A) Mg
B) Al
C) Fe
D) Cu
Correct Answer: A
Solution :
[a] \[Mg+2HN{{O}_{3}}(dil.)\to Mg{{(N{{O}_{3}})}_{2}}+{{H}_{2}}\] (Mg and Mn give \[{{\text{H}}_{2}}\] with dil \[HN{{O}_{3}}\])
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