Sulphur has highest oxidation state in a. `SO_(2)` b. `H_(2)SO_(4)` c. `Na_(2)S_(4)O_(6)` d. `Na_(2)S_(2)O_(3)`

Options

A. `SO_(2)`
B. `H_(2)SO_(4)`
C. `Na_(2)S_(2)O_(3)`
D. `Na_(2)S_(4)O_(6)`

Explanation

To determine which compound has sulfur in its highest oxidation state, we will calculate the oxidation state of sulfur in each of the given compounds: `SO2`, `H2SO4`, `Na2S4O6`, and `Na2S2O3`.

### Step 1: Calculate the oxidation state of sulfur in `SO2`
1. Let the oxidation state of sulfur be \( x \).
2. Oxygen has an oxidation state of \(-2\).
3. The compound is neutral, so the sum of oxidation states must equal zero:
\[ x + 2(-2) = 0 \]
4. Simplifying the equation:
\[ x - 4 = 0 \implies x = +4 \]
Thus, the oxidation state of sulfur in `SO2` is \( +4 \).

### Step 2: Calculate the oxidation state of sulfur in `H2SO4`
1. Let the oxidation state of sulfur be \( x \).
2. Hydrogen has an oxidation state of \( +1 \) and there are 2 hydrogen atoms, contributing \( +2 \).
3. Oxygen has an oxidation state of \(-2\) and there are 4 oxygen atoms, contributing \( -8 \).
4. The sum of oxidation states must equal zero:
\[ 2 + x + 4(-2) = 0 \]
5. Simplifying the equation:
\[ 2 + x - 8 = 0 \implies x - 6 = 0 \implies x = +6 \]
Thus, the oxidation state of sulfur in `H2SO4` is \( +6 \).

### Step 3: Calculate the oxidation state of sulfur in `Na2S4O6`
1. Let the oxidation state of sulfur be \( x \).
2. Sodium has an oxidation state of \( +1 \) and there are 2 sodium atoms, contributing \( +2 \).
3. Oxygen has an oxidation state of \(-2\) and there are 6 oxygen atoms, contributing \( -12 \).
4. The sum of oxidation states must equal zero:
\[ 2 + 4x + 6(-2) = 0 \]
5. Simplifying the equation:
\[ 2 + 4x - 12 = 0 \implies 4x - 10 = 0 \implies 4x = 10 \implies x = +2.5 \]
Thus, the oxidation state of sulfur in `Na2S4O6` is \( +2.5 \).

### Step 4: Calculate the oxidation state of sulfur in `Na2S2O3`
1. Let the oxidation state of sulfur be \( x \).
2. Sodium has an oxidation state of \( +1 \) and there are 2 sodium atoms, contributing \( +2 \).
3. Oxygen has an oxidation state of \(-2\) and there are 3 oxygen atoms, contributing \( -6 \).
4. The sum of oxidation states must equal zero:
\[ 2 + 2x + 3(-2) = 0 \]
5. Simplifying the equation:
\[ 2 + 2x - 6 = 0 \implies 2x - 4 = 0 \implies 2x = 4 \implies x = +2 \]
Thus, the oxidation state of sulfur in `Na2S2O3` is \( +2 \).

### Conclusion
Now, we compare the oxidation states of sulfur in all the compounds:
- In `SO2`, sulfur has an oxidation state of \( +4 \).
- In `H2SO4`, sulfur has an oxidation state of \( +6 \).
- In `Na2S4O6`, sulfur has an oxidation state of \( +2.5 \).
- In `Na2S2O3`, sulfur has an oxidation state of \( +2 \).

The highest oxidation state of sulfur is in `H2SO4`, which is \( +6 \).

### Final Answer
Thus, the correct answer is **b. `H2SO4`**.