Question

When $C{l}_2$ gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from:

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> AIPMT 2012

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

Answer 1

The correct option is C Zero to -1 and zero to +5

The reaction of chlorine gas with hot and concentrated sodium hydroxide solution is:

$3C{l}_2+6NaOH⟶NaCl{O}_3+5NaCl+3H_{2}O$

This is an example of disproportionation reaction.

Oxidation number of $Cl$ is 0 in $C{l}_2$, −1 in $NaCl$ and +5 in $NaCl{O}_3$
​
So, the oxidation number of chlorine changes from Zero to -1 and Zero to +5.

Hence, option (d) is the correct answer.