Oxidation states reflect electron distribution

Oxidation states, also known as oxidation numbers, are a way of representing the distribution of electrons in a compound as if it were composed of ions.

• In purely ionic compounds, the oxidation state is equal to the charge on the ion.
• In covalent compounds, oxidation states indicate the number of electrons shared or transferred when forming a bond.

For example, in the covalent molecule water (H₂O):

• If it were an ionic compound, oxygen would gain two electrons (as it is more electronegative than hydrogen) and each hydrogen would lose one.
• Therefore, the oxidation state of oxygen is assigned as -2 and each hydrogen as +1.

Oxidation states are a useful tool for keeping track of electrons in reactions, even though the bonding may not be purely ionic.

Rules for deducing oxidation states

To determine the oxidation states of atoms in a compound, there are several rules to follow:

1. The oxidation state of any free element is zero. This reflects that in elemental compounds like H_2_, the electronegativity of the atoms is equal so electrons are shared evenly, meaning each atom has an oxidation state of zero.
2. The sum of the oxidation states of all atoms in a neutral compound is zero.
3. The sum of the oxidation states of all atoms in a polyatomic ion equals the charge on the ion.
4. Fluorine always has an oxidation state of -1 in compounds.
5. Group 1 metals always have an oxidation state of +1 and group 2 metals are always +2.
6. Oxygen usually has an oxidation state of -2, except in peroxides, where it is -1, and in OF_2_ where it is +2.
7. Hydrogen usually has an oxidation state of +1, except in metal hydrides, where it is -1.

Worked example 1 - Determine the oxidation number of carbon in H_2_CO_3_.

Determine the oxidation number of carbon in carbonic acid, H_2_CO_3_.

Step 1: Assign known oxidation numbers

• Hydrogen (H) has an oxidation number of +1.
• Oxygen (O) usually has an oxidation number of -2.

Step 2: Calculate total oxidation number of O

There are 3 oxygen atoms, each with an oxidation number of -2.

Total oxidation number from oxygen = 3 x (-2) = -6

Step 3: Calculate total oxidation number of H

There are 2 hydrogen atoms, each with an oxidation number of +1.

Total oxidation number from hydrogen = 2 x (+1) = +2

Step 4: Determine oxidation number of carbon (C)

The total oxidation number of a neutral compound is 0. Let the oxidation number of carbon be x.

Total oxidation number of H + Oxidation number of C + Total oxidation number of O = 0

2 + x + (-6) = 0

x - 4 = 0

x = 4

Therefore, the oxidation number of carbon in H_2_CO_3_ is +4.

Worked example 2 - Determine the oxidation number of chromium in K_2_Cr_2_O_7_.

Determine the oxidation number of chromium in potassium dichromate, K_2_Cr_2_O_7_.

Step 1: Assign known oxidation numbers

• Potassium (K) is in group 1, so it has an oxidation number of +1.
• Oxygen (O) usually has an oxidation number of -2.

Step 2: Calculate total oxidation number of K

There are 2 potassium atoms, each with an oxidation number of +1.

Total oxidation number from potassium = 2 x (+1) = +2

Step 3: Calculate total oxidation number of O

There are 7 oxygen atoms, each with an oxidation number of -2.

Total oxidation number from oxygen = 7 x (-2) = -14

Step 4: Determine oxidation number of chromium (Cr)

The total oxidation number of a neutral compound is 0. Let the oxidation number of chromium be x.

Total oxidation number of K + Oxidation number of Cr + Total oxidation number of O = 0

2 + 2x + (-14) = 0.

Note: There are 2 chromium atoms, so the oxidation number is multiplied by 2.

2x - 12 = 0

x = $\frac{12}{2}$ = 6

Therefore, the oxidation number of chromium in K_2_Cr_2_O_7_ is +6.

Naming oxyanions

Oxyanions are polyatomic anions containing oxygen. When naming them, include the oxidation state of the central non-oxygen atom as a Roman numeral in brackets at the end of the name, with no space between the name and the bracket.

For example:

• MnO₄^- is called manganate(VII) because the oxidation state of Mn is +7. In the neutral compound KMnO₄, this becomes potassium manganate(VII), also known by its common name, potassium permanganate.
• The central atom can also be a non-metal, as in KClO₃, which is called potassium chlorate(V).

For common oxyanions, the Roman numeral is often omitted in the name, as shown in the table below.

Worked example 3 - Determining the formula of copper(II) nitrate

Determine the chemical formula of copper(II) nitrate.

Step 1: Identify oxidation numbers from the name

• Copper(II) indicates copper has an oxidation number of +2.
• The nitrate ion (NO_3_^-^) has a charge of -1.

Step 2: Set up ratio to balance charges

The overall charge of the compound must equal 0. Since copper has an oxidation number of +2 (Cu^2+^) and each nitrate ion has a charge of -1 (NO_3_^-^), we need two nitrate ions to balance one copper ion.

Step 3: Determine the formula

For the charges to balance to zero, one copper ion needs to pair with two nitrate ions.

Thus, the chemical formula for copper(II) nitrate is Cu(NO_3_)_2_.

Worked example 4 - Determining the name for the ClO_2_^-^ ion

Determine the systematic name of the ClO_2_^−^ ion.

Step 1: Assign known oxidation numbers

Oxygen (O) has an oxidation number of -2.

Step 2: Calculate total oxidation number of O

There are 2 oxygen atoms, each with an oxidation number of -2.

Total oxidation number from oxygen = 2 x (-2) = -4.

Step 3: Calculate the oxidation number of Cl

Let the oxidation number of chlorine be x. The sum of the oxidation numbers in the ion equals the overall charge of the ion = -1. Therefore,

x + (-4) = -1

x = -1 + 4

x = 3

Step 4: Determine the systematic name

Since the oxidation number of chlorine in ClO₂^- is +3, its systematic name is chlorate(III).