What is the oxidation number of $B r$ in the compound $R b B r O_{4}$ ?

- 1

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+ 7

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+ 1

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+ 4

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Solution

The correct option is A $+ 7$
$R b B r O_{4}$
Total charge $= 0.$
$R b$ is a $g r o u p$ $I$ element. So, its oxidation no. $= + 1$ .
Oxidation no. of oxygen $= - 2$ .
Let, Oxidation no. of $B r$ be $x$ .
So, $1 + x + 4 (- 2) = 0 1 + x - 8 = 0 x - 7 = 0 x = 7$
So, oxidation state of $B r o m i n e$ is $+ 7$ .

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