Short & Long Answer Question: The p-Block Elements (Group 15-18) | Chemistry Class 12 - NEET PDF Download

Q.1 Write the elements of group 15?

Ans. The elements of group 15 are Nitrogen (N), Phosphorous (P) , Arsenic (As) , Antimony (Sb) and Bismuth (Bi).
Q.2  Write chemical name & formulae of

a) Chile saltpetre

b) Indian saltpetre

Ans. (a) Chile saltpetre – Sodium nitrate – NaNO₃

(b) Indian saltpetre – Potasium nitrate –KNO₃
Q.3 What is special about the valence configuration of Group 15?

Ans. The valence configuration of 15 group is ns²np³ the s-orbital is completely filled and p-orbital is half filled . This half filled orbital gives extra stability to elements of this group.
Q.4  The atomic radii increases considerably from N to P but very little increase is observed from As to Bi. why?

Ans. There is a considerable increase in size from N to P as expected but due to the presence of completely filled d- orbitals which have very poor shielding effects, the increases in size is very little from As to Bi.
Q.5 Give reason for the following- the first ionization enthalpy of 15^th group elements is higher than 16^th group elements ?

Ans. Due to extra stability of half filled configuration, the first Ionisation enthalpy of 15^th

group elements is higher than 16^th group configuration ns²np³
Q.6 How does metallic character vary down the 15 group & why?

Ans. The metallic character increases down the group due to decrease in ionization enthalpy and increase in size of atom.
Q.7 What are the common oxidation states of this group?
Ans. The common oxidation states of the group are -3, +3 & +5.
Q.8 What is the maximum covalence shown by N? 

Ans. Nitrogen shows a maximum covalence of +4 because only four orbitals, one S and three P- orbitals are available for bonding in Nitrogen.
Q.9 Bi (v) is a stronger oxidizing agent than Bi(III). Why?

Ans. Bi is more stable in +3 oxidation state in comparison to +5 due to inert pair effect

therefore Bi (v) has a strong tendency to act as oxidizing agent.
Q.10 Give an example showing disproportionation of oxidation state of nitrogen?
Ans.
 
Here Nitrogen is getting oxidized to a higher oxidation state as well as reduced to a lower oxidation state.

Q.11 Complete and balance -
(i) (NH₄)₂SO₄ + 2NaOH →
(ii)  2FeCl₃(aq) + 3NH₄OH(aq)→
(iii) AgCl(s) + 2NH₃(aq)→
(iv) NaNO₃ + H₂SO₄→
(v) 3NO₂(g) + H₂O(l)→
(vi) Cu + 4HNO₃(conc)→

(vii) 4Zn + 10HNO3 (dil)→
(viii) [Fe(H₂O)₆]₂+NO→
(ix) I₂ 10HNO₃→
(x) S_s + 48 HNO₃(conc)→
Ans. (i) (NH₄)₂ SO₄ + 2 NaOH → 2NH₃ + 2H₂O + Na₂SO₄
(ii) 2FeCl₃(aq) + 3NH₄OH(aq) → Fe₂O₃.xH₂O(s) + 3NH₄Cl(aq)
(iii) AgCl(s) + 2NH₃(aq) → [Ag(NH₃]₂] Cl(aq)
(iv) NaNO₃ + H₂SO₄ → NaHSO₄ + HNO₃
(v) 3NO₂(g) + H₂O(l)→ 2HNO₃(aq) + NO(g)
(vi) Cu + 4HNO₃(conc) → Cu(NO₃)₂ + 5H₂O +N₂O
(vii) 4Zn + 10HNO₃(dil) → 4Zn(NO₃)₂ + 5H₂O + N₂O
(viii) [Fe(H₂O)₆]²⁺ + NO → [Fe(H₂O)₅(NO)]²⁺ + H₂O
(ix) I₂ + 10HNO₃ → 2HIO₃ + 10 NO₂ + 4H₂O
(x) S₈+48HNO₃(conc) → 8H₂SO₄ + 48NO₂ + 16H₂O
Q.12 What are the optimum conditions for maximum yield of ammonia?
Ans. The optimum conditions for the production of ammonia are - 200*10⁵ Pa or 200 atm pressure, 700K temperature , and presence of catalyst such as iron oxide with K₂O and Al₂O₃ as promoters.  
Q.13 Ammonia is a Lewis base. Why?

Ans. Due to the presence of lone pairs on nitrogen atom of ammonia, it can donate electron pair and acts as a lewis base.

