Question

Find out the hybridisations of $C^{1}, C^{2}$, and $C^{3}$ labelled carbon atoms in the given molecule

• A. $C^{1}-sp, \,\, C^{2} - sp^2, \,\,C^{3}sp^2$
• B. $C^{1}-sp, \,\, C^{2} - sp^2, \,\,C^{3}sp$
• C. $C^{1}-sp, \,\, C^{2} - sp, \,\,C^{3}sp$
• D. $C^{1}-sp^2, \,\, C^{2} - sp^2, \,\,C^{3}sp^2$

Answer 1

Answer: (B)

Among hydrocarbon bonds, the carbon atoms, which do not contain any pi $(\pi)$ bond show $s p^{3}$ hybridisation. The carbon atoms, that contain one pi $(\pi)$ bond show $s p^{2}$ and which contain two pi $(\pi)$ bonds show sp-hybridisation, because pi $(\pi)$ bonds do not get hybridised.
Thus,
$\left(C_{1}\right)=s p$-hybridisation
$\left(C_{2}\right)=s p^{2}$-hybridisation
$\left(C_{3}\right)=s p$-hybridisation