Q. The compound in which carbon atom uses only $sp^{3}$ - hybrid orbitals for bond formation is
A) HCOOH
B) $NH_{2}CONH_{2}$
C) $\left( CH_{3} \right)_{3}COH$
D) $CH_{3}CHO$
Explanation
(c)
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In the above compound all bonds are $\sigma$ bond and hence, carbon atom uses only $sp^{3}$- hybrid orbitals for bond formation.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.