How to find out unpaired electron in S2 molecule?
The molecular orbital energy-level diagram of $\ce{S2}$ is given below:-

(source)

We can see  how close the energy the 3s and 3p atomic orbitals are because their energy separation will determine whether the  $\ce{π_3_{p}_{x,y}}$,  or the $\ce{σ_{3}_{p}_{z}}$ molecular orbital is higher in energy. Because the ns–np energy gap increases as the nuclear charge increases , the $\ce{σ_{3}_{p}_{z}}$ molecular orbital will be lower in energy than the $\ce{π_3_{p}_{x,y}}$ pair.

Each sulfur atom contributes 6 valence electrons, for a total of 12 valence electrons. Ten valence electrons are used to fill the orbitals through $\ce{π_{3}_{p}_{z}}$and $\ce{π_3_{p}_{y}}$, leaving 2 electrons to occupy the degenerate $\ce{π^{⋆}_{3}_{p}_{z}}$ and $\ce{π^{⋆}_{3}_{p}_{y}}$ pair. From Hund’s rule, the remaining 2 electrons must occupy these orbitals separately with their spins aligned. With the numbers of electrons written as superscripts, the electron configuration of $\ce{S2}$ is $\ce{(σ_{3}_{s})^{2}(σ^{⋆}_{3}_{s})^{2}(σ_{3}_{p}_{z})^{2}(π_{3}_{p}_{z,y})^{4}(π_{3}_{p}^{⋆}_{x,y})^{2}}$ with 2 unpaired electrons at the antibonding $\ce{π^{⋆}}$ orbital and thus is paramagnetic. The bond order is (8 − 4) ÷ 2 = 2, so we predict an $\ce{S=S}$ double bond.