AmmoniaQ.14 Ammonia has higher boiling and melting points than expected. Why?

Ans. In solid and liquid states, ammonia molecules are associated by inter-molecular

hydrogen bonding. There fore ammonia has higher boiling and melting points.
Q.15  Give reasons for the following:-
a)Halogens have smallest atomic radii in their periods

b) The negative electron gain enthalpy of fluorine is less than that of chlorine.

c) All halogens are coloured.

d) The only possible oxidation state of fluorine -1.

e) Fluorine forms only one oxoacid.

f) The stability of hydrides follows the order HF>HCl>HBr>HI.

Ans.  (a) Due to maximum effective nuclear charge, halogens have smallest atomic radii.

(b) Due to small size of fluorine atom, there are strong inter electronic repulsions in the

small 2p orbital of fluorine and thus incoming electron does not experience much attraction and fluorine has less negative electron gain enthalpy than that of chlorine.

(c) Halogens absorb radiation in visible region which results in oxidation of electrons to

higher energy level by absorbing different quanta of radiation, they show different colours.

(d) Since fluorine is most electronegative element and is short of only one electron for

completing octet, it shows the only oxidation state of -1.

(e) Due to small size and high electro-negativity, Fluorine forms only one hypohalous acid.
(f) As the size of element increases down the group, the bond dissociation enthalpy for HX bond decreases making the bond weaker and weaker therefore the order of thermal stability is HI < HBr < HCl < HF.
Q.16 Complete and balance-
(i) 2F(g) + 2H2O(l)→
(ii) 4NaCl + MnO₂ + 4H₂SO₄ →
(iii) 4HCl + O₂ 
(iv) C₁₀H₁₆ + 8Cl₂ →
(v) 6NaOH + 3Cl₂→
Ans. (i) 2F₂(g) + 2H₂O(l) → 4H⁺(aq) + 4F^- + O₂(g)
(ii) 4NaCl + MnO₂ + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + 2H₂O + Cl₂
(iii) 4HCl + O₂   2Cl₂ + 2H₂O
(iv) C₁₀H₁₆ + 8Cl₂ → 16HCL + 10C
(v) 6NaOH + 3Cl₂  → 5NaCl + NaCiO₃ + 3H₂O

Q.17 Chlorine water on standing loses its yellow colour. Why?
Ans. On standing chlorine water forms HCl and Hypochlorous acid (HOCl) due to which it loses its colour. Cl₂ + H₂O → HCL + HOCI
Q.18 . Explain the bleaching action of chlorine?
Ans. The bleaching action of chlorine is due to its tendency to give nascent oxygen so that the substance gets oxidized.  Cl₂ + H₂O ---> 2HCl +[O]
Coloured substances + [O] → colourless substance.
Q.19 Write two uses of chlorine? 

Ans. Chlorine is used for

a) Bleaching wood pulp, cotton and textiles

b) Manufacturing dyes, drugs, refrigerants etc.

c) Sterilizing drinking water.
Q.20 Give reasons for the following? 
(a) Nitrogen does not show catenation. 
(b) PCl₅ exists but NCl₅ does not. 

(c) The stability of Hydrides follows the order- 
NH₃>PH₃>Ar₅H₃>SbH₃

(d) PH3 is a weaker base than NH₃ . 
(e) Molecular nitrogen is chemically inert.
Ans. (a) Nitrogen being small in size has high electron density. Due to strong inter electronic repulsions, N-N single bond is weak & nitrogen does not undergo catenation.
(b) Due to absence of d-orbitals, nitrogen cannot expand its oxidation state to +5 and NCl5 does not exist whereas in P due to presence of empty 3d orbital +5 oxidation state is attained.
(c) As we move down the group 15, atomic radii increases making the bond of element with Hydrogen weaker this decreases the stability of hydrides of heavier elements. Therefore the order of stability is.
NH₃>PH₃>AsH₃>SbH₃
(d) As Phosphorous atom is larger than N- atom, the lone pair of electrons is distributed over a large surface area of P-atom than N-atom. Therefore the tendency of P to donate the lone pair of electrons is less. 

