Chemistry - Complete P-block Question with Solution | TestHub

In ${\mathrm{S}}_2{\mathrm{O}}_3^{2-}$ since sulphur atoms are in double bond, the $\mathrm{S}-\mathrm{S}$ bond length $\left(2.01{\mathrm{A}}^{\circ }\right)$ is shorter than $\mathrm{S}-\mathrm{S}$ bond length in ${\mathrm{S}}_2{\mathrm{O}}_6^{2-}\left(2.15A^{\circ }\right)$

Since sulphur atom is involved in resonance structures, the $\mathrm{S}-\mathrm{O}$ bond length in ${\mathrm{S}}_2{\mathrm{O}}_3^{2-}\left(1.47{\mathrm{A}}^{\circ }\right)$ becomes more than $\mathrm{S}-\mathrm{O}$ bond length ${\mathrm{S}}_2{\mathrm{O}}_6^{2-}(1.43)$ in which the resonance is involved only between oxygen atoms. In ${\mathrm{S}}_2{\mathrm{O}}_3^{2-}$ one of the sulphur is in -2 oxidation state and can be oxidised to +6 or to 0 . So it can act as a strong reducing agent. In ${\mathrm{S}}_2{\mathrm{O}}_6^{2-}$ sulphur atoms are in +5 oxidation state. So its reduction power is less. The hybridization of the terminal sulphur in ${\mathrm{S}}_2{\mathrm{O}}_3^{2-}$ will be different.