Q. Underlined carbon is $sp^{3}$ hybridised in
A) $CH_{3}\underline{C}H = CH_{2}$
B) $CH_{3}\underline{C}H_{2}NH_{2}$
C) $CH_{3}\underline{C}ONH_{2}$
D) $CH_{3}CH_{2}\underline{C}N$
Explanation
(b)
$CH_{3} - {CH}_{2} - NH_{2}$
$sp^{3}\ \ \ \ \ \ sp^{3}$
$CH_{3} - CH = CH_{2}$
$sp^{3}\ \ \ \ \ sp^{2}\ \ \ \ \ \ sp^{2}$
$O$
$||$
$CH_{3} - C - NH_{2}$
$sp^{3}\ \ \ \ \ \ sp^{3}$
${CH}_{3} - CH_{2} - C \equiv N$
$sp^{3}\ \ \ \ \ sp^{2}\ \ \ \ \ \ sp$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.