A) \[N{{H}_{3}}>C{{H}_{3}}N{{H}_{2}}>{{C}_{6}}{{H}_{5}}N{{H}_{2}}\]
B) \[{{C}_{6}}{{H}_{5}}N{{H}_{2}}>N{{H}_{3}}>C{{H}_{3}}N{{H}_{2}}\]
C) \[C{{H}_{3}}N{{H}_{2}}>N{{H}_{3}}>{{C}_{6}}{{H}_{5}}N{{H}_{2}}\]
D) none of the above
Correct Answer: C
Solution :
(i) \[C{{H}_{3}}N{{H}_{2}}\] is more basic than \[N{{H}_{3}}\] due to presence of electrons donating methyl group. (ii) \[{{C}_{6}}{{H}_{5}}N{{H}_{2}}\] is less basic than \[N{{H}_{3}}\] due to presence of electron withdrawing phenyl group \[\therefore \]Correct order of basic nature is \[C{{H}_{3}}N{{H}_{2}}>N{{H}_{3}}>{{C}_{6}}{{H}_{5}}N{{H}_{2}}\]
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