How to Find Oxidation Numbers in Polyatomic Ions
Learn how to calculate oxidation numbers in polyatomic ions using known oxidation states, atom counts, and overall ionic charge. Includes worked examples, common exceptions, electron-transfer explanations, and HSC-style practice questions with solutions.
In sulfate, \(\mathrm{SO_{4}}^{2-}\), the whole ion has a charge of \(-2\). Yet sulfur has an oxidation number of \(+6\).
How can an atom have a positive oxidation number inside a negatively charged ion?
Predict the answer before reading on. Oxygen usually has an oxidation number of \(-2\). Four oxygen atoms therefore contribute \(-8\). For the entire ion to have a charge of \(-2\), sulfur must contribute \(+6\).
That is the key method for finding oxidation numbers in polyatomic ions: assign the known oxidation numbers, multiply each by the number of atoms present, and make their sum equal the overall charge of the ion.
The charge belongs to the entire ion. Oxidation numbers belong to individual atoms.
01The rule for polyatomic ions
A polyatomic ion contains two or more bonded atoms and carries an overall electrical charge.
Examples include:
- Sulfate: \(\mathrm{SO_{4}}^{2-}\)
- Nitrate: \(\mathrm{NO_{3}}^{-}\)
- Ammonium: \(\mathrm{NH_{4}}^{+}\)
- Carbonate: \(\mathrm{CO_{3}}^{2-}\)
- Permanganate: \(\mathrm{MnO_{4}}^{-}\)
- Dichromate: \(\mathrm{Cr_{2}O_{7}}^{2-}\)
Each atom has an assigned oxidation number, sometimes called its oxidation state.
The fundamental rule is:
The sum of all oxidation numbers in a polyatomic ion must equal the overall charge of that ion.
Mathematically:
\[ \sum_i n_i \times \mathrm{ON}_i=q \]
Where:
- \(n_i\) is the number of atoms of a particular element.
- \(\mathrm{ON}_i\) is the oxidation number assigned to that element.
- \(q\) is the overall charge of the ion.
Oxidation numbers and ionic charges have no units.
Think of the calculation as a group account. Each atom contributes an assigned positive or negative value, and the total must match the ion’s charge.
The analogy has a limit: oxidation numbers are formal bookkeeping values based on how bonding electrons are assigned. They are not necessarily the actual electrical charges on individual atoms.
Why the overall charge matters
Consider sulfate again:
\[ \mathrm{SO_{4}}^{2-} \]
There is:
- One sulfur atom.
- Four oxygen atoms.
- An overall charge of \(-2\).
Oxygen has an oxidation number of \(-2\).
Therefore:
\[ x+4(-2)=-2 \]
\[ x-8=-2 \]
\[ \boxed{x=+6} \]
Sulfur has an oxidation number of \(+6\).
Notice that the four oxygen atoms contribute \(-8\), not \(-2\). The ion’s overall charge is the sum of every atom’s oxidation number.
This is the calculation you will repeat throughout the topic.
02The oxidation number rules you need
Before solving an unfamiliar ion, identify which oxidation numbers you already know.
These rules allow you to calculate the remaining unknown value.
| Element or species | Usual oxidation number | Important condition |
|---|---|---|
| Free element | 0 | For example, \(\ce{O2}\), \(\ce{Cl2}\), and \(\ce{Fe}\) |
| Group 1 metals | +1 | In their compounds |
| Group 2 metals | +2 | In their compounds |
| Fluorine | −1 | In its compounds |
| Oxygen | −2 | Except in certain compounds, including peroxides |
| Hydrogen | +1 | When bonded to nonmetals |
| Hydrogen | −1 | In metal hydrides |
| Other halogens | Usually −1 | Can be positive when bonded to oxygen or fluorine |
| Single-atom ion | Equal to its charge | For example, \(\mathrm{Fe}^{3+}\) has oxidation number +3 |
For most HSC-style polyatomic ion questions, oxygen and hydrogen are the values you will use most frequently.
Two rules deserve particular attention:
Oxygen is usually \(-2\). This applies to sulfate, nitrate, carbonate, phosphate, and many other common ions.
