The correct option is A I2 & Na2S2O3
(0)I2+(+2)Na2S2O3→(+2.5)Na2S4O6+(−1)NaI.
(0)I2(oxidant)→(−1)NaI(reduction half).
(+2)Na2S2O3(reductant)→(2.5)Na2S4O6(oxidation half).
A species which gain electron in a redox reaction is called as oxidising agent.
(I2) gets converted into nitric oxide (I−). Iodine is getting reduced because it is gaining electrons, so I2 is the oxidizing agent.
A species which looses electron in a redox reaction is called as reducing agent.
Na2S2O3 gets converted to Na2S4O6 Na2S2O3 is getting oxidized since there is loss of electrons , so it is the reducing agent.
I2 and Na2S2O3 in this order is the correct answer.