Oxidation state of sulphur in anions $$SO_4^{2-}$$, $$S_2O_4^{2-}$$ and $$S_2O_6^{2-}$$ increases in the orders :

To determine the oxidation state of sulfur in the given anions, we recall that the oxidation state is the hypothetical charge on an atom if all bonds were ionic. The sum of oxidation states in a compound must equal the overall charge. Oxygen typically has an oxidation state of -2, except in peroxides, which is not the case here. Let the oxidation state of sulfur be denoted by $$x$$ for each anion.

First, consider the sulfate ion $$SO_4^{2-}$$. However, note that the options include $$SO_3^{2-}$$ (sulfite ion) instead of sulfate. Given the correct answer option and the context, it appears there might be a typo in the question, and we should consider sulfite ion $$SO_3^{2-}$$ instead. We will calculate for $$SO_3^{2-}$$, $$S_2O_4^{2-}$$, and $$S_2O_6^{2-}$$ as per the options.

Start with the sulfite ion $$SO_3^{2-}$$:

• It has one sulfur atom and three oxygen atoms.
• The overall charge is -2.
• So, oxidation state of sulfur + 3 × (oxidation state of oxygen) = overall charge.
• $$x + 3 \times (-2) = -2$$
• $$x - 6 = -2$$
• $$x = -2 + 6$$
• $$x = 4$$
• Thus, the oxidation state of sulfur in $$SO_3^{2-}$$ is +4.

Next, the dithionite ion $$S_2O_4^{2-}$$:

• It has two sulfur atoms and four oxygen atoms.
• The overall charge is -2.
• Assuming both sulfur atoms have the same oxidation state, we have 2 × (oxidation state of sulfur) + 4 × (oxidation state of oxygen) = overall charge.
• $$2x + 4 \times (-2) = -2$$
• $$2x - 8 = -2$$
• $$2x = -2 + 8$$
• $$2x = 6$$
• $$x = 3$$
• Thus, the oxidation state of sulfur in $$S_2O_4^{2-}$$ is +3 per sulfur atom.

Finally, the dithionate ion $$S_2O_6^{2-}$$:

• It has two sulfur atoms and six oxygen atoms.
• The overall charge is -2.
• Assuming both sulfur atoms have the same oxidation state, we have 2 × (oxidation state of sulfur) + 6 × (oxidation state of oxygen) = overall charge.
• $$2x + 6 \times (-2) = -2$$
• $$2x - 12 = -2$$
• $$2x = -2 + 12$$
• $$2x = 10$$
• $$x = 5$$
• Thus, the oxidation state of sulfur in $$S_2O_6^{2-}$$ is +5 per sulfur atom.

We have the oxidation states:

• $$S_2O_4^{2-}$$: +3
• $$SO_3^{2-}$$: +4
• $$S_2O_6^{2-}$$: +5

Arranging these in increasing order of oxidation state:

• +3 (from $$S_2O_4^{2-}$$) is less than +4 (from $$SO_3^{2-}$$), which is less than +5 (from $$S_2O_6^{2-}$$).
• So, the order is $$S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$$.

Comparing with the options:

• Option A: $$S_2O_6^{2-} < S_2O_4^{2-} < SO_3^{2-}$$ → +5 < +3 < +4, which is incorrect.
• Option B: $$SO_6^{2-} < S_2O_4^{2-} < S_2O_6^{2-}$$ → Note: $$SO_6^{2-}$$ is not standard and likely a typo; ignoring this as it does not match our anions.
• Option C: $$S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$$ → +3 < +4 < +5, which matches our order.
• Option D: $$S_2O_4^{2-} < S_2O_6^{2-} < SO_3^{2-}$$ → +3 < +5 < +4, but +5 is not less than +4, so incorrect.

Hence, the correct answer is Option C.