$${\text{Stability of alkene}} \propto \frac{1}{{{\text{Heat of hydrogenation of alkene}}}}$$
Greater the number of alkyl groups attached to the double bonded carbon atoms, more stable is the alkene. Hence, given alkene follow the following order of stability.

Hence, faster hydrogenation occurs in

For a compound to be aromatic it must have $$\left( {4n + 2} \right)\pi {e^ - }$$   and system should be planar where $$n$$ is an integer. (V) and (VI) both have $$4\,{e^ - }$$  and planar structure. Hence are Antiaromatic.

Acidic character of alkynes can be explained on the basis of $$sp$$ - hybridisation state of the carbon atom in alkynes. We know that an electron in $$s$$  orbital is more tightly held than in a $$P$$ orbital because $$s$$ - electrons are more closer to the nucleus. In $$sp$$ -hybridisation, $$s$$ - character is more 50% as compared to $$s{p^2}$$ - (33%) or $$s{p^3}$$ - (25%) hybrid orbitals. Due to very large $$s$$ - character, the electrons in $$sp$$ - hybrid orbitals are held tightly by the nucleus and are quite electronegative. Consequently, the electron pair of $$H - C \equiv $$   bond gets displaced more towards the carbon atom and help in the release of $${H^ + }\,ions.$$
For a example ethyne react with sodium metal and release the $${H^ + }\,ions.$$
$$CH \equiv CH + Na \to HC \equiv {C^ - }N{a^ + }$$

  ( $$o$$ - xylene ) which on ozonolysis give A, B and C.

is an example of electrophilic addition reaction.

TIPS/Formulae :
$$Anti$$ -Markovnikoff’s addition of $$HBr$$  is observed only with unsymmetrical alkenes, $$a,$$ $$b,$$ and $$d.$$
$$\eqalign{ & \mathop {C{H_3}CH = C{H_2}}\limits_{\left( {\text{a}} \right)} \cr & C{H_2} = \mathop {CHC{H_2}}\limits_{\left( {\text{b}} \right)} C{H_3} \cr & C{H_3}CH\mathop = \limits_{\left( {\text{c}} \right)} CHC{H_3} \cr & C{H_3}CH = \mathop {CH}\limits_{\left( {\text{d}} \right)} C{H_2}C{H_3} \cr} $$

Followed by Markownikoff's rule.

A parent chain with larger number of substituents is preferred.

Addition reaction occurs on a double bond.
The compound containing double bonds are also undergo polymerisation. So, the correct option are both (i) and (iv) e.g

$$\begin{array}{l}C{H}_2=C{H}_2+HOCl\to \underset{\text{(addition reaction)}}{C{H}_{2}OH-C{H}_{2}Cl}\\ nC{H}_2=C{H}_2\stackrel{\text{Polymerisation}}{\to }\underset{\text{Polyethene}}{{\sout{[}C{H}_2-C{H}_2\sout{]}}_n}\end{array}$$