Question

Assign oxidation states to all of the atoms in each of the following: a. \(\mathrm{PBr}_{3}\) b. \(\mathrm{C}_{3} \mathrm{H}_{8}\) c. \(\mathrm{KMnO}_{4}\) d. \(\mathrm{CH}_{3} \mathrm{COOH}\)

Short answer

Oxidation states in each compound: a. PBr₃: P = +3, Br = -1 b. C₃H₈: C = -8/3, H = +1 c. KMnO₄: K = +1, Mn = +7, O = -2 d. CH₃COOH: C₁ = -3, H = +1, C₂ = +1, O = -2, H = +1

Step-by-step solution

Rule 1: Oxidation state of any uncombined element is 0

All elements have an oxidation state of 0 in their pure, uncombined state.

Rule 2: Oxidation state of a monotonic ion is equal to its charge

For example, the oxidation state of Na in Na+ is +1, the oxidation state of O in O²⁻ is -2, etc.

Rule 3: Sum of oxidation states in a neutral compound is 0

The sum of the oxidation numbers of all the atoms in a neutral molecule will be equal to 0.

Rule 4: Sum of oxidation states in an ion is equal to its charge

The sum of the oxidation numbers of all the atoms in an ion will be equal to the charge of the ion.

Rule 5: Oxygen's oxidation state is -2 (except in peroxide)

Oxygen has an oxidation state of -2 in most cases, except in peroxides (O₂²⁻) where it has an oxidation state of -1.

Rule 6: Hydrogen's oxidation state is +1

Hydrogen will have an oxidation state of +1 when bonded with non-metals. Now, we will use the above rules to find the oxidation states in each given compound:

a. Oxidation state in \(\mathrm{PBr}_{3}\)

Using Rule 3, the sum of oxidation states is 0 for neutral molecules. Let's assume oxidation state of P is x and Br is y. So, x + 3y = 0. Bromine is a halogen, and its oxidation state is usually -1. Therefore, x + 3(-1) = 0. x = +3, so the oxidation state of P is +3. Oxidation states in PBr₃: P = +3 and Br = -1.

b. Oxidation state in \(\mathrm{C}_{3} \mathrm{H}_{8}\)

Using Rule 3, the sum of oxidation states is 0 for neutral molecules. Let's assume the oxidation state of C is x and H is y. So, 3x + 8y = 0. Using Rule 6, hydrogen's oxidation state is +1. Therefore, 3x + 8(1) = 0. 3x = -8 and x = -8/3, so the oxidation state of each C is -8/3. Oxidation states in C₃H₈: C = -8/3 and H = +1.

c. Oxidation state in \(\mathrm{KMnO}_{4}\)

Using Rule 4, the sum of oxidation states in an ion is equal to its charge. Let's assume the oxidation state of K is x, Mn is y, and O is z. So, x + y + 4z = 0 (since the compound is neutral). Using Rule 2, potassium's oxidation state is +1. Using Rule 5, oxygen's oxidation state is -2. Therefore, 1 + y + 4(-2) = 0. y = +7, so the oxidation state of Mn is +7. Oxidation states in KMnO₄: K = +1, Mn = +7, and O = -2.

d. Oxidation state in \(\mathrm{CH}_{3} \mathrm{COOH}\)

Using Rule 3, the sum of oxidation states is 0 for neutral molecules. The oxidation state of H is +1, and that of O is -2 (using Rules 6 and 5). For the first carbon (in CH₃): Let's assume its oxidation state is x. x + 3(+1) = 0 (since the sum of oxidation states in a neutral compound is 0). x = -3, so the first carbon has an oxidation state of -3. For the second carbon (in COOH): Let's assume its oxidation state is y. y + 1(-2) + 2(+1) = 0 (since the sum of oxidation states in a neutral compound is 0). y = +1, so the second carbon has an oxidation state of +1. Oxidation states in CH₃COOH: C₁ = -3, H = +1, C₂ = +1, O = -2, and H = +1.