Amongst the given options,the compound$(s)$ in which all the atoms are in one plane in all the possible conformations (if any),is (are):
$(A)$ $CH_2=CH-CH=CH_2$
$(B)$ $HC\equiv C-CH=CH_2$
$(C)$ $H_2C=C=O$
$(D)$ $H_2C=C=CH_2$

$(a)$ Discuss the concept of hybridization. What are its different types in a carbon atom.
$(b)$ What is the type of hybridization of carbon atoms marked with a star in the following structures:
$(a)$ $\overset{*}{C}H_2 = CH - C(=O)OH$
$(b)$ $CH_3 - \overset{*}{C}H_2 - OH$
$(c)$ $CH_3 - CH_2 - \overset{*}{C}H = O$
$(d)$ $\overset{*}{C}H_3 - CH = CH - CH_3$
$(e)$ $CH_3 - \overset{*}{C} \equiv CH$

The hybridisation of carbons of $C - C$ single bond of $HC \equiv C - CH = CH_2$ is

In the given structure,the number of $sp$ and $sp^{2}$ hybridized carbon atoms present respectively are:

Which among the following possesses a $sp$ carbon in its structure?

In the organic compound $CH_2=CH-CH_2-CH_2-C\equiv CH$,the pair of hybridized orbitals involved in the formation of the $C_2-C_3$ bond is: