Question

Oxidation state of sulphur in anions $SO^{2-}_{3}. S_{2}O^{2-}_{4}$ and $S_{2}O^{2-}_{6}$ increases in the orders:

• A. $S_{2}O^{2-}_{6}< S_{2}O^{2-}_{4}< SO^{2-}_{3}$
• B. $SO^{2-}_{6}< S_{2}O^{2-}_{4}< SO^{2-}_{6}$
• C. $S_{2}O^{2-}_{4}< SO^{2-}_{3}< S_{2}O^{2-}_{6}$
• D. $S_{2}O^{2-}_{4}< S_{2}O^{2-}_{6}< SO^{2-}_{3}$

Answer 1

The Correct Option is C

To determine the order in which the oxidation state of sulfur increases in each of the given anions: SO^{2-}_{3}, S_{2}O^{2-}_{4}, and S_{2}O^{2-}_{6}, we need to calculate the oxidation state of sulfur in each compound.

1. For SO^{2-}_{3} (sulfite ion):
   In the sulfite ion, the total charge is -2. Assuming the oxidation state of sulfur is x and knowing that the oxidation state of oxygen is -2, we can write the equation:
   \[ x + 3(-2) = -2 \] Solving for x: \[ x - 6 = -2 \] \[ x = +4 \] Thus, the oxidation state of sulfur in SO^{2-}_{3} is +4.
2. For S_{2}O^{2-}_{4} (disulfite ion):
   The total charge is -2. For two sulfur atoms, assuming x as the average oxidation state for one sulfur, we have:
   \[ 2x + 4(-2) = -2 \] Solving for x: \[ 2x - 8 = -2 \] \[ 2x = +6 \] \[ x = +3 \] Therefore, the oxidation state of sulfur in S_{2}O^{2-}_{4} is +3 (on average).
3. For S_{2}O^{2-}_{6} (dithionate ion):
   Again, the total charge is -2. Assuming x as the average oxidation state for each of the two sulfur atoms, we have:
   \[ 2x + 6(-2) = -2 \] Solving for x: \[ 2x - 12 = -2 \] \[ 2x = +10 \] \[ x = +5 \] Therefore, the oxidation state of sulfur in S_{2}O^{2-}_{6} is +5 (on average).

Now, if we compare the average oxidation states:

• S_{2}O^{2-}_{4}: +3
• SO^{2-}_{3}: +4
• S_{2}O^{2-}_{6}: +5

The oxidation state of sulfur increases in the order S_{2}O^{2-}_{4} < SO^{2-}_{3} < S_{2}O^{2-}_{6}. Therefore, the correct answer is:

S_{2}O^{2-}_{4} < SO^{2-}_{3} < S_{2}O^{2-}_{6}