Compare S-S bond length from the following molecules:

(a)- x > y
(b)- y > x
(c)- x = y
(d)- None of these

Hint: The bond length of the compound can be comparing the number of bonds and lone pair in the compound. The repulsion will be maximum when there are more numbers of lone pairs in the compound.

Complete step by step answer:
The given two compounds are ${{S}_{2}}O_{5}^{2-}$ and ${{S}_{2}}O_{4}^{2-}$. The compound ${{S}_{2}}O_{5}^{2-}$ will have one lone pair of electron and the compound ${{S}_{2}}O_{4}^{2-}$ will have two lone pair of electrons.
In structure (I), there are more numbers of oxygen atoms in the molecule and we know that the oxygen atom is more electronegative than the sulfur atom. So in structure (I), there will be a more partial positive charge on the sulfur atom than in structure (II) because the number of oxygen atoms is less in the structure (II).
We know that the repulsion due to the lone pair-lone pair is highest and lowest due to the bond pair-bond pair. There will be intermediate repulsion due to the bond pair-lone pair. So in structure (I), there is only one lone pair of the electron, therefore there will be bond pair-lone pair repulsion and in structure (II), there are two lone pairs of electrons, therefore there will be lone pair-lone pair repulsion. Due to more repulsion in structure (II), the S-S bond will increase as compared to the S-S bond in structure (I). So, y is greater than x.

Therefore, the correct answer is an option (b).

Note: It must be noted that as the repulsion increases the bond length increases and the bond angle between the atoms also increases, because both the factors depend on the repulsion in the molecule.