How do you determine the oxidation number of sulfur in each of the following substances?

• a) Barium sulfate, BaSO₄
• b) Sulfurous acid, H₂SO₃
• c) Strontium sulfide, SrS
• d) Hydrogen sulfide, H₂S

Determining the oxidation number of sulfur in various compounds involves understanding the oxidation states of the other elements present. Here’s how to find the oxidation number of sulfur in each substance:

Barium Sulfate (BaSO₄)

In BaSO₄, barium (Ba) has an oxidation number of +2. Oxygen (O) typically has an oxidation number of -2. Since there are four oxygen atoms, their total contribution is -8. To find sulfur's oxidation number (let's call it x), we set up the equation:

• +2 (Ba) + x (S) + 4(-2) (O) = 0

This simplifies to:

• +2 + x - 8 = 0
• x = +6

Thus, the oxidation number of sulfur in BaSO₄ is +6.

Sulfurous Acid (H₂SO₃)

In H₂SO₃, hydrogen (H) has an oxidation number of +1. There are two hydrogen atoms, contributing +2. Oxygen again is -2, and with three oxygen atoms, that totals -6. Setting up the equation:

• +2 (H) + x (S) + 3(-2) (O) = 0

This leads to:

• +2 + x - 6 = 0
• x = +4

Therefore, sulfur's oxidation number in H₂SO₃ is +4.

Strontium Sulfide (SrS)

In SrS, strontium (Sr) has an oxidation number of +2. Since sulfide (S) is in the form of sulfide ion, it typically has an oxidation number of -2. Thus, sulfur's oxidation number here is:

• x (S) = -2

So, sulfur in SrS has an oxidation number of -2.

Hydrogen Sulfide (H₂S)

For H₂S, similar to sulfurous acid, hydrogen has an oxidation number of +1. With two hydrogen atoms, that gives a total of +2. Setting up the equation:

• +2 (H) + x (S) = 0

This simplifies to:

• x = -2

Thus, the oxidation number of sulfur in H₂S is also -2.

In summary:

• Barium sulfate (BaSO₄): +6
• Sulfurous acid (H₂SO₃): +4
• Strontium sulfide (SrS): -2
• Hydrogen sulfide (H₂S): -2