Solve this :

88. The correct order of S-O bondlenght is :

$$\begin{gathered} 1) {\mathrm{SO}}_3^{2-}>{\mathrm{SO}}_4^{2-}>{\mathrm{SO}}_3>{\mathrm{SO}}_2 \\ 2) {\mathrm{SO}}_3^{2-}>{\mathrm{SO}}_4^{2-}>{\mathrm{SO}}_2>{\mathrm{SO}}_3 \\ 3) {\mathrm{SO}}_4^{2-}>{\mathrm{SO}}_3^{2-}>{\mathrm{SO}}_2>{\mathrm{SO}}_3 \\ 4) {\mathrm{SO}}_4^{2-}>{\mathrm{SO}}_3^{2-}>{\mathrm{SO}}_3>{\mathrm{SO}}_2 \end{gathered}$$

The solution is as follows:

1. The more is the bond order, the less is the bond length. SO₄^2- has bond order 1.33 and SO₃^2- has bond order 1.5. Thus, latter has less bond length as compared to the former.

2. In SO₂ and SO₃, SO₂ has higher bond length due to Bent's rule. Both have bond order 2, thus, they have lesser bond lengths.

3. So, correct option is (4).