Which of the following does not decolorize iodine?
A.\[N{a_2}S{O_3}\]
B\[N{a_2}{S_2}{O_3}\]
C.NaCl
D.NaOH

Hint: The colour of sold Iodine is almost a deep purple like colour. Even the vapours and solutions of iodine in solvents that are non – polar in nature, are purple in colour. On the other hand, aqueous solution of iodine has a yellowish – orange colour. This is because in aqueous solution, iodine forms a charge transfer complex.

Complete Step-by-Step Answer:
Before we move towards the solution of this question, let us understand some basic important concepts.
Iodine is also the least reactive of all the halogens and the most electropositive. Hence, the reactivity of iodine is lower than all the stable halogens. Even then, iodine reacts with a number of compounds and often causes the displacement of the key functional groups present in these compounds. Because of this, elemental iodine gets converted to being a functional group or an ionic or covalent constituent. Because of this, iodine undergoes decolorization.
The given compounds react with iodine as follows:
1.\[N{a_2}S{O_3}\] : Sodium sulphite reacts with iodine in an aqueous solution to form sodium sulphate along with hydrogen iodide. The chemical equation for this reaction can be given as –
 \[N{a_2}S{O_3} + {I_2} + {H_2}O \to N{a_2}S{O_4} + 2HI\]
Hence, there is a substitution involving iodine, and this causes the decolorization of iodine.
 2.\[N{a_2}{S_2}{O_3}\] : Sodium Thiosulphate reacts with iodine results in the formation of sodium tetrathionate along with sodium iodide. The chemical equation for this reaction can be given as –
 \[N{a_2}S{O_3} + {I_2} + {H_2}O \to N{a_2}S{O_4} + 2HI\]
Hence, there is a substitution involving iodine, and this causes the decolorization of iodine.
3.NaCl: The reactivity of chlorine is higher than iodine. Hence there is no reaction between sodium chloride and iodine.
4.NaOH: sodium hydroxide reacts with iodine to form sodium iodide along with sodium hypoiodite. The chemical equation for this reaction can be as –
 \[NaOH + {I_2} \to NaI + NaOI + {H_2}O\]
Hence, there is a substitution involving iodine, and this causes the decolorization of iodine.
Hence, we can observe that iodine does not react with NaCl. This causes iodine to not undergo decolorization.