Question

When $I_2$ gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of iodine changes from

• 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 1

Answer: (B)

$3\stackrel{0}{I_2}+\underset{\text{(hot and cone.)}}{6NaOH}\to 5Na\stackrel{-1}{I}+Na\stackrel{+5}{I}{O}_3+3H_{2}O$
This is an example of disproportionation reaction