(e) Molecular nitrogen (N₂) is inert because N N bond energy is very high due to small size of N- atom and presence of multiple bond.
Q.21 Complete and balance-
(a) NH₄Cl(aq) + NaNO₂(aq)→
(b) Ba(N₃)₂→
(c) 3Mg + N₂→
(d) N₂ + O₂
(e) 2NaN₃→
Ans. (a) NH₄Cl(aq) + NaNO₂(aq) → N₂(g) + 2H₂O(l) NaCl(aq)
(b) Ba(N₃)₂ → Ba + 3N₂
(c) 3Mg + N₂  Mg₃N₂
(d) N₂ + O₂   2NO(g)
(e) 2NaN₃  → 2Na+3N₂
Q.22 What are the two isotopes of nitrogen? Write two uses of dinitrogen.
Ans. The two isotopes of nitrogen are ¹⁴N and ¹⁵N
Q.23 Write the characteristics of pure Ozone?
Ans. Pure ozone is pale blue gas, dark blue liquid and violet black solid
Q.24 At what concentration ozone is harmful?
Ans. If the concentration of ozone increases above 100 ppm, breathing becomes uncomfortable resulting in headache and nausea and it becomes harmful.  
Q.25 Explain the oxidizing action of ozone? 
Ans. Ozone has a very strong tendency to liberate nascent oxygen according to the equation:- O3 → O2+O
Therefore it acts as a strong oxidizing agent.
Q.26 How is ozone estimated quantitatively?
Ans. When ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (PH = 9.2) iodine is liberated. This iodine can be titrated against a standard solution of sodium thiosulphate to estimate the amount of ozone.  
2I-(aq)+H₂O + O₃ → 2OH^- (aq) + I₂(s) + O₂(g)
Q.27 Complete and balance:-
(i) PbS(s) + 4O₃(g)→
(ii) NO(g) + O₃(g) →
(iii) 4FeS₂(s) + 11O₂(g)→
(iv) 2NaOH + SO₂ →
(v) 2Fe³⁺ + SO₂ +2H₂O →
(vi) Cu + 2H₂SO₄(conc) →
Ans. (i) PbS(s) + 4O₃(g) → PbSO₄(s) +4O₂(g)
(ii) NO(g) + O₃(g) → NO₂(g) + O₂(g)
(iii) 4FeS₂(s) + 11O₂(g) → Na₂SO₃ + H₂O
(iv) 2NaOH + SO₂ → Na₂SO₃ + H₂O
(v) 2Fe³⁺ + SO2 + 2H₂ → O2Fe²⁺ + SO^2-₄ + 4H⁺
(vi) Cu + 2H₂SO₄(conc) → CuSO₄ + SO₂ + 2H₂O
Q.28 Give a test to detect the presence of SO₂ gas?
Ans. When SO₂ (g) is passed through a violet coloured acidified potassium permanganate solution, it gets decolourised.
5SO₂ + 2MnO^-₄ + 2H₂O → 5SO^2-₄ + 4H⁺ + 2Mn²⁺
Q.29 Which aerosols and oxides deplete ozone layer

Ans. Freon used in aerosol sprays deplete the ozone layer. Nitrogen oxides emitted from the exhaust systems of supersonic jet aeroplanes deplete the ozone layer.
Q.30 Write the members of 16 Group. 
Ans. The elements of group 16 are: Oxygen(O), Sulphur (S), Selenium (Se), Tellurium (Te) and Polonium (Po)
Q.31 Give the general electronic configuration of 16 Group.
Ans. The general electronic configuration of 16 group is ns²np⁴ .
Q.32 Discuss the geometry of SF₄.
Ans. In SF₄ the hybridisation of sulphur is sp³d. The structure is trigonal bi-pyramidal in which one of the equatorial positions is occupied by a one pair of electrons. The geometry is called see-saw geometry.
Q.33 Enlist some uses of dioxygen.
Ans. Oxygen is used-
(a) In oxyacetylene welding.
(b) In manufacture of steel.
(c) In hospitals and mountaineering as oxygen cylinders.
 Q.34 Write different isotopes of oxygen.
Ans. The isotopes of oxygen are  ¹⁶O, ¹⁷O, ¹⁸O.
Q.35 Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe
Ans. Pt is a noble metal and does not react very easily. All other elements, Zn, Ti, Fe, are quite reactive. Hence, oxygen does not react with platinum (Pt) directly.  
Q.36 Why is ICl more reactive than I₂ ?
Ans. ICl is more reactive than I₂ because I-Cl bond in ICl is weaker than I-I bond in I₂ .  
Q.37 Balance the following equation: 
XeF₆ + H₂O → XeO₂F₂ + HF 
Ans. Balanced equation: XeF₆ + H₂O → XeO₂F₂ + 4HF
Q.38 How is ammonia manufactured industrially? 
Ans. Ammonia is prepared on a large-scale by the Haber's process.
N₂(g) + 3H₂(g) → 2NH₃(g)
Q.39 Why does R₃P = O exist but R₃N = O does not (R = alkyl group)? 
Ans. N(unlike P) lacks the d-orbital. This restricts nitrogen to expand its coordination number beyond four. Hence,R₃N = O  does not exist.
Q.40 . Explain why fluorine forms only one oxoacid, HOF.
Ans. Fluorine forms only one oxoacid i.e., HOF because of its high electronegativity and small size.
Q.41 Which one of the following does not exist?  
(i) XeOF₄ (ii) NeF₂ (iii) XeF₂ (iv) XeF₆
Ans. NeF₂ does not exist.

