The oxidation state of $C l$ in $C l_{2} O_{7}$ is:

+ 7

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+ 5

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+ 3

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- 7

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Solution

The correct option is A $+ 7$
We know $C l$ can show different oxidation states from $- 1$ to $+ 7$ .
In $C l_{2} O_{7}$ let oxidation number of $C l$ be $x$ .
So $2 x + 7 (- 2) = 0 \Rightarrow 2 x = 14 \Rightarrow x = + 7$

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