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What is the oxidation number of sulfur in (i) sulfur dioxide (SO₂), (ii) the sulfate ion (SO₄²⁻)? - Leaving Cert Chemistry - Question d - 2019

Question d

What is the oxidation number of sulfur in (i) sulfur dioxide (SO₂), (ii) the sulfate ion (SO₄²⁻)?

Worked Solution & Example Answer:What is the oxidation number of sulfur in (i) sulfur dioxide (SO₂), (ii) the sulfate ion (SO₄²⁻)? - Leaving Cert Chemistry - Question d - 2019

Step 1

(i) sulfur dioxide (SO₂)

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Answer

To determine the oxidation number of sulfur in sulfur dioxide (SO₂), we start by considering the oxidation state of oxygen, which is typically -2. In SO₂, there are two oxygen atoms, contributing a total of -4 to the oxidation states. Let the oxidation number of sulfur be denoted as x. The sum of the oxidation states in a neutral compound must equal zero:

$x + 2(-2) = 0$

Solving for x gives:

$x - 4 = 0$ $x = +4$

Thus, the oxidation number of sulfur in SO₂ is +4.

Step 2

(ii) the sulfate ion (SO₄²⁻)

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Answer

Now, for the sulfate ion (SO₄²⁻), we again begin by recognizing the oxidation state of each oxygen atom as -2. Since there are four oxygen atoms, they contribute a total of -8. Let's denote the oxidation state of sulfur as y. For the sulfate ion, which has an overall charge of -2, the sum of the oxidation states must equal the charge of the ion:

$y + 4(-2) = -2$

Solving for y gives:

$y - 8 = -2$ $y = +6$

Therefore, the oxidation number of sulfur in the sulfate ion (SO₄²⁻) is +6.

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