Q.42 Complete and Balance-
a) P₄ + 8SICl₂ →
b) 3CH₃CooH + PCl₃ →
c) P₄ + 10SO₂Cl₂ →
d) POCl₃ + 3H₂O →
e) Sn + PCl₅ →
f) 4AgNO₃ + 2H₂O +H₃PO₂ →
Ans. Complete and Balance
a) P₄ + 8SOCl₂ → 4PCl₃ + 4SO₂ + 2S₂Cl₂
b) 3CH₃COOH + PCI₃ → 3CH₃COCI + H₃PO₃
c) P₄ + 10SO₂CI₂ → 4PCI₅ + 10SO₂

d) POCI₃ + 3H₂O → H₃PO₄ + 3HCI
e) Sn + PCI₅ → SnCl₄ + 2PCI₃

f) 4AgNO₃ + 2H₂O + H₃PO₂ → 4Ag + 4HNO₃ + H₃PO₄
Q.43 All five bonds in PCI₃ are not equal. Give an equation in support of this statement.
Ans. When heated, PCI₅ loses a chlorine molecule this shows that two P- CL bonds are weaker and hence longer than others.  

Q.44 Explain the chemistry behind brown ring test for detection of nitrate ions.
Ans. The brown ring test for nitrate ions depends on the ability of Fe²⁺ to reduce nitrates to nitric oxide, which reacts with Fe²⁺ to form a brown coloured complex.
NO^-₃ +3Fe²⁺ + 4H⁺ → NO + 3Fe³⁺ + 2H₂O
[Fe(H₂O)₆]²⁺ + NO → [Fe(H₂O)₅ NO]²⁺+ H₂O
Q.45 Write the various steps for preparation of sulphuric acid by contact process?
Ans. Contact process for sulphuric acid:-  
Step 1: Burning of sulphur in air to give SO2
S+O₂SO₂
Step 2: Conversion of SO₂ to SO₃ by reacting it with oxygen in presence of V₂O₅ .

Step 3: Absorption of SO₃ in H₂SO₄ to give of oleum (H₂S₂O₇)
SO₃ + H₂SO₄ → H₂S₂O₇(oleum)
Step 4: Dilution of oleum with water to get H₂SO₄ of desired concentration

H₂S₂O₇ + H₂O → 2H₂SO₄
Q.46 Name different sulphates formed by sulphuric acid?
 Ans. The two type of sulphates are –
(i) Normal sulphate eg. Na₂SO₄ , CuSO₄
(ii) acid sulphate eg.NaHSO₄.
Q.47.Why are pentahalides more covalent than trihalides?  
Ans. In pentahalides, the oxidation state is +5 and in trihalides, the oxidation state is +3. Since the metal ion with a high charge has more polarizing power, pentahalides are more covalent than trihalides.
Q.48 Why is BiH₃ the strongest reducing agent amongst all the hydrides of Group 15 elements?  

Ans. As we move down a group, the atomic size increases and the stability of the hydrides of group 15 elements decreases. Since the stability of hydrides decreases on moving from NH₃ to BiH₃ , the reducing character of the hydrides increases on moving from NH₃ to BiH₃ .
Q.49 Why is N₂ less reactive at room temperature?
Ans. The two N atoms in N₂ are bonded to each other by very strong triple covalent bonds. The bond dissociation energy of this bond is very high. As a result, N₂ is less reactive at room temperature.  
Q.50. How does ammonia react with a solution of Cu²⁺?
Ans. NH₃ acts as a Lewis base. It donates its electron pair and forms a linkage with metal ion.

Q.51. What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂?
Ans. White phosphorous dissolves in boiling NaOH solution (in a CO₂ atmosphere) to give phosphine,PH₃ .  

