A) 1
B) 2
C) 3
D) 4
Correct Answer: C
Solution :
\[\frac{5.1}{17}=0.3\]mole of \[N{{H}_{3}}\] No of electrons in \[N{{H}_{3}}={{e}^{-}}\operatorname{s}\,in\,N+3\times {{e}^{-}}s\]in H =7+3=10 No. of electron = \[10\times 0.3{{N}_{A}}=3{{N}_{A}}\]
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