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The oxidation number of N and Cl in `NOClO_(4)` respectively areA. `+2 and +7`B. `+3 and +7`C. `-3 and +5`D. `+2 and -7` |
Answer» Correct Answer - B `NOClO_(4)-=(NO^(+))(ClO_(4)^(-))` Let O.N. of N in `NO^(+)` is x `:. x + (-2)=+1` `x=3` Let O.N of Cl in `ClO_(4)^(-)` is x `:. x + 4 (-2)=-1 or x = +7`. |
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