A) \[N{{H}_{3}}\]
B) \[C{{H}_{3}}N{{H}_{2}}\]
C) \[N{{F}_{3}}\]
D) \[N{{(Si{{H}_{3}})}_{3}}\]
Correct Answer: B
Solution :
The basic nature of compounds depend upon their tendency to donate thy electron. In\[N{{F}_{3}}\]and\[{{(Si{{H}_{3}})}_{3}}N\]the lone pair of nitrogen transfer to the vacant orbitals of F and\[Si\] respectively. Hence, the electrons are not available for donation and therefore, these compounds are not basic. \[N{{H}_{3}}\]and\[C{{H}_{3}}N{{H}_{2}}\]both are basic in nature but \[C{{H}_{3}}\]group is an electron donating group. It increases the electron density on N and hence, the electron donating tendency of N increases. Therefore,\[C{{H}_{3}}N{{H}_{2}}\]is the most basic in the given compounds.
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