What is the oxidation number of sulfur (S) in Al2(SO3)3?
0-2
0-6
O-4
+4
+6
The correct answer and explanation is :
The correct answer is: +4
Explanation (300+ words):
To determine the oxidation number of sulfur (S) in Al₂(SO₃)₃, we need to break down the compound and apply some basic rules of oxidation numbers.
Step 1: Understand the compound
The compound Al₂(SO₃)₃ is aluminum sulfite. It contains:
- Aluminum ions (Al³⁺)
- Sulfite ions (SO₃²⁻)
There are 2 aluminum atoms and 3 sulfite ions in one formula unit.
Step 2: Rules of oxidation numbers
Here are the key oxidation number rules we’ll use:
- The oxidation number of Aluminum (Al) in a compound is +3.
- The oxidation number of Oxygen (O) is usually -2.
- The sum of oxidation numbers in a polyatomic ion is equal to the charge of the ion.
- In a neutral compound, the sum of all oxidation numbers must equal zero.
Step 3: Focus on the SO₃²⁻ ion
We’re interested in the oxidation number of sulfur (S) in the sulfite ion (SO₃²⁻).
Let the oxidation number of sulfur be x.
Each oxygen contributes -2, and there are 3 oxygen atoms:
[
x + 3(-2) = -2
]
[
x – 6 = -2
]
[
x = +4
]
So, the oxidation number of sulfur in SO₃²⁻ is +4.
Step 4: Confirm with the entire compound
In Al₂(SO₃)₃:
- 2 aluminum atoms = 2(+3) = +6
- 3 sulfite ions = 3(–2) = –6
Total charge: +6 (from Al) + (–6) (from sulfite) = 0, as expected for a neutral compound.
Summary:
- The oxidation number of sulfur in the sulfite ion (SO₃²⁻) is +4.
- Therefore, in Al₂(SO₃)₃, sulfur has an oxidation number of +4.
- Correct answer: +4 ✅
This process helps ensure we understand redox chemistry, ionic charges, and oxidation states within polyatomic ions and complex compounds.