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)

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