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When $C l_{2}$ gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from

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zero to $+ 1$ and zero to $- 5$

zero to $- 1$ and zero to $+ 5$

zero to $- 1$ and zero to $+ 3$

zero to $+ 1$ and zero to $- 3$

answer is B.

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Detailed Solution

$3 \overset{0}{C} l_{2} + \underset{(hot and conc.)}{6 N a O H} \to 5 N a \overset{- 1}{C l} + N a \overset{+ 5}{C l} O_{3} + 3 H_{2} O$
This is an example of disproportionation reaction and oxidation state of chlorine change from 0 to –1 and +5.

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