A) \[C{{H}_{3}}N{{H}_{2}}<N{{H}_{3}}<{{(C{{H}_{3}})}_{2}}NH\]
B) \[{{(C{{H}_{3}})}_{2}}NH<N{{H}_{3}}<C{{H}_{3}}N{{H}_{2}}\]
C) \[N{{H}_{3}}<C{{H}_{3}}N{{H}_{2}}<{{(C{{H}_{3}})}_{2}}NH\]
D) \[C{{H}_{3}}N{{H}_{2}}<{{(C{{H}_{3}})}_{2}}NH<N{{H}_{3}}\]
Correct Answer: C
Solution :
\[N{{H}_{3}},\,\,C{{H}_{3}}N{{H}_{2}},\,\,{{(C{{H}_{3}})}_{2}}+O{{H}^{-}}\] \[{{(C{{H}_{3}})}_{2}}NH+{{H}_{2}}O{{(C{{H}_{3}})}_{2}}\overset{+}{\mathop{N}}\,{{H}_{2}}+O{{H}^{-}}\] \[{{I}^{+}}\]effect maximum, stabilization more, \[H-\]bond stability high, hence most basic \[C{{H}_{3}}N{{H}_{2}}+{{H}_{2}}O\overset{+}{\mathop{C{{H}_{3}}N{{H}_{3}}}}\,+O{{H}^{-}}\] \[N{{H}_{3}}+{{H}_{2}}ONH_{4}^{+}+O{{H}^{-}}\] \[{{K}_{1}}>{{K}_{2}}>{{K}_{3}}\] Hence, basicity order is \[{{(C{{H}_{3}})}_{2}}NH>C{{H}_{3}}N{{H}_{2}}>N{{H}_{3}}\]
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