MCQ
$sp^3-$ hybridised nitrogen is present in
  • A

  • B

  • $C{H_2} = CH - \mathop N\limits^ \oplus  {H_3}$
  • D
    $C{H_2} = CH - \mathop N\limits^{ \bullet  \bullet } {H_2}$

Answer

Correct option: C.
$C{H_2} = CH - \mathop N\limits^ \oplus  {H_3}$
c
$A) \,sp ^2$ hybridized Nitrogen

$B) \,sp ^2$ hybridized Nitrogen due to participation of lone pair into ring.

$C) \,CH _2= CH - NH _3+\left( sp ^3\right.$ hybridised nitrogen $)$

$D) \,\left( sp ^2\right.$ hybridised) as nitrogen is in conjugation with double bond. So, option $C)$ is correct.

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