To determine the oxidation state of sulfur (S) in the ion [tex]\( \text{HSO}_4^{2-} \)[/tex], we follow a step-by-step process.
1. Identify the oxidation states of known atoms in the molecule:
- Hydrogen (H) typically has an oxidation state of +1.
- Oxygen (O) typically has an oxidation state of -2.
2. Let the oxidation state of sulfur (S) be denoted by [tex]\( x \)[/tex].
3. Write out the total oxidation state equation for the ion:
- The ion [tex]\( \text{HSO}_4^{2-} \)[/tex] has an overall charge of -2.
- The sum of the oxidation states of all atoms in the ion should be equal to the ion's charge.
4. Set up the oxidation state equation:
- There is 1 hydrogen atom: [tex]\( (+1) \)[/tex]
- There are 4 oxygen atoms: [tex]\( 4 \times (-2) = -8 \)[/tex]
- There is 1 sulfur atom: [tex]\( x \)[/tex]
The sum of these oxidation states should equal the charge of the ion:
[tex]\[
+1 + x + 4(-2) = -2
\][/tex]
5. Solve the oxidation state equation:
- Rewrite the equation:
[tex]\[
1 + x - 8 = -2
\][/tex]
- Simplify:
[tex]\[
x - 7 = -2
\][/tex]
- Solve for [tex]\( x \)[/tex]:
[tex]\[
x = -2 + 7
\][/tex]
[tex]\[
x = 5
\][/tex]
Therefore, the oxidation state of sulfur (S) in [tex]\( \text{HSO}_4^{2-} \)[/tex] is [tex]\( +5 \)[/tex].
The answer is:
C. +5
