The oxidation state of sulfur in sulfur dioxide SO2 is
The Correct Answer and Explanation is:
The oxidation state of sulfur in sulfur dioxide (SO₂) is +4.
Explanation:
To determine the oxidation state of sulfur in SO₂, we need to follow a systematic approach based on the rules for assigning oxidation states:
- Rule for oxygen: Oxygen typically has an oxidation state of -2 in most compounds (except in peroxides or when bonded to fluorine).
- Sum of oxidation states: The sum of oxidation states in a neutral compound is 0.
In the case of SO₂:
- The molecule is neutral, so the sum of the oxidation states must equal zero.
- Oxygen, being bonded to sulfur, has an oxidation state of -2. Since there are two oxygen atoms, the total contribution of oxygen to the oxidation state is: 2×(−2)=−42 \times (-2) = -42×(−2)=−4
- Let the oxidation state of sulfur be xxx. According to the rule, the sum of the oxidation states of all atoms in SO₂ must be zero: x+2(−2)=0x + 2(-2) = 0x+2(−2)=0 x−4=0x – 4 = 0x−4=0 x=+4x = +4x=+4
Thus, the oxidation state of sulfur in SO₂ is +4.
This means that sulfur has lost four electrons in forming the bonds with oxygen atoms in SO₂. Sulfur in this state is not as oxidized as it would be in sulfur trioxide (SO₃), where it has an oxidation state of +6. Therefore, SO₂ represents an intermediate oxidation state for sulfur.