Hydrogen is usually \(+1\). This is particularly useful when working with ions such as ammonium, \(\mathrm{NH_{4}}^{+}\).
However, these are rules with exceptions. You must check the chemical formula before assigning values automatically.
03A reliable five-step method
When you encounter an unfamiliar polyatomic ion, follow this process.
Step 1: Write down the overall charge.
Read the superscript at the top right of the formula.
For example:
\[ \mathrm{SO_{4}}^{2-} \quad q=-2 \]
\[ \mathrm{NO_{3}}^{-} \quad q=-1 \]
Do not confuse the overall charge with an individual atom’s oxidation number.
Step 2: Assign the oxidation numbers you already know.
Start with the familiar rules.
Usually, this means assigning oxygen \(-2\) or hydrogen \(+1\).
Step 3: Let the unknown oxidation number equal \(x\).
If you are finding sulfur, use \(x\) for sulfur.
If there are two sulfur atoms, their combined contribution is \(2x\), assuming both have the same oxidation number.
Step 4: Make the sum equal the overall ion charge.
Multiply each oxidation number by the number of atoms of that element.
Then set the sum equal to the charge.
Step 5: Solve and check.
Rearrange the equation to find \(x\).
Substitute your answer back into the original expression to confirm that the total equals the ion’s charge.
If it doesn’t, check your signs and atom counts.
04Worked examples: applying the method
Worked example: Find nitrogen in nitrate
Determine the oxidation number of nitrogen in the nitrate ion, \(\mathrm{NO_{3}}^{-}\).
Step 1
Oxygen has an oxidation number of \(-2\).
There are three oxygen atoms, so their combined contribution is:
\[ 3(-2)=-6 \]
Step 2
There is one nitrogen atom.
The total oxidation number must equal the ion’s charge of \(-1\):
\[ x+3(-2)=-1 \]
Step 3
\[ x-6=-1 \]
\[ x=+5 \]
Step 4
\[ (+5)+3(-2)=-1 \]
The calculation matches the overall charge.
Therefore:
\[ \boxed{\mathrm{N}=+5} \]
Nitrogen has an oxidation number of \(+5\) in nitrate.
Notice that nitrate still has an overall charge of \(-1\). The \(+5\) belongs to nitrogen, not the entire ion.
Worked example: Find chromium in dichromate
Determine the oxidation number of chromium in the dichromate ion, \(\mathrm{Cr_{2}O_{7}}^{2-}\).
This example introduces an extra step: there are two chromium atoms, not one.
Step 1
Oxygen is \(-2\).
There are seven oxygen atoms, giving:
\[ 7(-2)=-14 \]
Step 2
There are two chromium atoms, so their combined contribution is \(2x\).
The total must equal the ion’s charge of \(-2\):
\[ 2x+7(-2)=-2 \]
Step 3
\[ 2x-14=-2 \]
\[ 2x=12 \]
\[ x=+6 \]
Step 4
\[ 2(+6)+7(-2) \]
\[ =12-14 \]
\[ =-2 \]
Therefore:
\[ \boxed{\mathrm{Cr}=+6} \]
Each chromium atom has an oxidation number of \(+6\).
A common mistake is to stop at \(2x=12\) and report \(+12\). That is the combined contribution of both chromium atoms, not the oxidation number of one chromium atom.
05The important exception: oxygen is not always −2
You might now be tempted to assign \(-2\) to every oxygen atom you encounter.
That usually works, but there is an important exception: peroxides.
Consider the peroxide ion:
\[ \ce{O2^2-} \]
There are two oxygen atoms, and the overall charge is \(-2\).
Let the oxidation number of each oxygen atom equal \(x\).
\[ 2x=-2 \]
\[ \boxed{x=-1} \]
Each oxygen atom has an oxidation number of \(-1\), not \(-2\).
Why?
The two oxygen atoms are bonded to each other. Because they are identical elements, the bonding electrons in the O-O bond are shared equally when assigning oxidation numbers.
