$\mathrm{Cl}–\mathrm{O}$ bond length in ${{\mathrm{ClO}}_2}^{–}$,${\mathrm{ClO}}^{-}$, ${\mathrm{ClO}}_3^{-}$ and  ${\mathrm{ClO}}_4^{-}$ follows the order :

Bond length is inversely proportional to the bond order. This depicts that if bond order is high, then bond length will be shorter.

Bond order can be calculated using the following expression:

$\mathrm{Bond} \mathrm{order}=\mathrm{Valency} \mathrm{of} \mathrm{peripheral} \mathrm{atoms}+\frac{\mathrm{charge} \mathrm{on} \mathrm{the} \mathrm{acid} \mathrm{radical}}{\mathrm{Number} \mathrm{of} \mathrm{peripheral} \mathrm{atoms}}$

Peripheral atom is the oxygen atom. 

Bond order of ${\mathrm{ClO}}_2^{-}$ is:

$\begin{array}{ccc}\mathrm{Bond} \mathrm{order}& =& 2-\frac{1}{2}\\ & =& 1.5\end{array}$

Bond order of ${\mathrm{ClO}}^{-}$ is:

$\begin{array}{ccc}\mathrm{Bond} \mathrm{order}& =& 2-\frac{1}{1}\\ & =& 1\end{array}$

Bond order of ${\mathrm{ClO}}_3^{-}$ is:

$\begin{array}{ccc}\mathrm{Bond} \mathrm{order}& =& 2-\frac{1}{3}\\ & =& 1.66\end{array}$

Bond order of ${\mathrm{ClO}}_4^{-}$ is:

$\begin{array}{ccc}\mathrm{Bond} \mathrm{order}& =& 2-\frac{1}{4}\\ & =& 1.75\end{array}$

Therefore, the correct bond length order is:

${\mathrm{ClO}}_4^{-}<{\mathrm{ClO}}_3^{-}<{\mathrm{ClO}}_2^{-}<{\mathrm{ClO}}^{-}$