Violet colour of iodine is decolorized by :

Iodine has a large size and easily loose its valence electrons. Therefore, iodin acts as a strong oxidizing agent. In the first reaction, NaOH reacts with iodine to form NaI and NaIO₃ which causes the decolorization of iodine.

$3{\mathrm{I}}_2 + 6\mathrm{NaOH} \to 5\mathrm{NaI} +{\mathrm{NaIO}}_3 + 3{\mathrm{H}}_2\mathrm{O}$

In the second reaction, Na₂S₂O₃ forms sodium iodide when reacts with iodine and cause the decolorization of iodine.

${\mathrm{I}}_2 + 2{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3 \to 2\mathrm{NaI} + {\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6$

 In the third rection, the nitric acid form HIO₃ with iodine and causes decolorization.

$10{\mathrm{HNO}}_3 + {\mathrm{I}}_2 \to 2{\mathrm{HIO}}_3 + 10{\mathrm{NO}}_2 + 4 {\mathrm{H}}_2\mathrm{O}$

 Therefore, all given reactions cause the decolorization of iodine.