Q.

Find the sum of oxidation number of nitrogen in (NH4)2SO4. N2H4 and N2O4.

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answer is -1.

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Detailed Solution

Let O.N. of N be x In NH42SO4;x=-3 In N2H4;x=-2 In N22O44;x=+4Sum of oxidation numbers of N = – 3 + (– 2) + 4 = – 1
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