Compare the two species:
| Species | Oxidation number of oxygen |
|---|---|
| Oxide ion, \(\mathrm{O}^{2-}\) | −2 |
| Peroxide ion, \(\ce{O2^2-}\) | −1 per oxygen |
The important lesson is to check whether a known oxidation-number rule has an exception before using it.
A further point: average oxidation numbers
There is another situation worth recognising.
If a polyatomic ion contains two or more atoms of the same element, the sum rule can give their average oxidation number. That average does not necessarily describe every individual atom if they occupy different bonding environments.
For example, in the thiosulfate ion, \(\mathrm{S_{2}O_{3}}^{2-}\):
\[ 2x+3(-2)=-2 \]
\[ 2x=4 \]
\[ x=+2 \]
The average oxidation number of sulfur is \(+2\).
However, the two sulfur atoms are not in equivalent bonding environments. Therefore, the average does not automatically mean that each individual sulfur atom has oxidation number \(+2\).
For formula-only questions, use the sum rule to obtain the average. Only assign different values to individual atoms when the question provides enough structural information.
06Using oxidation numbers to identify electron transfer
Finding oxidation numbers is not just an isolated calculation. It gives you a way to track what happens to electrons during a redox reaction.
The key relationships are:
| Change in oxidation number | Process | Electron accounting |
|---|---|---|
| Increases | Oxidation | Electrons lost |
| Decreases | Reduction | Electrons gained |
| Stays the same | No oxidation-state change | No net electron transfer indicated |
For example:
\[ \ce{Fe^2+ -> Fe^3+} \]
Iron changes from \(+2\) to \(+3\).
Its oxidation number increases by one, meaning it loses one electron in oxidation-state bookkeeping.
Now consider:
\[ \ce{MnO4^- -> Mn^2+} \]
We first need to calculate manganese’s oxidation number in permanganate.
Let it equal \(x\):
\[ x+4(-2)=-1 \]
\[ x-8=-1 \]
\[ x=+7 \]
Manganese changes from \(+7\) to \(+2\).
Its oxidation number decreases by five, meaning each manganese atom gains five electrons.
This gives us a useful way to compare the electron changes.
Worked example: Permanganate reacting with iron(II)
In acidic solution, permanganate ions react with iron(II) ions to produce manganese(II) ions and iron(III) ions.
The balanced ionic equation is:
\[ \ce{MnO4^- + 5Fe^2+ + 8H^+ -> Mn^2+ + 5Fe^3+ + 4H2O} \]
Identify the oxidation and reduction processes, calculate the total number of electrons transferred, and verify that the equation conserves atoms and charge.
Step 1
Let the oxidation number equal \(x\).
\[ x+4(-2)=-1 \]
\[ x=+7 \]
In the product, manganese is \(\mathrm{Mn}^{2+}\), so its oxidation number is \(+2\).
The change is:
\[ +7\rightarrow +2 \]
Manganese’s oxidation number decreases by five.
Each manganese atom therefore gains five electrons. Manganese is reduced.
Step 2
Iron begins as \(\mathrm{Fe}^{2+}\) and becomes \(\mathrm{Fe}^{3+}\).
\[ +2\rightarrow +3 \]
Each iron ion loses one electron.
Iron is oxidised.
Step 3
The equation contains one permanganate ion, so one manganese atom gains five electrons.
There are five iron(II) ions, each losing one electron.
Therefore:
\[ \text{Electrons gained}=1\times5=5 \]
\[ \text{Electrons lost}=5\times1=5 \]
The electrons gained and lost match.
This is essential. Electrons cannot simply disappear from the overall reaction.
Step 4
| Element | Reactants | Products |
|---|---|---|
| Mn | 1 | 1 |
| Fe | 5 | 5 |
| O | 4 | 4 |
| H | 8 | 8 |
Every element has the same number of atoms on both sides.
Step 5
On the reactant side:
\[ -1+5(+2)+8(+1)=+17 \]
On the product side:
\[ (+2)+5(+3)=+17 \]
The total charge is \(+17\) on both sides.
The equation therefore conserves both atoms and charge.
The important distinction is that the oxidation number of manganese in permanganate is always \(+7\), regardless of whether the solution is acidic or alkaline.
