Metal Oxidation States Practice- Problems and Solutions

What You Need to Know About Metal Oxidation States

Oxidation states are one of those foundational concepts that trip up chemistry students constantly. You memorize the rules, you think you understand them, and then you hit a compound like MnO₄^- or Fe₃O₄ and everything falls apart.

This guide cuts through the confusion. You'll get practice problems with real solutions, a comparison table of common metal oxidation states, and a method that actually works for assigning oxidation numbers to any metal in any compound.

The Quick Rules for Assigning Oxidation States

Before diving into problems, you need these rules locked in your memory. No exceptions.

• Free elements have an oxidation state of zero (Na, O₂, Fe, etc.)
• Monatomic ions have oxidation states equal to their charge (Na⁺ = +1, O^2- = -2)
• Oxygen is usually -2 (except in peroxides where it's -1, and with fluorine)
• Hydrogen is +1 (except in metal hydrides where it's -1)
• The sum of oxidation states equals the charge of the compound
• Alkali metals are always +1 in compounds
• Alkaline earth metals are always +2 in compounds

That's it. These seven rules solve 90% of oxidation state problems. Memorize them.

Common Metal Oxidation States Reference Table

Use this table when you're stuck. It shows the most stable oxidation states for transition metals you'll encounter most often.

Transition metals are tricky because they have multiple possible oxidation states. That's why you can't just memorize one number—you have to work it out using the rules.

How to Determine Oxidation States: The Method That Works

Step 1: Identify the easy atoms first

Assign oxidation states to O (-2), H (+1), and alkali/alkaline earth metals (+1 or +2) immediately. These rarely change.

Step 2: Set up your equation

Write out the sum of all oxidation states equals the charge of the compound. For neutral compounds, this equals zero.

Step 3: Solve for the unknown

Plug in your known values and solve for the metal's oxidation state. Basic algebra.

Step 4: Check your work

Verify your answer makes sense. Does the oxidation state match a known stable state for that metal? If you get something weird like +7 for iron, you made a mistake.

Practice Problems and Solutions

Work through each problem before checking the solution. No peeking.

Problem 1: Find the oxidation state of Mn in KMnO₄

Solution:

K is +1 (alkali metal). O is -2 (there are 4 oxygens, so -8 total). Let x = oxidation state of Mn.

+1 + x + (-8) = 0
x = +7

The oxidation state of Mn in KMnO₄ is +7. This is why permanganate is such a strong oxidizing agent.

Problem 2: Find the oxidation state of Fe in Fe₂O₃

Solution:

O is -2 (3 oxygens = -6 total). There are two iron atoms, so 2x.

2x + (-6) = 0
2x = +6
x = +3

Iron is in the +3 oxidation state in Fe₂O₃ (hematite ore). This is the more stable form of iron oxide.

Problem 3: Find the oxidation state of Cr in K₂Cr₂O₇

Solution:

Two K atoms = +2. Seven O atoms = -14. Let x = oxidation state of Cr. There are two Cr atoms.

+2 + 2x + (-14) = 0
2x = +12
x = +6

Each chromium atom is in the +6 oxidation state. Dichromate is a powerful oxidizer because of this.

Problem 4: Find the oxidation state of Pb in PbO₂

Solution:

O is -2 (2 oxygens = -4 total). Let x = oxidation state of Pb.

x + (-4) = 0
x = +4

Lead is +4 in PbO₂. This is the lead in lead-acid battery anodes.

Problem 5: Find the oxidation state of Cu in Cu₂O

Solution:

O is -2. Two Cu atoms, so 2x.

2x + (-2) = 0
2x = +2
x = +1

Each copper is +1 in Cu₂O. This is the red oxide of copper, used in some pigments and ceramics.

Problem 6: Find the oxidation state of Mn in MnO₂

Solution:

O is -2 (2 oxygens = -4 total). Let x = oxidation state of Mn.

x + (-4) = 0
x = +4

Manganese is +4 in MnO₂. This is the manganese dioxide used in dry cell batteries.

Problem 7: Find the oxidation state of Hg in Hg₂Cl₂

Solution:

Cl is -1 (2 chlorines = -2 total). This is tricky—Hg₂Cl₂ contains the Hg₂²⁺ dimer ion.

Each Hg has a +1 oxidation state, giving the dimer a +2 charge which balances the -2 from chlorine.

The oxidation state of Hg in Hg₂Cl₂ is +1. This compound is called calomel.

Common Mistakes to Avoid

• Forgetting the coefficient matters — Na₂O means 2 sodium atoms. Count every atom.
• Missing polyatomic ion charges — SO₄^2- has a -2 charge that doesn't come from the sulfur alone.
• Confusing oxidation state with ionic charge — Fe₂O₃ isn't made of Fe³⁺ ions. It's covalent with shared electrons counted differently.
• Ignoring the compound charge — For ions like MnO₄^-, the sum equals -1, not 0.
• Assuming metals always have the same oxidation state — Transition metals are variable. Work it out every time.

Quick Drill: Self-Test Questions

Try these without looking at the table or rules. Answers below.

1. What is the oxidation state of Zn in ZnCl₂?
2. What is the oxidation state of Sn in SnCl₄?
3. What is the oxidation state of Cr in Cr₂O₃?
4. What is the oxidation state of Mn in MnCl₂?
5. What is the oxidation state of Pb in PbCl₂?

Answers:

1. Zn = +2
2. Sn = +4
3. Cr = +3
4. Mn = +2
5. Pb = +2

Got those wrong? Go back to the rules section and work through Problem 1 again using the same method.

When This Gets Complicated

Some compounds break the simple rules. Here's how to handle them:

• Peroxides (O₂^2-) — Each oxygen is -1, not -2. Examples: H₂O₂, Na₂O₂.
• Superoxides (O₂^-) — Each oxygen is -½. Examples: KO₂, NaO₂.
• Metal hydrides — Hydrogen is -1 when bonded to active metals (NaH, CaH₂).
• Interhalogen compounds — Work from the known halogen charges.

These exceptions are limited. Once you recognize the pattern, they're just as solvable as the simple problems.

The Bottom Line

Oxidation state problems are algebra problems in disguise. Identify what you know, set up the equation, solve for the unknown. That's the entire process.

Most mistakes come from three sources: not counting atoms correctly, forgetting the compound charge, or not knowing basic oxidation rules. Fix those three things and you'll solve any oxidation state problem they throw at you.

Work through the practice problems until you can do them without the rules. Then you're ready.