In concentrated sulphuric acid, H2SO4, what is the oxidation number of sulphur (S) in the molecule?

Introduction / Context:
This question assesses your understanding of oxidation numbers in molecular compounds, specifically in sulphuric acid, H2SO4. Oxidation state calculations are crucial in redox reactions, industrial chemistry and in predicting oxidising power of acids such as H2SO4.

Given Data / Assumptions:

• The formula of the compound is H2SO4.
• We are asked to find the oxidation number of sulphur (S).
• We use standard oxidation number rules: hydrogen in most compounds is +1, oxygen in oxides is -2.
• H2SO4 is electrically neutral, so the sum of oxidation numbers must be zero.

Concept / Approach:
Oxidation number rules state that hydrogen is usually assigned +1 and oxygen is usually assigned -2 in covalent compounds. For a neutral molecule, the algebraic sum of oxidation numbers of all atoms is zero. By assigning known values to hydrogen and oxygen, we can solve for the unknown oxidation number of sulphur in H2SO4. This simple algebraic approach is standard for such problems.

Step-by-Step Solution:
1) Write the chemical formula H2SO4 and identify the number of each atom: 2 H, 1 S and 4 O.2) Assign oxidation numbers based on rules: hydrogen = +1 and oxygen = -2 in this typical oxyacid.3) Let the oxidation number of sulphur be x.4) Set up the total sum of oxidation numbers for a neutral molecule: 2*(+1) + x + 4*(-2) = 0.5) Simplify the equation: 2 + x - 8 = 0, which gives x - 6 = 0, so x = +6.6) Therefore, the oxidation number of sulphur in H2SO4 is +6.

Verification / Alternative check:
Substitute x = +6 back into the sum: total = 2*(+1) + (+6) + 4*(-2) = 2 + 6 - 8 = 0, which satisfies the neutrality condition. This value also matches the known chemistry of sulphuric acid, where sulphur is often described as being in its highest common oxidation state of +6. In redox reactions, concentrated sulphuric acid behaves as an oxidising agent, consistent with a high positive oxidation state for sulphur.

Why Other Options Are Wrong:
+2: This oxidation state corresponds to sulphur in compounds like H2S, not in H2SO4.
-6: A highly negative oxidation state would require sulphur to be combined with elements more electropositive than itself, which is not the case here with oxygen.
-2: This is the oxidation state of sulphur in sulphides, such as Na2S or H2S, not in an oxyacid with four oxygens.
0: A zero oxidation state would mean elemental sulphur, which is not what we have in H2SO4.

Common Pitfalls:
Students often confuse the oxidation state of sulphur in different sulphur containing compounds because sulphur exhibits many valencies. Mixing up H2SO4 with H2S or SO2 leads to incorrect answers. Another mistake is forgetting that there are four oxygen atoms and miscalculating the total contribution from oxygen. Carefully counting atoms and applying the oxidation number rules step by step prevents such errors and makes these problems straightforward.

Final Answer:
The oxidation number of sulphur in sulphuric acid, H2SO4, is +6.