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 Na OH ( " conc")overset(Delta)to `
` overset( -1)( 5Na Cl ) + overset(+5)( NaCIO_3) + 3H_2 O`
Here, O.N. of CI changes from 0 in `Cl_2` to-1 in NaCl and form in `Cl_2` to + 5 in `NaClO_(3)`