Analisando caso a caso: • $\ce{KClO_2}$: $\color{#3368b8}{+3}$\begin{matrix}\begin{array}{|c|c|c|}\hline \ce{K} & \ce{Cl} & \ce{O} \\ \hline +1 & x & -2 \\ \hline \end{array} &\Rightarrow& (+1) + x + 2(-2) = 0 &\therefore& x = +3 \end{matrix}• $\ce{Ca(ClO)_2}$: $\color{#3368b8}{+1}$\begin{matrix}\begin{array}{|c|c|c|}\hline \ce{Ca} & \ce{Cl} & \ce{O} \\ \hline +2 & y & -2 \\ \\ \hline \end{array} &\Rightarrow& (+1) + 2y + 2(-2) = 0 &\therefore& y = +1 \end{matrix}• $\ce{Mg(ClO_3)_2}$: $\color{#3368b8}{+5}$ \begin{matrix}\begin{array}{|c|c|c|}\hline \ce{Mg} & \ce{Cl} & \ce{O} \\ \hline +2 & w & -2 \\ \hline \end{array} &\Rightarrow& (+2) + 2w + 6(-2) = 0 &\therefore& w = +5 \end{matrix}• $\ce{Ba(ClO_4)_2}$: $\color{#3368b8}{+7}$\begin{matrix}\begin{array}{|c|c|c|}\hline \ce{Ba} & \ce{Cl} & \ce{O} \\ \hline +2 & z & -2 \\ \hline \end{array} &\Rightarrow& (+2) + 2z + 8(-2) = 0 &\therefore& z = +7 \end{matrix}\begin{matrix} Letra \ (E) \end{matrix}