What is the oxidation state of each element in SO4(2-)?
The correct answer and explanation is:
To determine the oxidation state of each element in the sulfate ion (SO₄²⁻), follow these steps:
Step 1: Understand the charge of the ion
The sulfate ion has a 2- charge, written as SO₄²⁻. This means the total oxidation states of all the elements in the ion must sum up to -2.
Step 2: Assign oxidation states to oxygen
In most compounds, oxygen has an oxidation state of -2. Since there are four oxygen atoms in the sulfate ion, the total oxidation state of all oxygen atoms combined will be: 4×(−2)=−84 \times (-2) = -8
Step 3: Assign the oxidation state to sulfur
Let the oxidation state of sulfur be xx. Since the sum of the oxidation states of all elements must equal the charge of the ion (-2), we can set up the following equation: x+(−8)=−2x + (-8) = -2
Solving for xx, we get: x=+6x = +6
Step 4: Double-check the result
To ensure the calculation is correct, add up the oxidation states of sulfur and oxygen:
- The oxidation state of sulfur is +6.
- The oxidation state of each oxygen atom is -2, and there are four oxygen atoms, contributing a total of -8.
The sum of the oxidation states is: +6+(−8)=−2+6 + (-8) = -2
This matches the overall charge of the sulfate ion, confirming that the oxidation state of sulfur is +6 and the oxidation state of oxygen is -2.
Conclusion:
In the sulfate ion (SO₄²⁻):
- The oxidation state of sulfur is +6.
- The oxidation state of oxygen is -2.
This method of assigning oxidation states ensures that the overall charge of the ion is accurately accounted for.