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What is the oxidation number of $\mathrm{Br}$ in $\mathrm{KBrO}_4$ ?
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The correct answer is:
$+7$
Let the oxidation no. of $\mathrm{Br}$ be $x$.
$\begin{aligned} \text { In } \mathrm{KBrO}_4+1+x+4(-2) & =0,-7+x=0 \\ x & =+7\end{aligned}$
$\begin{aligned} \text { In } \mathrm{KBrO}_4+1+x+4(-2) & =0,-7+x=0 \\ x & =+7\end{aligned}$
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