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  4. Oxidation number of N in (NH4)2SO4 is

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Q. Oxidation number of $N$ in $(NH_{4})_{2}SO_{4}$ is

Redox Reactions

Solution:

$(NH_{4})_{2}SO_{4} \rightleftharpoons 2NH^{+}_{4} +SO_{4}$
$\overset{*}{N}H^{+}_{4}$
$x+4=+1; x=11-4=-3$