The oxidation states of iodine in HIO₄, H₃IO₅ and H₅IO₆ are respectively

Calculation the oxidation states of I ,we get

$$\begin{gathered} {\mathrm{HIO}}_4 \\ 1 +\mathrm{x} +4 (-2)=0 \\ \mathrm{x} -7=0 \\ \mathrm{X} =+7 \end{gathered}$$

$$\begin{gathered} {\mathrm{H}}_3{\mathrm{IO}}_5 \\ 3 +\mathrm{x} +5(-2)=0 \\ \mathrm{x} -7 =0 \\ \mathrm{X}= +7 \end{gathered}$$

$$\begin{gathered} {\mathrm{H}}_5{\mathrm{IO}}_6 \\ 5 +\mathrm{x} +6(-2)=0 \\ \mathrm{x}-7=0 \\ \mathrm{x}=+7 \end{gathered}$$

Thus, the correct answer is +7,+7,+7