(a) In NCl₃, Cl has vacant d-orbitals to accept the lone' pair of electrons donated by O-atom of  \({ H }_{ 2 }O\) molecule but in NF₃, F does not have d-orbitals. Thus, NCl₃ undergoes hydrolysis but NF₃ does not\(N{ Cl }_{ 3 }+3{ H }_{ 2 }O\longrightarrow { NH }_{ 3 }+3HOCl\ ;\ { NF }_{ 3 }+{ H }_{ 2 }O\longrightarrow No\ reaction\)
(b) Compounds in which one of elements exists in three different oxidation states can show disproportionation reactions. For exampl \(3H\overset { +3 }{ N{ O }_{ 2 } } \longrightarrow H{ \overset { +5 }{ NO } }_{ 3 }+2\overset { +2 }{ NO } +{ H }_{ 2 }O\ ;\ 4{ H }_{ 3 }\overset { +3 }{ { PO }_{ 3 } } \xrightarrow { \Delta } 3{ H }_{ 3 }\overset { +5 }{ { PO }_{ 4 } } +\overset { -3 }{ { PH }_{ 3 } } \)

Since the element forms an oxide with formula, R₂O_5, it must be an element of group 15. Now group 15 has two metalloids, As and Sb. Since the acidic character of pentoxides of group 15 elements decreases while the basic character increases down the group, therefore, \({ As }_{ 2 }{ O }_{ 5 }\) is acidic while  \({ Sb }_{ 2 }{ O }_{ 5 }\) is amphoteric. Thus, the metalloid of group 15 elements which forms an acidic pentoxide (\({ As }_{ 2 }{ O }_{ 5 }\)) is As.

(a)
  \(4N{ H }_{ 3 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \xrightarrow [ 1100K ]{ Pt } 4NO\left( g \right) +6{ H }_{ 2 }O\left( l \right) ;\ 2NO\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2{ NO }_{ 2 }\left( g \right) \)
\(\\ 3{ NO }_{ 2 }\left( g \right) +{ H }_{ 2 }O\left( l \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right) +NO\left( g \right) ;\ { NH }_{ 3 }\left( g \right) +HN{ O }_{ 3 }\left( aq \right) \longrightarrow { NH }_{ 4 }{ NO }_{ 3 }\left( aq \right) \)
\(\\ { NH }_{ 4 }{ NO }_{ 3 }\left( aq \right) \xrightarrow [ at\ room\quad temperature ]{ Vacuum\ evaporation } { NH }_{ 4 }{ NO }_{ 3 }\left( s \right) ;\ { NH }_{ 4 }{ NO }_{ 3 }\left( s \right) \xrightarrow { 523K } { N }_{ 2 }O\left( g \right) +2{ H }_{ 2 }O\left( l \right) \)
(b) Animals need large amount of energy to move around and maintain the body temperature. Therefore, to obtain the required energy, it is much easier for them to break weaker double bond (493-4 kJ \({mol }^{ -1 }\) ) of  \({ O }_{ 2 }\) than breaking the much stronger triple bond (941·4 kJ \({ mol}^{ -1 }\) ) of \({ N }_{ 2 }.\) 

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.

Due the presence of a lone pair of electrons on N and P, both \({ NH }_{ 3 }\quad and\quad { PH }_{ 3 }\) act as Lewis bases and accept a proton to form an additional N-H and P-H bonds respectively

However, due to smaller size of N over P, N-H bond thus formed is much stronger than the P-H bond. Therefore, \({ NH }_{ 3 }\) has higher proton affinity than \({ PH }_{ 3 }\) In other words, \({ NH }_{ 3 }\) is more basic than \({ PH }_{ 3 }\) .