(A)              

In nitride ion $\left({\mathrm{N}}_3^{-}\right)$, nitrogen is in its minimum possible oxidation state (-3) hence it cannot be further reduced. Thus ${\mathrm{N}}_3^{-}$ cannot act as an oxidizing agent.

(B)        

$\stackrel{0}{\mathrm{C}}{\mathrm{l}}_2\left(\mathrm{g}\right)+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+\stackrel{+1}{\mathrm{Cl}}{\mathrm{O}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

a = 1, b = 2, c = 1, d = 1

(D)         

In acidic medium the half reaction is balanced as follows:

(A) Balance atoms undergoing reduction i.e. chromium:

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\left(\mathrm{aq}\right)\to 2{\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)$

(B) Balance oxygen using ${\mathrm{H}}_2\mathrm{O}$

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\left(\mathrm{aq}\right)\to 2{\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+7{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

(C) Balance hydrogen using ${\mathrm{H}}^{+}$

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\left(\mathrm{aq}\right)+14{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to 2{\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+7{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

(D) Balance charge using electrons:

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\left(\mathrm{aq}\right)+14{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+6{\mathrm{e}}^{-}\to 2{\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+7{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

On comparing with given reaction we get,

$\mathrm{X}=14, \mathrm{Y}=6, \mathrm{Z}=7, \mathrm{A}={\mathrm{Cr}}^{3+}$

(D) 

$2\stackrel{+2}{{\mathrm{S}}_2}{\mathrm{O}}_3^{2-}+{\mathrm{I}}_2\to \stackrel{+2.5}{{\mathrm{S}}_4}{\mathrm{O}}_6^{2-}+2{\mathrm{I}}^{-}$

$\stackrel{+2}{{\mathrm{S}}_2}{\mathrm{O}}_3^{2-}+5{\mathrm{Br}}_2+5{\mathrm{H}}_2\mathrm{O}\to 2\stackrel{+6}{\mathrm{S}}{\mathrm{O}}_4^{2-}+4{\mathrm{Br}}^{-}+10{\mathrm{H}}^{+}$

${\mathrm{Br}}_2$ is oxidizing S of thiosulphate to a higher oxidation state than ${\mathrm{I}}_2.$ This illustrates that ${\mathrm{Br}}_2$ is a stronger oxidizing agent than ${\mathrm{I}}_2$.