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.