However, the product formed when permanganate is reduced can depend on the reaction conditions. In acidic solution, manganese commonly forms \(\mathrm{Mn}^{2+}\). Do not assume the same product in a different medium.
If you want to develop this into complete redox equations, see Oxidation numbers in redox reactions: HSC Chemistry Guide.
For the next stage, see How to Balance a Redox Half-Equation in Acidic Solution.
07Common mistakes to avoid
Before practising, make sure you can recognise these mistakes.
| Mistake | Why it is wrong | Correct approach |
|---|---|---|
| Assigning the ion’s charge to its central atom | The charge belongs to the entire ion | Make the sum of oxidation numbers equal the ion charge |
| Forgetting subscripts | Every atom contributes to the total | Multiply by the number of atoms present |
| Assuming oxygen is always −2 | Peroxides are an important exception | Check the chemical species first |
| Dividing the overall charge equally between atoms | Oxidation numbers are assigned using established rules | Assign known values before solving for unknowns |
| Reporting the combined value for several atoms | The question usually asks for the value per atom | Divide the combined contribution by the number of equivalent atoms |
| Changing subscripts to balance an equation | This changes the chemical species | Use coefficients to change the number of particles |
| Ignoring the reaction medium | Different conditions can produce different redox products | Use the products and conditions stated in the question |
One final distinction is particularly important.
A coefficient tells you how many whole particles are involved. A subscript tells you how many atoms are inside each particle.
For example:
\[ \ce{3SO4^2-} \]
means three sulfate ions.
Each sulfate ion still contains one sulfur atom and four oxygen atoms. Its oxidation numbers do not change just because a coefficient has been placed in front.
08Questions and solutions
The following questions move from straightforward oxidation-number calculations to more demanding redox reasoning. The final question combines oxidation numbers, electron transfer, and conservation of atoms and charge.
Question 1
Calculate the oxidation number of:
a. Carbon in carbonate, \(\mathrm{CO_{3}}^{2-}\).
b. Phosphorus in phosphate, \(\mathrm{PO_{4}}^{3-}\).
Show your working and verify both answers using the overall ionic charges.
Solution 1
Carbon has an oxidation number of \(+4\) in carbonate, and phosphorus has an oxidation number of \(+5\) in phosphate.
Part a: Carbonate
Oxygen has an oxidation number of \(-2\).
Let carbon equal \(x\).
\[ x+3(-2)=-2 \]
\[ x-6=-2 \]
\[ \boxed{x=+4} \]
Check:
\[ (+4)+3(-2)=-2 \]
This matches the carbonate ion’s overall charge.
Part b: Phosphate
Again, oxygen has an oxidation number of \(-2\).
Let phosphorus equal \(x\).
\[ x+4(-2)=-3 \]
\[ x-8=-3 \]
\[ \boxed{x=+5} \]
Check:
\[ (+5)+4(-2)=-3 \]
This matches the phosphate ion’s overall charge.
The important technique in both questions is to use the number of oxygen atoms before solving for the central atom.
Question 2
Two polyatomic ions are provided:
\[ \mathrm{NH_{4}}^{+} \qquad \mathrm{NO_{3}}^{-} \]
a. Determine the oxidation number of nitrogen in each ion.
b. Explain why nitrogen can have different oxidation numbers in these two species.
c. A student argues that because nitrogen changes from \(-3\) to \(+5\) when comparing the two ions, an electron-transfer reaction must have occurred. Explain why this conclusion is not justified by the information provided.
Solution 2
Nitrogen has an oxidation number of \(-3\) in ammonium and \(+5\) in nitrate. However, the presence of both ions does not, by itself, demonstrate that a redox reaction has occurred.
Part a: Ammonium
Hydrogen has an oxidation number of \(+1\).
Let nitrogen equal \(x\).
\[ x+4(+1)=+1 \]
\[ x+4=1 \]
\[ \boxed{x=-3} \]
Part b: Nitrate
Oxygen has an oxidation number of \(-2\).
