A) 8
B) 5
C) 3
D) 7
Correct Answer: D
Solution :
[d] \[\overset{+5}{\mathop{N}}\,O_{3}^{-}\xrightarrow{{}}{{\overset{-2}{\mathop{N}}\,}_{2}}{{H}_{4}}\] So, for reduction of 1 mole of \[NO_{3}^{-}\] number of electrons required is 7.
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