$ClO^-_4$ does not disproportionate because in this oxo anion chlorine is present in its highest oxidation state $(+7)$. The disproportionation reactions for the other three oxo anions of chlorine are as follows :
$\overset{\text{+1}}{3C}lO^-\rightarrow \overset{\text{-1}}{2C}l^- + \overset{\text{+5}}{Cl}O^{-}_3$
$\overset{\text{+3}}{6C}lO^{-}_2 \xrightarrow{hv} \overset{\text{+5}}{4C}lO^{-}_3 + \overset{\text{-1}}{2C}l^{-}$
$\overset{\text{+5}}{4C}lO^{-}_3 \rightarrow \overset{\text{-1}}{C}l^- + \overset{\text{+7}}{3C}lO^{-}_4$