When `Cl_(2)` reacts with hot and concentrated soldium, 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)+underset("(hot & conc.)")(6NaOH)rarr Naoverset(-1)Cl+Naoverset(+5)ClO_(3)+3H_(2)O`
In this case oxidation state of chlorine changes from 0 to `-1 and +5` and hence is an example of disproportionation reaction.