Let nitrogen equal \(x\).
\[ x+3(-2)=-1 \]
\[ x-6=-1 \]
\[ \boxed{x=+5} \]
Part c: Interpreting the difference
Nitrogen has different oxidation numbers because it is part of two different chemical species.
In ammonium, nitrogen is bonded to hydrogen atoms. In nitrate, nitrogen is bonded to oxygen atoms.
The oxidation number depends on the chemical environment and the relative assignment of bonding electrons.
The tempting conclusion is that a change from \(-3\) to \(+5\) automatically proves electron transfer.
But these are two separate species, not necessarily the reactant and product of a reaction.
To identify oxidation and reduction, you need to know how oxidation numbers change between reactants and products in an actual chemical reaction.
A difference between two oxidation numbers is not, on its own, evidence that a reaction occurred.
Question 3
In acidic solution, dichromate ions react with iron(II) ions to form chromium(III) and iron(III) ions.
The balanced ionic equation is:
\[ \ce{Cr2O7^2- + 14H^+ + 6Fe^2+ -> 2Cr^3+ + 7H2O + 6Fe^3+} \]
a. Calculate the oxidation number of chromium in dichromate.
b. Determine the oxidation-number change of chromium and calculate the total number of electrons gained.
c. Determine the oxidation-number change of iron and calculate the total number of electrons lost.
d. Verify that the equation conserves total charge.
e. Explain why the hydrogen ions do not contribute to the oxidation-number calculation for chromium.
Solution 3
Chromium has an oxidation number of \(+6\) in dichromate and is reduced to \(+3\), gaining six electrons in total. Iron is oxidised from \(+2\) to \(+3\), losing six electrons in total.
Part a: Chromium in dichromate
Oxygen has an oxidation number of \(-2\).
Let the oxidation number of each chromium atom equal \(x\).
There are two chromium atoms and seven oxygen atoms:
\[ 2x+7(-2)=-2 \]
\[ 2x-14=-2 \]
\[ 2x=12 \]
\[ \boxed{x=+6} \]
Part b: Chromium reduction
Chromium changes from:
\[ +6\rightarrow +3 \]
The oxidation number decreases by three.
Each chromium atom gains three electrons.
However, the dichromate ion contains two chromium atoms.
Therefore:
\[ 2\times3=6\text{ electrons gained} \]
Chromium is reduced.
Part c: Iron oxidation
Iron changes from:
\[ +2\rightarrow +3 \]
Each iron(II) ion loses one electron.
There are six iron(II) ions in the equation.
Therefore:
\[ 6\times1=6\text{ electrons lost} \]
Iron is oxidised.
The total electron accounting is consistent:
\[ 6\text{ electrons gained} = 6\text{ electrons lost} \]
Part d: Charge conservation
Calculate the total charge on the reactant side:
\[ -2+14(+1)+6(+2) \]
\[ =-2+14+12 \]
\[ =+24 \]
Now calculate the product-side charge:
\[ 2(+3)+6(+3) \]
\[ =6+18 \]
\[ =+24 \]
Both sides have a total charge of \(+24\).
The atoms are also balanced: two chromium atoms, seven oxygen atoms, fourteen hydrogen atoms, and six iron atoms appear on each side.
Part e: The role of hydrogen ions
The hydrogen ions are present because the reaction occurs in acidic solution.
They contribute to the overall charge and hydrogen balance, but they are not part of the dichromate ion itself.
When calculating chromium’s oxidation number, use only the atoms within \(\mathrm{Cr_{2}O_{7}}^{2-}\) and its overall charge of \(-2\).
Including the fourteen hydrogen ions in that calculation would incorrectly treat separate reactant species as part of the dichromate ion.
Question 4
In acidic solution, permanganate ions react with oxalate ions to produce manganese(II) ions and carbon dioxide.
The balanced ionic equation is:
\[ \ce{2MnO4^- + 5C2O4^2- + 16H^+ -> 2Mn^2+ + 10CO2 + 8H2O} \]
a. Determine the average oxidation number of carbon in oxalate.
b. Determine the oxidation number of carbon in carbon dioxide.
c. Calculate the number of electrons lost by each oxalate ion.
d. Calculate the total number of electrons transferred from all five oxalate ions.
e. Show that the manganese reduction accounts for the same number of electrons.
f. Verify the total charge on both sides.
