Step 1: Understand the question.
Chlorine reacts with cold dilute NaOH and also with hot concentrated NaOH. We must find which statement about these two reactions is NOT correct.
Step 2: Write the cold dilute reaction.
\[ Cl_2 + 2NaOH \rightarrow NaCl + NaClO + H_2O \] Here chlorine is both reduced to $Cl^-$ in NaCl and oxidised to $ClO^-$ in hypochlorite.
Step 3: Write the hot concentrated reaction.
\[ 3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O \] Here chlorine goes to $Cl^-$ and to chlorate, $ClO_3^-$.
Step 4: Check the true statements.
Both reactions are disproportionation because the same element is both oxidised and reduced. NaCl appears in both. Hypochlorite forms in cold dilute conditions. These three statements are correct.
Step 5: Test the perchlorate claim.
Hot concentrated NaOH gives chlorate ($ClO_3^-$), not perchlorate ($ClO_4^-$). So the statement that perchlorate forms is wrong.
Step 6: Pick the incorrect statement.
The wrong statement is that perchlorate ion is formed in hot, concentrated conditions.
\[ \boxed{\text{Perchlorate ion is formed in hot, concentrated conditions}} \]