Please see photo for question. **Question: Molecular Orbital Theory and Bond Lengths**Based on their MO diagrams, arrange the three diatomic species, S₂⁻, S₂⁺, S₂, in order of increasing bond lengths (shortest to longest).A. S₂⁺ < S₂ < S₂⁻ B. S₂⁻ > S₂ > S₂⁺ C. S₂⁺ < S₂⁻ < S₂ D. S₂⁻ > S₂⁺ > S₂ E. S₂ < S₂⁻ < S₂⁺ Explore the species to understand their molecular configurations through MO theory. Analyze how the electron arrangements influence the bond order and, hence, the bond lengths.

Molecular orbitals are formed by the Linear Combination of Atomic Orbitals (LCAO). Number of molecular orbitals formed is always equal to the number of atomic orbitals combined. There are three types of molecular orbitals that is Bonding molecular orbital, Non-bonding molecular orbital and Anti-bonding molecular orbital.Electronic configuration of S2 molecule is σ1s2σ*1s2 σ2s2σ*2s2 σ2px2π2py2 π2pz2 π*2py2 π*2pz2 σ*2p2 σ3s2σ*3s2 σ3px2π3py2 π3pz2 π*2py1 π*2pz1. B.O=bonding electrons-antibonding electrons2 Bond order in S2 molecule is 2. Bond order in case of S2+ is 2.5. Bond order in case of S2- is 1.5. Bond length is inversely proportional to the bond order of the molecule. Hence, the order of the increasing bond length is given as: S2+ < S2 < S2- So, the correct option is (a).