Question

oxidation number of Ru and Os in RuO4 and OsO4.a) + IV b)+VII c) + VI d)+VIII​

Answer 1

Answer:

we know that the sun is oxidation number of all elements in a compound is zero and that of O is always -2 excel in case of super oxides, peroxides and OF2

so let the oxidation number of Ru be x then in RuO4,

$x + 4( - 2) = 0$

or

$x - 8 = 0$

or

$x = + 8$

similarly for OsO4, we get that the oxidation number of Os is +8

so the answer will be d) +VIII

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