The given reactions show the behavior of thiosulphate with iodine and bromine:
Let's analyze the reactions:
Thiosulphate with Iodine: In the first reaction, iodine \((I_2)\) is reduced to iodide \((I^-)\), while thiosulphate \((S_2O_3^{2-})\) is oxidized to tetrathionate \((S_4O_6^{2-})\). This indicates that iodine acts as an oxidizing agent.
Thiosulphate with Bromine: In the second reaction, bromine \((Br_2)\) is reduced to bromide \((Br^-)\). Here, the thiosulphate \((S_2O_3^{2-})\) is completely oxidized to sulfate \((SO_4^{2-})\). This shows that bromine acts as a stronger oxidizing agent compared to iodine.
Conclusion:
The difference in behavior is due to the relative strengths of iodine and bromine as oxidizing agents. Bromine is a stronger oxidant than iodine, which is why it can oxidize thiosulphate to a greater extent, all the way to sulfate.
Based on this analysis, the correct statement that justifies the dual behavior of thiosulphate is:
Bromine is a stronger oxidant than iodine.
In the reaction with I$_2$, the oxidation state of sulphur changes from $+2$ to $+2.5$.
In the reaction with Br$_2$, the oxidation state of sulphur changes from $+2$ to $+6$.
Thus, both I$_2$ and Br$_2$ are oxidants, but Br$_2$ is stronger as it increases the oxidation state of sulphur further.
Final Answer:
Bromine is a stronger oxidant than iodine.
Given below are two statements:
Statement (I): The first ionization energy of Pb is greater than that of Sn.
Statement (II): The first ionization energy of Ge is greater than that of Si.
In light of the above statements, choose the correct answer from the options given below:
The product (A) formed in the following reaction sequence is:

Let \( C_{t-1} = 28, C_t = 56 \) and \( C_{t+1} = 70 \). Let \( A(4 \cos t, 4 \sin t), B(2 \sin t, -2 \cos t) \text{ and } C(3r - n_1, r^2 - n - 1) \) be the vertices of a triangle ABC, where \( t \) is a parameter. If \( (3x - 1)^2 + (3y)^2 = \alpha \) is the locus of the centroid of triangle ABC, then \( \alpha \) equals: