Oxidation Numbers in Ions

Introduction

A polyatomic ion can carry an overall charge while its atoms have different assigned oxidation numbers. Sulfate is SO₄²⁻: its charge is −2, but sulfur is not −2. The oxygen atoms account for much of the formal charge sum, leaving sulfur at +6 under the usual rule. Keeping net ion charge separate from each atom's oxidation number is the central skill of this page.

Core explanation

A monatomic ion's oxidation number equals its charge. Fe³⁺ has iron at +3, and S²⁻ has sulfur at −2. For a polyatomic ion, the oxidation numbers of all atoms sum to the ion's net charge. If an ion carries a −1 charge, the sum is −1, not zero. Multiply each element's assigned oxidation number by its atom count before summing.

In sulfate, SO₄²⁻, oxygen is normally −2. Four oxygens contribute 4(−2) = −8. Let sulfur be x: x − 8 = −2, so x = +6. In sulfite, SO₃²⁻, three oxygens contribute −6; x − 6 = −2 gives sulfur +4. Comparing the two ions shows that an element's formal state can change even when the total ionic charge stays the same.

In nitrate, NO₃⁻, N + 3(−2) = −1, giving N = +5. In nitrite, NO₂⁻, N + 2(−2) = −1 gives N = +3. These are common electron-accounting patterns in redox reactions. It would be incorrect to assign nitrogen −1 simply because the whole ion has charge −1. The ion charge belongs to the group, not automatically to one atom.

In ammonium, NH₄⁺, hydrogen is normally +1 because it is bonded to a nonmetal. N + 4(+1) = +1 gives nitrogen −3. The same ion contains a positively charged whole group and a formally negative nitrogen oxidation state. This is not a contradiction; oxidation number is a formal allocation and net charge is the sum of all assigned contributions.

Parentheses in formulas require care. In a neutral compound (NH₄)₂SO₄, there are two ammonium ions of charge +1 each and one sulfate ion of charge −2. The sulfur oxidation number remains +6 and nitrogen −3 in the respective ions. Alternatively, count every atom in the full formula: two N, eight H, one S and four O. The charge sum is 2(−3) + 8(+1) + 6 + 4(−2) = 0. Grouping known polyatomic ions often simplifies the work.

An atom's oxidation number can be different from its oxidation number in another ion with the same element. Manganese is +7 in permanganate MnO₄⁻ because x − 8 = −1, whereas chromium is +6 per atom in dichromate Cr₂O₇²⁻ because 2x − 14 = −2. The factor of two before chromium is essential. These ions appear often in acidic redox balancing, so their values are worth deriving rather than memorising without a check.

Some ions contain peroxo O–O bonds, for which oxygen is not −2. A charge-sum equation using the wrong assumed oxygen value can produce a plausible but incorrect central-atom number. Identify structural exceptions before solving. Formal oxidation numbers may also average over inequivalent sites if the formula alone lacks structural information, so do not overinterpret one average as a measured charge on every atom.

Step-by-step reasoning

1. Write the complete ion formula and its net charge. 2. Assign usual values to atoms where rules apply, checking known exceptions. 3. Multiply each value by the number of that atom in the ion. 4. Set the sum equal to the ion charge and solve unknown values. 5. Re-add all contributions to verify the net charge.

Visual explanation

Draw a sulfate bracket around S and four O labels, with “overall −2” written outside. Under it list S = x and 4O = −8, then write x − 8 = −2 and x = +6. Use a second bracket for NH₄⁺ to show N = −3 even though the bracket carries +1.

Real-world analogy

A team's net score can be −2 even if one member scores +6 and others total −8. The group score is the sum, not each member's score. Oxidation-number arithmetic works that way, though atomic values are formal chemical assignments rather than individual measured scores.

Real-world example

Wastewater chemistry may involve nitrate and nitrite ions. Their overall charges are both −1, yet nitrogen has oxidation numbers +5 and +3. Transforming nitrate to nitrite is therefore a reduction of nitrogen despite no change in the written net charge of the ion type.

Why?

Why is sulfate sulfur +6 rather than −2? Oxygen contributes four times −2 = −8, and the total ion is only −2. Sulfur's +6 contribution is needed to bring the formal sum back to −2.

Common misconception

“The oxidation number of the central atom equals the polyatomic ion's charge.” The charge applies to the whole ion. Each atom contributes a formal value, and their atom-count-weighted sum equals the charge.

Worked example

Find chromium's average oxidation number in Cr₂O₇²⁻. Oxygen normally contributes 7(−2) = −14. Let each equivalent chromium contribution be x: 2x − 14 = −2. Then 2x = 12 and x = +6. Check: 2(+6) + 7(−2) = 12 − 14 = −2. Dropping the chromium subscript would incorrectly give +12. The +6 is formal electron accounting, not an isolated Cr⁶⁺ ion floating inside dichromate.

Quick check

1. What is nitrogen's oxidation number in NO₂⁻ when oxygen is −2? Answer: +3, because N + 2(−2) = −1.

Exam focus

Set the charge-sum equation equal to the ion's net charge and include subscripts. Distinguish monatomic ions, whose oxidation number equals charge, from polyatomic ions, whose individual atom values usually differ from the group's charge.

Advanced insight

Ions such as superoxide O₂⁻ can give fractional average oxidation numbers, −1/2 per oxygen under a symmetry-based assignment. This does not mean each oxygen carries a literal half-electron charge. Bonding and electronic structure determine how the formal average should be interpreted.

Summary

Monatomic ions have oxidation number equal to ionic charge; polyatomic ions have atom-count-weighted oxidation numbers summing to the group's charge. Sulfate, nitrate, ammonium and dichromate illustrate why the central atom's value must be solved rather than copied from the ion charge.

Practice questions

1. Find sulfur's oxidation number in SO₃²⁻. Answer: +4, because S + 3(−2) = −2. 2. Find nitrogen's oxidation number in NH₄⁺. Answer: −3, because N + 4(+1) = +1. 3. What is manganese's oxidation number in MnO₄⁻? Answer: +7, because Mn + 4(−2) = −1.