Question

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

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

Answer 1

Answer: (D)

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

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

Oxidation state of $Cl$ in $C{l}_2$ is 0

Oxidation state of $Cl$ in $NaCl$ is -1 ($1+x=0$)

Oxidation state of $Cl$ in $NaCl{O}_3$ is +5 ($1+x+3\times (-2)=0$)

Therefore it changes from Zero to $-1and+5$.

Hence, option D is correct.