(i) High pressure (2 bar),
(ii) 720 K temperature,
(iii) V₂O₅ as catalyst.

In presence of moisture, \({ Cl }_{ 2 }\) releases nascent oxygen which converts coloured material to colourless material. Thus, bleaching by \({ Cl }_{ 2 }\) is due to oxidation and hence permanent.
\({ Cl }_{ 2 }+{ H }_{ 2 }O\ \longrightarrow \ 2HCl+\left[ O \right] \)
\(Coloured\ material+\left[ O \right] \longrightarrow Colourless\ material\)
In contrast, in presence of moisture, \({ SO }_{ 2 }\) liberates nascent hydrogen which reduces coloured material to colourless material. Thus, bleaching with \({ SO }_{ 2 }\) is due to reduction. When colourless material is exposed to air, it gets oxidised and the colour returns. Thus, bleaching by \({ SO }_{ 2 }\) is temporary.

Its structure is  Since it contains only two ionizable H-atoms which are present as OH groups, it behaves as a dibasic acid.

Formation of ozone has positive value of \(\triangle\)_f H^o, that is why it is less thermodynamically stable than O₂ .

(i) OF₂ and CIF are neutral molecules isoelectronic with CIO^-.
(ii) Bi(V) is stronger oxidising agent due to inert pair effect as Bi(III) is more stable as compared to Sb(III).

(a) (i)

(ii) 
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\('A'\quad is\quad { S }_{ 8 }.\quad 'B'is{ SO }_{ 2 }(g).\)
\({ S }_{ 8 }+{ 8O }_{ 2 }\overset { heat }{ \rightarrow } { 8SO }_{ 2 }(g)\)
'B' decolorizes \(KMnS{ O }_{ 4 }\) 
\(2KMn{ O }_{ 4 }+5S{ O }_{ 2 }+{ 2H }_{ 2 }O\longrightarrow { 2H }_{ 2 }{ SO }_{ 4 }+2Mn{ SO }_{ 4 }+{ K }_{ 2 }{ SO }_{ 4 }\)
'B' turns lime water milky due to formation of  
\(Ca{ (OH) }_{ 2 }(aq)+{ SO }_{ 2 }(g)\longrightarrow CaSO_{ 3 }(s)+{ H }_{ 2 }O(l)\)
'B' is obtained by roasting of sulphideores
\(2ZnS(s)+{ 3O }_{ 2 }\longrightarrow 2ZnO(s)+{ 2SO }_{ 2 }(g)\)
 'B' reduces in aqueous solution.
\(2{ Fe }^{ 3+ }+{ SO }_{ 2 }+{ 2H }_{ 2 }O\longrightarrow { Fe }^{ 2+ }+{ SO }_{ 4 }^{ 2- }+{ 4H }^{ + }\)

\({ P }_{ 4 }={ 6CI }_{ 2 }\longrightarrow { 4PCI }_{ 3 }\)
\(\left[ { PCI }_{ 3 }+{ 6H }_{ 2 }O\longrightarrow { H }_{ 3 }{ PO }_{ 3 }+3HCI \right] \times 4\)
\({ P }_{ 4 }+{ 6CI }_{ 2 }+{ 12H }_{ 2 }O\longrightarrow { 4H }_{ 3 }{ PO }_{ 3 }+{ 12HCI }\)
1 mole of white phosphorus produces 12 moles of HCI.
\(\frac { 62 }{ 124 }\)mole of white phosphorus produces \(12\times \frac { 62 }{ 124 } =6\)moles of HCI.
Mass of 6 moles of HCI= 6 X 36.5
= 219.0 g of HCI.

They have same general electronic configuration ns²np^6.
They show +2, +4, +6, -2 oxidation states except oxygen which cannot show higher positive oxidation states.
All of them form hydrides having formula H₂X where 'X' is group 16 element.  Acidic character and reducing power of hydrides increases down the group due to decrease in bond dissociation energy.

(a)  (i) \({ { [PCI }_{ 4 }] }^{ + }\quad { { [PCI }_{ 6 }] }^{ - }\)in solid state.
(ii)

(b)  (i) It is because  \({ NH }_{ 3 }\) molecules are associated with intermolecular H-bonding.
(ii) Helium is small in size and has very high ionization energy, least polarising power therefore no chemical compound of helium is known.
(iii) \({ Bi }^{ 3+ }\) more stable than  \({ Bi }^{ 5+ }\) due to inert pair effect, therefore,   \({ Bi }^{ 5+ }\)gains two electrons to form \({ Bi }^{ 3+ }\) 
\({ Bi }^{ 5+ }+{ 2e }^{ - }\longrightarrow { Bi }^{ 3+ }\)

(a) (i) \(3Cu+{ 8HNO }_{ 3 }(dil.)\longrightarrow { { 3 }Cu(NO_{ 3 }) }_{ 2 }+2NO+{ 4H }_{ 2 }O\)
(ii) \({ XeF }_{ 4 }+{ O }_{ 2 }{ F }_{ 2 }\overset { { -130°C } }{ \longrightarrow } { XeF }_{ 6 }+{ O }_{ 2 }\)
(b) (i) The property of catenation depends upon the strength of the element-element bond. Since P-P bond strength (213 kJ \({ mol }^{ -1 }\) ) is much more than N-N bond strength (159 kJ \({ mol }^{ -1 }\)), phosphorus has marked catenation properties than nitrogen.
(ii) Due to small size and high electronegativity, oxygen atom forms \(p\pi -p\pi \)  double bond, 0= 0. The intermolecular forces in oxygen are weak van der Waal's forces and therefore, oxygen exists as a gas. On the other hand, sulphur does not form stable \(p\pi -p\pi \)  bonds and do not exists as \({ S }_{2}\) It is linked by single bonds and form polyatomic complex molecules having eight atoms per molecule \({ S }_{ 8 }\) and have puckered ring structure. Therefore, S atoms are strongly held together and it exists as a solid.
(iii) All the halogens are coloured. This is due to absorption of radiations in the visible region which results in the excitation of outer electrons to higher energy levels. By absorbing different quanta of radiations, they display different colours. The fluorine atom is the smallest and the force of attraction between the nucleus and the outer electrons is very large. As a result, it requires large excitation energy and absorbs violet light (high energy) and therefore, appears pale yellow. On the other hand, iodine needs very less excitation energy and absorbs yellow light of low energy. Thus, it appears dark violet. Similarly, we can explain the greenish yellow colour of chlorine and reddish brown colour of iodine.

Neither N nor F have d-orbits to accept electrons donated by H₂O for hydrolysis.