The oxidation state of sulphur on H2SO4 acid is

Analysis

To find the oxidation state of sulphur (S) in sulphuric acid ($H_2SO_4$), we apply the rule that the sum of the oxidation states of all atoms in a neutral compound must equal zero.

Concept

• Hydrogen (H): In most compounds, hydrogen has an oxidation state of +1.
• Oxygen (O): In most compounds, oxygen has an oxidation state of -2.
• Sulphur (S): This is the unknown element we need to solve for. Let its oxidation state be represented by $x$.

Step-by-Step Solution

1. Write the formula: The compound is $H_2SO_4$.
2. Set up the equation: $(2 \times \text{Oxidation state of H}) + (1 \times \text{Oxidation state of S}) + (4 \times \text{Oxidation state of O}) = 0$
3. Substitute the known values: $(2 \times +1) + (x) + (4 \times -2) = 0$
4. Simplify the expression: $2 + x - 8 = 0$
5. Solve for $x$: $x - 6 = 0$ $x = +6$

The oxidation state of sulphur in $H_2SO_4$ is +6.

Conclusion

Comparing our result to the given options, the correct choice is B.

Exam Tip

Always remember that for neutral molecules, the sum is zero, but if you were dealing with a polyatomic ion (like $SO_4^{2-}$), the sum would be equal to the charge of the ion (in this case, -2). This simple distinction helps you solve oxidation state questions for both molecules and radicals!