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

Options

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

Answer

B

Explanation

`3overset(0)(Cl_(2)) + 6 NaOH (conc.) overset(Delta)rarr 5 Naoverset(-1)(Cl) + Naoverset(+5)(ClO_(3))+ 3H_(2)O`
Here, 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)`.