The oxidation states of S atom in `S_(4)O_(6)O^(2-)` from left to right respectively are
`O^(-)-underset(O)underset(||)overset(O)overset(||)S-S-S-underset(O)underset(||)overset(O)overset(||)S-O^(-)`

Options

A. `+6,0,0,+6`
B. `+3,+1,+1,+3`
C. `+5,0,0,+5`
D. `+4,+1,+1,+4`

Answer

To determine the oxidation states of sulfur (S) atoms in the ion \( S_4O_6O^{2-} \) from left to right, we can follow these steps:

### Step 1: Understand the Structure
The given ion \( S_4O_6O^{2-} \) consists of 4 sulfur atoms and 6 oxygen atoms. The overall charge of the ion is -2.

### Step 2: Assign Oxidation States to Oxygen
In most compounds, oxygen has an oxidation state of -2. However, in this case, we have one oxygen atom that is part of the ion and carries a -1 charge. Thus, we can assign oxidation states to the oxygen atoms as follows:
- 5 oxygen atoms: -2 each
- 1 oxygen atom: -1

### Step 3: Calculate Total Contribution of Oxygen
The total contribution of the oxygen atoms to the overall charge can be calculated as:
\[ (5 \times -2) + (-1) = -10 - 1 = -11 \]

### Step 4: Set Up the Equation for Sulfur
Let the oxidation states of the sulfur atoms be \( x_1, x_2, x_3, x_4 \). The total contribution from sulfur and oxygen must equal the overall charge of the ion (-2):
\[ x_1 + x_2 + x_3 + x_4 - 11 = -2 \]
This simplifies to:
\[ x_1 + x_2 + x_3 + x_4 = 9 \]

### Step 5: Analyze the Structure
From the structure of the ion, we can see that:
- The first sulfur atom (leftmost) is likely to be in a higher oxidation state due to its bonding with multiple oxygens.
- The middle sulfur atoms are likely to be in a lower oxidation state due to their bonding with each other and fewer oxygens.
- The last sulfur atom (rightmost) is also likely to be in a higher oxidation state similar to the first one.

### Step 6: Assign Oxidation States
Based on the structure and the calculations:
1. The first sulfur atom \( S_1 \) can be assigned an oxidation state of +5.
2. The second sulfur atom \( S_2 \) is in a zero oxidation state (due to the non-metal bond).
3. The third sulfur atom \( S_3 \) is also in a zero oxidation state.
4. The fourth sulfur atom \( S_4 \) can be assigned an oxidation state of +5.

Thus, the oxidation states from left to right are:
- \( +5, 0, 0, +5 \)

### Final Answer
Therefore, the oxidation states of sulfur in \( S_4O_6O^{2-} \) from left to right are:
**+5, 0, 0, +5**