Q.52 Write a balanced equation for the hydrolytic reaction of PCI₅ in heavy water.
Ans. All the bonds that are present in PCI₅ are not similar. It has three equatorial and two axial bonds. The equatorial bonds are stronger are stronger than the axial ones. Therefore, when PCI₅ is heated strongly, it decomposes to form PCI₃ .
Q.53 What happens when PCI₅ is heated ?
Ans. PCI₅ + D₂O → POCI₃ + 2DCI₂
POCI₃ + 3D₂O → D₃PO₄ + 3DCI
Therefore, the net reaction can be written as
PCI₅ + 4D₂O → D₃PO₄ + 5DCI
Q.54 List the important sources of sulphur.
Ans. Sulphur mainly exists in combined form in the earth's crust primarily as sulphates [gypsum (CaSO₄. 2H₂O ), Epsom salt (MgSO₄. 7H₂O), baryte (BaSO₄)] and sulphides [galena (PbS), zinc blends (ZnS), copper pyrites (CuFeS₂)].
Q.55 Why is H₂O a liquid and H₂S a gas?  
Ans. H₂O has oxygen as the central atom. Oxygen has smaller size and higher electronegativity as compared to sulphur. Therefore, there is extensive hydrogen bonding in H₂O , which is absent in H₂S . Molecules of H₂S are held together only by weak Van der Waal's forces of attraction.
Hence, H₂O exists as a liquid while H₂S as a gas.
Q.56 Complete the following reactions:
(i) C₂H₄ + O₂→
(ii) 4Al + 3O₂ →
Ans. (i)
(ii)

Therefore, ozone acts as a powerful oxidising agent.
Q.57 Why does O₃ act as a powerful oxidizing agent? 
Ans. Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen. Nascent oxygen, being a free radical, is very reactive.
 Therefore, ozone acts as a powerful oxidizing agent.

Q.58 What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt?
Ans. SO₂ acts as a reducing agent when passed through an aqueous solution containing Fe(III) salt. It reduces Fe(III) to Fe(II) i.e., ferric ions to ferrous ions.
2Fe₂₊ + SO₂ + 2H₂O → 2Fe²⁺ + SO^2-₄ + 4H⁺ 
Q.59 Give two examples to show the anomalous behaviour of fluorine.
Ans. Anomalous behaviour of fluorine
 (i) It forms only one oxoacid as compared to other halogens that form a number of oxoacids.
(ii) Ionisation enthalpy, electronegativity, and electrode potential of fluorine are much higher than expected.
Q.60 Sea is the greatest source of some halogens. Comment.
Ans. Sea water contains chlorides, bromides, and iodides of Na, K, Mg, and Ca. However, it primarily contains NaCl. The deposits of dried up sea beds contain sodium chloride and carnallite, KCL.MGCl₂.6H₂O . Marine life also contains iodine in their systems. For example, sea weeds contain upto 0.5% iodine as sodium iodide. Thus, sea is the greatest source of halogens.
Q.61 Give the reason for bleaching action of C12.
Ans. When chlorine reacts with water, it produces nascent oxygen. This nascent oxygen then combines with the coloured substances present in the organic matter to oxide them into colourless substances.
Cl₂ + H₂O → 2HCl+[O]
Coloured substances + [O]→Oxidized colourless substance
Q.62 Name two poisonous gases which can be prepared from chlorine gas.
Ans. Two poisonous gases that can be prepared from chlorine gas are
(i) Phosgene (COCI₂)
(ii) Mustard gas (CICH₂CH₂SCH₂CH₂CI)
Q.63 Why has it been difficult to study the chemistry of radon?
Ans. It is difficult to study the chemistry of radon because it is a radioactive substance having a half-life of only 3.82 days. Also, compounds of radon such as RnF₂ have not been isolated. They have only been identified.
Q.64 Why is helium used in diving apparatus?  
Ans. Air contains a large amount of nitrogen and the solubility of gases in liquids increases with increase in pressure. When sea divers dive deep into the sea, large amount of nitrogen dissolves in their blood. When they come back to the surface, solubility of nitrogen decreases and it separates from the blood and forms small air bubbles. This leads to a dangerous medical condition called bends. Therefore, air in oxygen cylinders used for diving is diluted with helium gas. This is done as He is sparingly less soluble in blood.
Q.65 Why does the reactivity of nitrogen differ from phosphorus?
Ans. Nitrogen is chemically less reactive. This is because of the high stability of its molecule, N₂. In, N₂  the two nitrogen atoms form a triple bond. This triple bond has very high bond strength, which is very difficult to break. It is because of nitrogen's small size that it is able to form pπ-pπ bonds with itself. This property is not exhibited by atoms such as phosphorus. Thus, phosphorus is more reactive than nitrogen.
Q.66 Why does NX₃ form hydrogen bond but PH₃ does not?
Ans. Nitrogen is highly electronegative as compared to phosphorus. This causes a greater attraction of electrons towards nitrogen in NX₃ than towards phosphorus in PH₃ . Hence, the extent of hydrogen bonding in PH₃ is very less as compared to NX₃ .  
Q.67 How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.

