Let x be the oxidation number of C.
∴ O.N. of C in cyanogen, (CN)2 = 2(x – 3) = 0 or x = +3
O.N. of C in cyanide ion, CN– = x – 3 = –1 or x = +2
O.N. of C in cyanate ion, CNO– = x – 3 – 2 = –1 or x = +4
The four informations about the reaction are :
(i) The reaction involves decomposition of cyanogen, (CN)2 in the alkaline medium to cyanide ion, CN– and cyanate ion, CNO–.
(ii) The O.N. of C decreases from +3 in (CN)2 to + 2 in CN– ion and increases from +3 in (CN)2 to +4 in CNO– ion. Thus, cyanogen is simultaneously reduced to cyanide ion and oxidised to cyanate ion.
(iii) It is an example of disproportionation redox reaction.
(iv) Cyanogen is a pseudohalogen while cyanide ion is a pseudohalide ion.