: a) SO3 6) SO32 (‘”1 ‘”1 – c) HSO4 S8
The Correct Answer and Explanation is:1
Here are the oxidation numbers for sulfur in each of the specified chemical species.
- a) In SO₃, the oxidation number of sulfur is +6.
- b) In SO₃²⁻, the oxidation number of sulfur is +4.
- c) In S₈, the oxidation number of sulfur is 0.
- c) In H₂SO₄, the oxidation number of sulfur is +6.
Explanation
The oxidation number, or oxidation state, is a conceptual tool used to track electrons in chemical reactions. It represents the hypothetical charge an atom would have if all its bonds to different elements were 100% ionic. The assignment of oxidation numbers follows a set of established rules.
a) Sulfur Trioxide (SO₃)
This is a neutral molecule, so the sum of the oxidation numbers of all atoms must equal zero. Oxygen is almost always assigned an oxidation number of -2, except in peroxides or when bonded to fluorine. Here, each of the three oxygen atoms has an oxidation number of -2, for a total contribution of 3 × (-2) = -6. To maintain neutrality for the molecule, the single sulfur atom must have an oxidation number that balances this negative charge.
Let S be the oxidation number of sulfur:
S + 3(-2) = 0
S – 6 = 0
S = +6
b) Sulfite Ion (SO₃²⁻)
This is a polyatomic ion with an overall charge of 2-. Therefore, the sum of the oxidation numbers must equal -2. As in the previous case, each oxygen atom is assigned an oxidation number of -2, contributing a total of 3 × (-2) = -6. The oxidation number of sulfur must be such that the total sum equals the ion’s charge.
Let S be the oxidation number of sulfur:
S + 3(-2) = -2
S – 6 = -2
S = +4
c) Octasulfur (S₈)
This molecule represents the elemental form of sulfur. A fundamental rule of oxidation states is that any atom in its pure elemental form has an oxidation number of zero. Since all atoms in the S₈ molecule are sulfur, there is no difference in electronegativity, and no hypothetical charge transfer occurs. Therefore, the oxidation number of sulfur is 0.
c) Sulfuric Acid (H₂SO₄)
Sulfuric acid is a neutral compound, so the sum of all oxidation numbers is zero. According to the rules, hydrogen is assigned an oxidation number of +1 when bonded to nonmetals, and oxygen is assigned -2.
Let S be the oxidation number of sulfur:
2(H) + S + 4(O) = 0
2(+1) + S + 4(-2) = 0
+2 + S – 8 = 0
S – 6 = 0
S = +6thumb_upthumb_down
