When $C{l}_2$  reacts with cold dilute $NaOH$ the difference in the oxidation state of chlorine atom in the products is ‘X’ and with hot $Conc.NaOH$  it is ‘Y’ then  $(Y-X)=?$

$\begin{array}{l}C{l}_2+NaOH(Cold,dil)\to Na\underset{-}{\overset{-1}{Cl}}+NaO\underset{-}{\overset{+1}{Cl}}+H_{2}O\\ diff\left(X\right)=+1-(-1)=2\\ C{l}_2+NaOH(Hot,conc)\to Na\underset{-}{\overset{-1}{Cl}}+Na\underset{-}{\overset{+5}{Cl}}{O}_3+H_{2}O\\ diff\left(Y\right)=+5-(-1)=6\end{array}$