To solve this problem, we need to identify the ions that contain an S–S bond.$\newline$Let's analyze each option:$\newline$Option 1: $S_4O_6^{2-}, S_2O_3^{2-} \newline$• $S_4O_6^{2-}$ (tetrathionate ion) contains two S–S bonds.$\newline$• $S_2O_3^{2-}$ (thiosulfate ion) contains one S–S bond.$\newline$This option contains S–S bonds in both ions.$\newline$Option 2: $S_2O_7^{2-}, S_2O_8^{2-} \newline$• $S_2O_7^{2-}$ (disulfate ion) does not contain an S–S bond.$\newline$• $S_2O_8^{2-}$ (peroxodisulfate ion) does not contain an S–S bond.$\newline$This option does not contain S–S bonds in either ion.$\newline$Option 3: $S_4O_6^{2-}, S_2O_7^{2-} \newline$• $S_4O_6^{2-}$ contains S–S bonds.$\newline$• $S_2O_7^{2-}$ does not contain an S–S bond.$\newline$This option does not contain S–S bonds in both ions.$\newline$Option 4: $S_4O_7^{2-}, S_2O_3^{2-} \newline$• $S_4O_7^{2-}$ (tetrathionate ion) contains S–S bonds.$\newline$• $S_2O_3^{2-}$ contains an S–S bond.$\newline$This option contains S–S bonds in both ions.$\newline$Therefore, the correct option is Option 4.