Step 1: Recall the rule for disproportionation.
Disproportionation is when one species both gives away electrons and takes them in at the same time, so the same element gets oxidised and reduced together. This is only possible when the element sits in a middle oxidation state, with room to go both up and down.
Step 2: Find chlorine's oxidation state in each ion.
In $\text{ClO}^-$ chlorine is $+1$, in $\text{ClO}_2^-$ it is $+3$, in $\text{ClO}_3^-$ it is $+5$, and in $\text{ClO}_4^-$ it is $+7$. The top stable state for chlorine is $+7$ and the lowest is $-1$.
Step 3: Check the $+7$ ion.
In $\text{ClO}_4^-$ chlorine is at its maximum $+7$. It cannot be oxidised any further, so it has nowhere to go up. With no upward room, it cannot disproportionate. Rejected.
Step 4: Check the $+1$ ion.
In $\text{ClO}^-$ chlorine is $+1$, an intermediate state. It can drop down to $-1$ (chloride) and rise up to higher states, so it can disproportionate. Included.
Step 5: Check the $+3$ and $+5$ ions.
In $\text{ClO}_2^-$ chlorine is $+3$ and in $\text{ClO}_3^-$ it is $+5$. Both are intermediate states with room to go both up toward $+7$ and down toward lower states, so both can disproportionate. Included.
Step 6: Collect the answer.
The species in intermediate states, namely $+1$, $+3$, and $+5$, can disproportionate, while only the $+7$ perchlorate cannot.
\[ \boxed{(i)\ \text{ClO}^-,\ (ii)\ \text{ClO}_2^-,\ \text{and}\ (iii)\ \text{ClO}_3^- \text{ only}} \]