Solution 4
Carbon has an oxidation number of \(+3\) in oxalate and \(+4\) in carbon dioxide. Each oxalate ion loses two electrons, giving ten electrons lost across five oxalate ions. The two manganese atoms gain ten electrons in total.
Part a: Carbon in oxalate
Oxygen has an oxidation number of \(-2\).
Let the average oxidation number of carbon equal \(x\).
There are two carbon atoms and four oxygen atoms.
The overall charge is \(-2\).
\[ 2x+4(-2)=-2 \]
\[ 2x-8=-2 \]
\[ 2x=6 \]
\[ \boxed{x=+3} \]
Each carbon atom has oxidation number \(+3\) in the oxalate ion because the two carbon atoms are equivalent.
Part b: Carbon in carbon dioxide
Carbon dioxide is neutral.
Let carbon equal \(x\).
\[ x+2(-2)=0 \]
\[ x-4=0 \]
\[ \boxed{x=+4} \]
Part c: Electrons lost by each oxalate ion
Carbon changes from:
\[ +3\rightarrow +4 \]
Each carbon atom loses one electron.
However, each oxalate ion contains two carbon atoms.
Therefore:
\[ 2\times1=2 \]
Each oxalate ion loses two electrons.
This is an important distinction. Counting one electron per carbon atom would underestimate the total electron transfer if you forgot that there are two carbon atoms in each ion.
Part d: Total electrons lost
There are five oxalate ions.
\[ 5\times2=10 \]
Therefore:
\[ \boxed{10\text{ electrons lost}} \]
Part e: Electrons gained by manganese
First calculate the oxidation number of manganese in permanganate:
\[ x+4(-2)=-1 \]
\[ x=+7 \]
Manganese changes from \(+7\) to \(+2\).
Each manganese atom gains five electrons.
There are two manganese atoms:
\[ 2\times5=10 \]
Therefore:
\[ \boxed{10\text{ electrons gained}} \]
The electron accounting matches:
\[ 10\text{ electrons lost} = 10\text{ electrons gained} \]
Part f: Charge conservation
On the reactant side:
\[ 2(-1)+5(-2)+16(+1) \]
\[ =-2-10+16 \]
\[ =+4 \]
On the product side:
\[ 2(+2)=+4 \]
The total charge is conserved.
The balanced equation also conserves every element:
- Manganese: 2 atoms.
- Carbon: 10 atoms.
- Oxygen: 28 atoms.
- Hydrogen: 16 atoms.
The result demonstrates why oxidation numbers are useful beyond identifying individual atoms. They allow you to check whether the electron transfer in an entire reaction is consistent with the coefficients in its balanced equation.
Question 5
Under hot, concentrated alkaline conditions, hypochlorite ions can undergo a disproportionation reaction, producing chloride and chlorate ions.
The unbalanced ionic equation is:
\[ \ce{ClO^- -> Cl^- + ClO3^-} \]
A student makes the following argument:
Every chlorine atom begins in hypochlorite with the same oxidation number. Therefore, every chlorine atom must undergo the same oxidation-number change. It is impossible for hypochlorite to produce both chloride and chlorate without another reactant supplying or accepting electrons.
Assess this argument.
Your response must:
a. Determine the oxidation number of chlorine in all three species.
b. Identify which chlorine atoms undergo oxidation and which undergo reduction.
c. Derive the correct coefficients using electron conservation.
d. Verify conservation of atoms and charge.
e. Explain whether one individual chlorine atom is both oxidised and reduced, or whether different chlorine atoms follow different pathways.
Solution 5
The student is incorrect. Hypochlorite can act as both an oxidising agent and a reducing agent because different chlorine atoms from the same reactant species undergo different oxidation-number changes.
The balanced equation is:
\[ \boxed{\ce{3ClO^- -> 2Cl^- + ClO3^-}} \]
Part a: Determine the oxidation numbers
Start with hypochlorite:
\[ \mathrm{ClO}^{-} \]
Oxygen has an oxidation number of \(-2\).
