When `CI_(2)` gas reacts with hot and concentrated sodium hydroxide solution ,the oxidation number of chlorine changes from:

Options

A. Zero to + and zero to +5
B. Zero to -1 and zero to +5
C. Zero to -1 and zero to +3
D. Zero two + and zero to -3

Answer

A

Explanation

`3overset(0)(Cl_2)+6NaOH(conc.) overset(Delta)tooverset(-1)(5NaCl) + overset(+5)(NaClO_3)+3H_2O`
`therefore` O.N of Cl changes from 0 in `Cl_2` to -1 in NaCl and from 0 in `Cl_2` to +5 in `NaClO_3`.