In \( \text{H}_2\text{SO}_4 \), hydrogen is assigned an oxidation state of +1, and oxygen an oxidation state of -2. Let the oxidation state of sulfur be denoted by \( x \).For a neutral compound, the sum of all oxidation states must equal zero. This leads to the equation:\[2 \times (+1) + x + 4 \times (-2) = 0\]Simplifying the equation:\[2 + x - 8 = 0\]Solving for \( x \):\[x = +6\]Therefore, the oxidation state of sulfur in \( \text{H}_2\text{SO}_4 \) is +6.