Answer the following questions:
$(i)$ Give hybridization of Carbon atoms in the following compound: $CH_3-CH=CH-CH_2-C\equiv CH$
$(ii)$ Give the number of $\sigma$ and $\pi$ bonds in the above compound.
$(iii)$ Give the $IUPAC$ name of the above compound.
$(iv)$ Draw the geometrical isomers of the above compound.
$(v)$ Why are $cis$ and $trans$ isomers not possible in alkynes?
$(vi)$ Which bond possesses more bond length between $C=C$ and $C\equiv C$?

(N/A) $(i)$ Hybridization of carbons in $CH_3-CH=CH-CH_2-C\equiv CH$:
$CH_3$ $(sp^3)$,$-CH=$ $(sp^2)$,$=CH-$ $(sp^2)$,$-CH_2-$ $(sp^3)$,$-C\equiv$ $(sp)$,$\equiv CH$ $(sp)$.
$(ii)$ Total $\sigma$ bonds = $13$,Total $\pi$ bonds = $3$.
$(iii)$ $IUPAC$ name: $Hex-4-en-1-yne$.
$(iv)$ Geometrical isomers exist due to the $C=C$ double bond:
$Cis$-isomer: $H_3C-C(H)=C(H)-CH_2-C\equiv CH$ (with $CH_3$ and $CH_2-C\equiv CH$ on the same side).
$Trans$-isomer: $H_3C-C(H)=C(H)-CH_2-C\equiv CH$ (with $CH_3$ and $CH_2-C\equiv CH$ on opposite sides).
$(v)$ $Cis$ and $trans$ isomers are not possible in alkynes because the carbon atoms involved in the triple bond are $sp$ hybridized and have a linear geometry ($180^\circ$ bond angle),preventing the formation of different steric arrangements around the triple bond.
$(vi)$ Bond length of $C=C$ $(134 \ pm)$ $>$ Bond length of $C\equiv C$ $(120 \ pm)$.