The oxidation number of Nitrogen (N) in is: — Redox Reactions and Volumetric Analysis Chemistry Question
Question
The oxidation number of Nitrogen (N) in $(NH_4)_2SO_4$ is:
Answer: B
💡 Solution & Explanation
The compound dissociates into $NH_4^+$ and $SO_4^{2-}$ ions. In the ammonium ion $NH_4^+$ , let the oxidation state of N be x. Since H is +1, $x + 4(+1) = +1 \Rightarrow x = -3$ .
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