Both oxygen and sulphur exist as diatomic molecules $O_{2} & S_{2}$ in gaseous state but at room temperature oxygen remains as diatomic molecule while sulphur converts into $S_{8}$ rings. This is because

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$O - O$ single bonds are weaker due to repulsion of non–bonding electron pairs on small oxygen atoms

Paramagnetic $S_{2}$ molecules convert into diamagnetic, $S_{8}$ molecular rings and give stability.

When 4 Sulphur gaseous $S_{2}$ molecules converts into solid S8 molecules energy is released due to conversion of 4 weak $S - S π$ bond to 4 extra strong $σ$ bonds and get stability.

Due to conversion into $S_{8}$ rings molecular size increases which increases stability.

answer is A, B.

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Detailed Solution

When solid sulphur is heated it converts into vapour by absorbing energy to break strong 4 sigma bonds in $S_{8}$ ring. In $S_{2}$ molecule a weak $π$ bond is formed, due to less effective lateral overlap of p orbitals between bigger sulphur atoms. At room temperature when $S_{2}$ molecule converts into S8 rings the 4 weak, $π$ bonds breakup and form strong sigma bonds liberating energy thus giving stability to $4 S - S$ molecule. $O - O$ single bond is weaker due to repulsion between non–bonding electron pairs on small oxygen atoms. But the $π$ bond is stronger in $O_{2}$ due to effective overlap of P-orbitals in $O_{2}$ because of small size of oxygen atoms. During conversion of $S_{2}$ molecules to $S_{8}$ rings though paramagnetic character decreases and molecular size increases they have no role in the stability