Ans. An aqueous solution of ammonium chloride is treated with sodium nitrite.  

NO and HNO3are produced in small amounts. These are impurities that can be removed on passing nitrogen gas through aqueous sulphuric acid, containing potassium dichromate.
Q.68 Why does nitrogen show catenation properties less than phosphorus?
Ans. Catenation is much more common in phosphorous compounds than in nitrogen compounds. This is because of the relative weakness of the N-N single bond as compared to the P-P single bond. Since nitrogen atom is smaller, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening the N-N single bond.
Q.69 Give the disproportionation reaction of H₃PO₃.
Ans. On heating, orthophosphorus acid H₃PO₃ disproportionates to give orthophosphoric acid H₃PO₃ and phosphine (PH₃) . The oxidation states of P in various species involved in the reaction are mentioned below.  

Q.70 Which aerosols deplete ozone?
Ans. Freons or chlorofluorocarbons (CFCs) are aerosols that accelerate the depletion of ozone. In the presence of ultraviolet radiations, molecules of CFCs break down to form chlorine-free radicals that combine with ozone to form oxygen.
Q.71 Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.  
Ans. Both chlorine and oxygen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding. This is because in comparison to chlorine, oxygen has a smaller size and as a result, a higher electron density per unit volume.
Q.72 Write two uses of CIO₂.
Ans. Uses of CIO₂ :
(i) It is used for purifying water.

(ii) It is used as a bleaching agent.  
Q.73 Why are halogens coloured?
Ans. Almost all halogens are coloured. This is because halogens absorb radiations in the visible region. This results in the excitation of valence electrons to a higher energy region. Since the amount of energy required for excitation differs for each halogen, each halogen displays a different colour.
Q.74 Write the reactions of F₂ and Cl₂ with water.
Ans.(i)
  
(ii) 

Q.75 How can you prepare Cl₂ from HCl and HCl from Cl₂ ? Write reactions only.
Ans.(i) Cl₂can be prepared from HCl by Deacon's process.
 

(ii) HCl can be prepared from Cl2 on treating it with water

Q.76 What inspired N. Bartlett for carrying out reaction between Xe and PtF_{6 .}
Ans. Neil Bartlett initially carried out a reaction between oxygen and PtF₆ . This resulted in the formation of a red compound, O⁺₂[PtF₅]^-.
Later, he realized that the first ionization energy of oxygen (1175 kJ/mol) and Xe (1170 kJ/mol) is almost the same. Thus, he tried to prepare a compound with Xe and PtF₆ . He was successful and a red-coloured compound, was formed.

Q.77 Write balanced equations for the following: 
(i) NaCl is heated with sulphuric acid in the presence of MnO₂ . 
(ii) Chlorine gas is passed into a solution of NaI in water. 
Ans. (i)
(ii)
Q.78 With what neutral molecule is CIO^- isoelectronic? Is that molecule a Lewis base? 
Ans. CIO^- is isoelectronic to ClF. Also, both species contain 26 electrons in all as shown. Total electrons CIO^- = 17 + 8 + 1 = 26
In ClF = 17 + 9 = 26
ClF acts like a Lewis base as it accepts electrons from F to form CIf₃.
Q.79 How XeO₃ are and XeOF₄ prepared? 
Ans. (i) XeO₃  can be prepared in two ways as shown.
 

(ii) XeOF₄ can be prepared using XeF₆ .

Q.80 Why do noble gases have comparatively large atomic sizes?
Ans. Noble gases do not form molecules. In case of noble gases, the atomic radii corresponds to van der Waal's radii. On the other hand, the atomic radii of other elements correspond to their covalent radii. By definition, van der Waal's radii are larger than covalent radii. It is for this reason that noble gases are very large in size as compared to other atoms belonging to the same period.