Question

Choose the correct option regarding change in oxidation number of the underlined atom in the following conversion.

$2NaOH+\underset{–}{B{r}_2}\to NaBr+NaBrO+H_{2}O$

• A. Decreases to $-1$ w.r.t $NaBr$ and increases to $-3$ w.r.t $NaBrO$
• B. Decreases to $-1$ w.r.t $NaBr$ and increases to $+2$ w.r.t $NaBrO$
• C. Decreases to $-1$ w.r.t $NaBr$ and increases to $+1$ w.r.t $NaBrO$
• D. Decreases to $-2$ w.r.t $NaBr$ and increases to $+1$ w.r.t $NaBrO$

Answer 1

The correct option is C Decreases to $-1$ w.r.t $NaBr$ and increases to $+1$ w.r.t $NaBrO$

The oxidation number of bromine in $B{r}_2,NaBr$ and $NaBrO$ is $0,-1$ and $+1$ respectively.
Hence, during the reaction, the oxidation number of bromine in $B{r}_2$ decreases to $-1$ w.r.t $NaBr$ and increases to $+1$ w.r.t $NaBrO$.
Thus, the correct option will be C.