What is the oxidation state of sulfur in SO4^2-?
Understand the Problem
The question is asking for the oxidation state of sulfur (S) in the sulfate ion (SO4^2-). To find this, we can use the fact that the overall charge of the ion is -2 and the known oxidation state of oxygen, which is typically -2. We will set up an equation to solve for the oxidation state of sulfur.
Answer
The oxidation state of sulfur in the sulfate ion (SO4^2-) is \( +6 \).
Answer for screen readers
The oxidation state of sulfur in the sulfate ion (SO4^2-) is ( +6 ).
Steps to Solve
- Identify the oxidation states of known elements The oxidation state of oxygen (O) is usually -2. In the sulfate ion, there are 4 oxygen atoms. Therefore, the total contribution of oxygen to the oxidation state is:
$$ 4 \times (-2) = -8 $$
- Set up the equation for the total charge The overall charge of the sulfate ion (SO4^2-) is -2. We can set up the equation:
$$ \text{Oxidation state of S} + \text{Total contribution from O} = \text{Charge of the ion} $$
In this case, we express it as:
$$ x + (-8) = -2 $$
where ( x ) is the oxidation state of sulfur.
- Solve for the oxidation state of sulfur Rearranging the equation, we have:
$$ x - 8 = -2 $$
Now, add 8 to both sides:
$$ x = -2 + 8 $$
This simplifies to:
$$ x = 6 $$
Thus, the oxidation state of sulfur in the sulfate ion is 6.
The oxidation state of sulfur in the sulfate ion (SO4^2-) is ( +6 ).
More Information
The oxidation state of +6 for sulfur in sulfate is important in various chemical reactions, particularly in acid-base chemistry and redox reactions. Sulfate plays a key role in the environment and in biological processes.
Tips
- A common mistake is forgetting to account for all oxygen atoms. Make sure to multiply the oxidation state of oxygen by the number of oxygen atoms present.
- Another mistake is miscalculating the final value when rearranging the equation. Double-check arithmetic operations.
