide reacts with chlorine forming sodium chloride and sodium hypochlorite.
2NaOH + Cl2 → NaCl + NaClO + H2O

Question 60.
Complete the following equation.
2NaOH + Cl2
(Cold and dil)
Answer:
2NaOH + Cl2 → NaCl + NaCIO + H2O

Question 61.
How hot and concentrated sodium hydroxide reacts with chlorine gas?
Answer:
Hot and Conc, sodium hydroxide reacts with chlorine gas forming sodium chloride and sodium chlorate.
6NaOH + 3Cl2 → 5 NaCl + NaCIO 3 + 3H2O

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 62.
Give the reaction of chlorine with slaked lime.
Answer:
Chlorine reacts with slaked lime forming bleaching powder.
2Ca(OH)2 + 2Cl2 → Ca(OCl)2 + CaCl2 + 2H2O

Question 63.
Give the composition of bleaching powder.
Answer:
Ca(OCl)2.CaCl2.Ca(OH)2.2H2O

Question 64.
Give three examples to show that chlorine is an oxidising agent.
Answer:
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 15
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 16

Question 65.
Complete the following reactions.
1. 2 NaOH + Cl2
(Cold and dilute)
2. 2FeSO4 + H2SO4 + Cl2
3. Cl2 + 3F2
Answer:
1. 2NaOH + Cl2 → NaCl + NaCIO + H2O
2.
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 15
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 16
3. Cl2 + 3F2 → 2ClF3

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 66.
Complete the equation
SO2 + Cl2 + 2H2O →
Answer:
SO2 + Cl2 + 2H2O → H2SO4 + 2HCl

Question 67.
Which halogen has highest electron gain enthalpy or electron affinity?
Answer:
Chlorine.

Question 68.
Show that chlorine is a bleaching agent.
Answer:
Moist chlorine is a bleaching agent. Bleaching action is due to oxidation and bleaching action of chlorine is permanent.
H2O + Cl2 → 2HCl + [O]
Colour substance + [O] → Colourless substance.

Question 69.
What is an aqua regia? Give its one use.
Answer:
Aqua regia is a mixture of 1 : 3 by volume of Cone. HNO3 and Cone. HCl.
Aqua regia is used to dissolve noble metals like Au and Pt.
Au + 4H+ + NO3 + 4Cl → AuCl4 + 2H2O + NO

Question 70.
Give the structure of
(a) Hypochlorous acid
(b) Chlorous acid
(c) Chloric acid
(d) Perchloric acid.
Answer:
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 17

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 71.
Which is the strongest, acid among the hydrogen halides? Give one reason.
Answer:
HI is the strongest acid among the hydrogen halides because it has low bond dissociation enthalpy.

Question 72.
Give reason “BrF5 is more reactive than Br2“.
Answer:
Br – F bond is weaker than Br – Br bond.

Question 73.
What is the shape of CIF3 BrF5 IF7 molecules.
Answer:
CIF3 → T – Shaped;
BrF5 → Square pyramidal
IF7 → Pentagonal bipyramidal.

Question 74.
Interhalogen compounds are more reactive than halogens. Give reason
Answer:
In interhalogen compounds, X – X’ bond is weaker than X – X bond in halogens

Question 75.
Why are the elements of Group-18 known as noble gases?
Answer:
Group 18 elements have their valence shell orbitals are completely filled. Hence they are inert towards chemical activity. Therefore group-18 elements an known as noble gases.

Question 76.
What is the commercial sources of helium?
Answer:
Natural gas.

Question 77.
Complete the following equatIon.
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 18
Answer:
42He

Question 78.
Noble gases have very low boiling points. Why?
Answer:
In noble gases interatomic forces are weak dispersion forces and therefore they are liquified at very low temperatures. Hence they have low boiling points.

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 79.
Give reason for chemical inertness of noble gases.
Answer:
Stable ns2np6 valence shell configuration.

Question 80.
Name the noble gas which does not have general noble gas configuration ns2np6
Answer:
Helium.

Question 81.
Name the noble gas obtained as a decay product of 226R.
Answer:
Helium.

Question 82.
Complete the following equation.
Answer:
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 19

Question 83.
Write the general electronic configuration of noble gases.
Answer:
ns2np6

Question 84.
Noble gases are less reactive. Give two reasons.
Answer:
Inertness to chemical reactivity is due to the following reasons:

  1. Noble gases have completely filled ns2np6 electronic configuration in their valence shell except helium (1s2).
  2. They have high ionisation enthalpy and positive electron gain enthalpy.

Question 85.
Noble gases are chemically inert. Give one reason.
Answer:
Stable octet configuration (ns2np6)

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 86.
Mention the noble gas element used in cancer therapy.
Answer:
Radon.

Question 87.
Name the noble gas compound prepared by the Bartlet.
Answer:
Xe[PtF6]. Xenon hexafluridoplantinate(v) (Xennon platinum hexafluoride).

Question 88.
How xenon platinum hexafluoride is prepared?
Answer:
Red coloured xenon platinum hexafluoride is prepared by mixing PtF6 with xenon.

Question 89.
How is XeF6 prepared?
Answer:
XeF6 is prepared by the interaction of XeF4 and O2F2.
XeF4 + O2F2 → XeF6 + O2

Question 90.
What is the shape of XeF2, XeF4 and XeF6 molecules?
Answer:
XeF2 → Linear shape.
XeF4 → Square planar.
XeF6 → Distorted octahedral.

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 91.
Write the structure of
(a) XeF2
(b) XeF4
(c) XeO3
(d) XeOF4
Answer:
2nd PUC chemistry Question Bank Chapter 7 The 'P'- Block Elements 20

Question 92.
Identify the product A in the following reaction.
XeF2 + 3H2O → A + 6HF
Answer:
XeO3

Question 93.
Complete the following equation and balance.
XeF2 + H2O → _____ + _____ + O2
Answer:
XeF2 + 2H2O → 2Xe + 4HF + O2

Question 94.
How is XeO3 is prepared?
Answer:
Hydrolysis of XeF6 gives XeO3
XeF6 + 3H2O → XeO3 + 6HF

Question 95.
How is XeOF4 and XeO2F2 prepared?
Answer:
Partial hydrolysis of XeF6 gives XeOF4 and XeO2F2
XeF6 + H2O → XeOF4 + 2HF
XeF6 + 2H2O → XeO2F2 + 4HF

Question 96.
Give two uses of Helium.
Answer:

  1. Helium is used in wheather balloons.
  2. Oxygen diluted with helium used in diving appratus.
  3. Liquid helium is used as cryogenic agent for carrying out various experiments at low temperatures.

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 97.
Give two uses of Neon.
Answer:

  1. Neon is used in discharge tubes and fluorescent bulbs.
  2. Neon bulbs are used in botonical gardens and in green houses,

Question 98.
Give two uses of Argon.
Answer:

  1. Argon is used to provide inert atomsphere in high temperature metallurgical process.
  2. If is used for filing electric bulbs.

Question 99.
Which noble gas is most abundant, in atmospheric dry air?
Answer:
Argon(Ar)

Question 100.
Give reason ICI is more reactive than I2
Answer:
ICI bond is weaker than I2 bond.

2nd PUC Chemistry Question Bank Chapter 7 The 'P'- Block Elements

Question 101.
HF’ is liquid but other hydrogen halides are gas. Give reason.
Answer:
Due to strong intermolecular hydrogen bonding.

Question 102.
Which noble gas does not occur in atmosphere?
Answer:
Radon.