Let chlorine equal \(x\).
\[ x+(-2)=-1 \]
\[ \boxed{x=+1} \]
Now consider chloride:
\[ \mathrm{Cl}^{-} \]
This is a single-atom ion, so its oxidation number equals its charge:
\[ \boxed{\mathrm{Cl}=-1} \]
Finally, consider chlorate:
\[ \mathrm{ClO_{3}}^{-} \]
Let chlorine equal \(x\).
\[ x+3(-2)=-1 \]
\[ x-6=-1 \]
\[ \boxed{x=+5} \]
The three oxidation numbers are:
| Species | Chlorine oxidation number |
|---|---|
| \(\mathrm{ClO}^{-}\) | +1 |
| \(\mathrm{Cl}^{-}\) | −1 |
| \(\mathrm{ClO_{3}}^{-}\) | +5 |
Part b: Identify oxidation and reduction
There are two possible changes from the original oxidation number of \(+1\).
First, chlorine can become chloride:
\[ +1\rightarrow -1 \]
The oxidation number decreases by two.
Each chlorine atom following this pathway gains two electrons.
This is reduction.
Second, chlorine can become chlorate:
\[ +1\rightarrow +5 \]
The oxidation number increases by four.
Each chlorine atom following this pathway loses four electrons.
This is oxidation.
Part c: Determine the coefficients
The electron changes must balance.
One chlorine atom becoming chlorate loses four electrons.
One chlorine atom becoming chloride gains two electrons.
Therefore, two chlorine atoms must be reduced for every one chlorine atom oxidised:
\[ 2\times2=4 \]
Two chloride ions account for four electrons gained, matching the four electrons lost by one chlorine atom forming chlorate.
This gives three hypochlorite ions in total:
- Two form chloride ions.
- One forms a chlorate ion.
The balanced equation is:
\[ \boxed{\ce{3ClO^- -> 2Cl^- + ClO3^-}} \]
Part d: Verify atom and charge conservation
Check chlorine atoms:
\[ 3=2+1 \]
There are three chlorine atoms on each side.
Check oxygen atoms:
\[ 3=3 \]
The three oxygen atoms in the reactants are all present in the chlorate product.
Now check the total charge.
Reactant side:
\[ 3(-1)=-3 \]
Product side:
\[ 2(-1)+(-1)=-3 \]
The total charge is \(-3\) on both sides.
Both atoms and charge are conserved.
Part e: Explain the misconception
The student’s mistake is assuming that because all three chlorine atoms begin in chemically identical hypochlorite ions, they must all undergo the same change.
That does not follow.
The reaction involves a population of hypochlorite ions. Some chlorine atoms undergo reduction, while others undergo oxidation.
A single chlorine atom does not simultaneously gain and lose electrons. Different chlorine atoms follow different pathways.
The original hypochlorite species therefore has two roles:
- It acts as an oxidising agent when its chlorine is reduced to chloride.
- It acts as a reducing agent when its chlorine is oxidised to chlorate.
This type of redox reaction is called disproportionation.
The key conceptual point is that oxidation and reduction must balance across the reaction as a whole, not necessarily within each individual reactant particle.
09The method to remember
When finding oxidation numbers in polyatomic ions, keep this sequence in mind:
- Write down the ion’s overall charge.
- Assign the oxidation numbers you already know.
- Multiply each value by the number of atoms present.
- Set the sum equal to the overall ion charge.
- Solve for the unknown oxidation number.
- Check that your values reproduce the correct charge.
When the ion appears in a redox reaction, take the method one step further.
Compare the oxidation number before and after the reaction. An increase indicates oxidation, while a decrease indicates reduction. Multiply the change by the number of atoms involved to account for the total electrons transferred.
Finally, check that the complete ionic equation conserves both atoms and charge, and pay attention to whether the reaction occurs in acidic or alkaline conditions.
Once you can move confidently between an ion’s formula, its overall charge, and the oxidation numbers of its atoms, unfamiliar polyatomic ions become much less intimidating. More importantly, you can use those values to explain the electron transfer that drives redox reactions.