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When $Cl_2$ gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from
A
Zero to +1 and zero to −3
B
Zero to +1 and zero to −5
C
Zero to −1 and zero to +5
D
Zero to −1 and zero to +3
Detailed Solution
$3Cl_2 + 6NaOH$ (hot, conc.) $\rightarrow 5NaCl + NaClO_3 + 3H_2O$
In $Cl_2$ the oxidation number of Cl is 0.
In NaCl it is −1, and in $ClO_3^-$ it is +5.
So chlorine changes from zero to −1 and from zero to +5 (disproportionation).
In $Cl_2$ the oxidation number of Cl is 0.
In NaCl it is −1, and in $ClO_3^-$ it is +5.
So chlorine changes from zero to −1 and from zero to +5 (disproportionation).