The Sum Rule for Polyatomic Ions

Introduction

The oxidation numbers of atoms in a polyatomic ion add to the ion's overall charge, not to zero. This is the neutral-compound sum rule with a different target. It allows a central atom's formal oxidation number to be calculated from the ion formula, charge and usual values of the other atoms.

Core explanation

In sulfate, SO₄²⁻, oxygen is usually −2. Four oxygen atoms contribute −8. Let sulfur be x. Then x + 4(−2) = −2, so x − 8 = −2 and x = +6. The sulfur is not −2; −2 is the charge of the entire five-atom ion. The answer checks because +6 − 8 = −2.

For nitrate, NO₃⁻, three oxygens contribute −6. Let nitrogen be x: x − 6 = −1, so x = +5. In nitrite, NO₂⁻, two oxygens contribute −4: x − 4 = −1, so nitrogen is +3. Comparing nitrate and nitrite gives a two-unit difference in nitrogen's formal oxidation number, even though both ions have overall charge −1. Total ion charge alone cannot reveal the atom-level change.

For carbonate, CO₃²⁻, carbon satisfies x + 3(−2) = −2, so carbon is +4. This equals carbon's number in neutral CO₂, despite the species and charge being different. Therefore converting CO₂ to carbonate in an acid–base process does not by itself imply carbon redox. Comparing the central atom's formal number is more informative than comparing molecular charge.

In ammonium, NH₄⁺, hydrogen is usually +1. Four H atoms contribute +4, so nitrogen satisfies x + 4 = +1 and has −3. Ammonia NH₃ also has nitrogen −3. Protonating ammonia to ammonium changes the species's charge but not nitrogen's oxidation number. This is an especially useful contrast: a change in whole-species charge does not automatically mean electron transfer.

Charge notation is easy to misread. NO₃⁻ has one negative charge on the whole ion, not one negative charge per oxygen. CO₃²⁻ has two negative units in total, not two on carbon. Put the ionic charge at the right side of the algebraic equation once. Then multiply oxidation numbers by their atom subscripts on the left.

As with neutral compounds, usual oxidation-number shortcuts have exceptions. Oxygen in peroxides is not −2, and complex ions can have unusual assignments. The sum rule itself remains valid. If a proposed set of values fails to sum to the ion's charge, review the assumptions and species formula rather than forcing an answer.

Step-by-step reasoning

1. Read the full polyatomic formula and its signed overall charge. 2. Assign usual oxidation numbers to elements with known values. 3. Multiply each by its atom count. 4. Add the unknown contribution and set the sum equal to ion charge. 5. Solve, substitute back, and compare with related species if classifying redox.

Visual explanation

Draw a large bracket around SO₄ with a 2− label outside. Inside, put one box for S marked x and four O boxes each marked −2. Under the bracket show x + 4(−2) = −2. The placement of the charge outside the bracket visually separates whole-ion charge from sulfur's individual value.

Real-world analogy

A team's net score may be −2 even when one player contributes +6 and four others contribute −2 each. The team total does not identify one player's contribution. A polyatomic ion's charge similarly constrains the sum but is not automatically any one atom's oxidation number.

Real-world example

Nitrate and nitrite ions occur in environmental and biological nitrogen chemistry. They have the same net charge, −1, yet nitrogen is +5 in nitrate and +3 in nitrite under usual oxygen assignments. Their interconversion can therefore involve a nitrogen redox change even when the written ion charges look identical.

Why?

Why set the sum equal to ionic charge? Oxidation numbers allocate electrons formally among the atoms but must preserve the net charge of the whole species. A one-negative ion must have atomic assignments summing to −1, regardless of how many atoms it contains.

Common misconception

“Nitrogen is −1 in nitrate because NO₃⁻ has charge −1.” The −1 belongs to all four atoms together. With three oxygens at −2, nitrogen must be +5 to make the total −1.

Worked example

Calculate chromium's oxidation number in chromate, CrO₄²⁻, under the usual oxygen −2 rule. Four oxygens total −8. Let chromium be x; x − 8 = −2, so x = +6. Check +6 + 4(−2) = −2. This formal number does not mean a free Cr⁶⁺ ion is present. In a later redox equation, compare chromium's product oxidation number to +6 to classify its change.

Quick check

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

Exam focus

Put the ion charge on the right side of the sum equation exactly once. Multiply usual atomic values by subscripts. Distinguish central-atom oxidation number from the overall charge and verify by substitution.

Advanced insight

The carbonate–carbon dioxide and ammonium–ammonia comparisons show that proton or charge changes can occur without redox. Formal oxidation-state tracking separates acid–base transformations from electron-transfer transformations even when both change species formulas.

Summary

For a polyatomic ion, weighted oxidation numbers sum to the ion's charge. Sulfur is +6 in sulfate, nitrogen +5 in nitrate and +3 in nitrite. Whole-ion charge is not a single atom's number; algebra and the usual assignment rules determine the atomic values.

Practice questions

1. Calculate sulfur's number in SO₄²⁻. Answer: +6, because x + 4(−2) = −2. 2. Calculate carbon's number in CO₃²⁻. Answer: +4, because x + 3(−2) = −2. 3. Calculate nitrogen's number in NH₄⁺. Answer: −3, because x + 4(+1) = +1. 4. Does conversion of NH₃ to NH₄⁺ change nitrogen's oxidation number? Answer: No. Nitrogen remains −3, so the protonation itself is not nitrogen redox.