General Chemistry 202-SN1-RE • CEGEP, Montreal

Revision sheet: redox, oxidation numbers and nomenclature (202-SN1-RE)

This revision sheet covers redox reactions and inorganic nomenclature in CEGEP general chemistry: naming ionic and molecular compounds, acids and hydrates, calculating an oxidation number, recognizing a redox reaction among the five reaction types, balancing by electron balance and handling a combustion. It is written for CEGEP students in General Chemistry 202-SN1-RE (formerly 202-NYA).

The thread: an oxidation number is not a real charge, it is electron bookkeeping, and the same bookkeeping names the compounds and balances the reactions. That is why it can be fractional, and why it is enough to balance an equation that inspection cannot handle.

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The thread of the chapter

An oxidation number is not a real charge, it is electron bookkeeping. The same bookkeeping names the compounds and balances the reactions.

This chapter is part of 202-SN1-RE, General Chemistry

The essentials

Naming a compound: three families, three methods

• BINARY IONIC, a metal and a nonmetal: the cation first, then the anion with the ending -ide (oxygen gives oxide). NO numerical prefix, ever: the charges fix the subscripts.
• Metal with a variable charge: Stock notation, the charge in Roman numerals in parentheses. $FeCl_{3}$ is iron(III) chloride, $Cu_{2}O$ is copper(I) oxide.
• POLYATOMIC IONIC: the ion keeps its name, and goes in parentheses as soon as there is more than one. $Ca(NO_{3})_{2}$ is calcium nitrate.
• MOLECULAR, two nonmetals: Greek prefixes are compulsory, mono di tri tetra penta hexa hepta. Mono- is omitted on the first element named, never on oxygen: carbon monoxide, carbon dioxide.
• An ionic formula is always the EMPIRICAL formula of the lattice: cross the charges, then SIMPLIFY. $CaO$ , never $Ca_{2}O_{2}$ .

On the left, a lattice of ions, where the formula is only a ratio: no prefix. On the right, two distinct molecules of the same pair of elements: only the prefix tells them apart.

A one-second test to choose the method: is there a metal in the formula? Yes, no prefix. No, prefixes.

The oxygen series and the acids

• Series of a single element: the reference ion takes -ate, one oxygen fewer gives -ite, two fewer gives hypo- plus -ite, one more gives per- plus -ate.
• Chlorine: $ClO^{-}$ hypochlorite, $ClO_{2}^{-}$ chlorite, $ClO_{3}^{-}$ chlorate, $ClO_{4}^{-}$ perchlorate. Each oxygen removed lowers the oxidation number by TWO units, from $+7$ to $+1$ .
• Acids: an anion ending in -ide gives an acid named hydro-...-ic, an anion ending in -ate gives an acid ending in -ic, an anion ending in -ite gives an acid ending in -ous.
• Sulfate gives sulfuric, sulfite gives sulfurous: one letter in the name for one oxygen in the formula. It is the most common trap on a test.
• Hydrate: the dot in $CuSO_{4}\cdot 5\,H_{2}O$ is not a multiplication, but the water counts in the molar mass. Heated to constant mass, the salt becomes anhydrous.

Each oxygen added raises chlorine by two units. The names ending in -ite give acids ending in -ous, in orange; the names ending in -ate give acids ending in -ic, in blue.

The two reference points to memorize for each column are the -ate ion and its formula. The other three follow by adding or removing oxygens, with nothing more to learn.

Oxidation numbers and the electron balance

• The rules apply IN ORDER: element in its elemental form $0$ ; monatomic ion, its charge; fluorine always $-1$ ; oxygen $-2$ except peroxides and fluorides; hydrogen $+1$ except metal hydrides; the sum is $0$ in a neutral compound and the CHARGE in an ion.
• An oxidation number is the real charge only for a MONATOMIC ion. Elsewhere it is an accounting convention: the two electrons of each bond go to the more electronegative atom. It can therefore be fractional, and it is then only an average.
• Oxidized means the oxidation number goes UP, so electrons are lost. Reduced means it goes down, so electrons are gained. The OXIDIZING AGENT is the one that gets reduced, the REDUCING AGENT the one that gets oxidized.
• In every redox reaction, the number of electrons lost EQUALS the number gained. This equality is what balances the equations that inspection cannot handle.
• An oxidizing agent may be only partly reduced within its own compound: of $8\,HNO_{3}$ , only $2$ are reduced, the other $6$ remaining as spectator nitrates.

The six rules in their order of priority. Go down, stop at the first one that applies, and the exceptions to rules 4 and 5 are the only ones to memorize.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Putting a numerical prefix on an ionic compound

1 mark per name, and the grader concludes that the logic is not mastered

What not to write

“ $CaBr_{2}$ : it is calcium dibromide.”

What to write

“It is calcium bromide. The charges $Ca^{2+}$ and $Br^{-}$ force the subscript $2$ : saying it would be redundant.”

Why: A prefix exists only where it carries information. Between a metal and a nonmetal, neutrality fixes a single possible formula; between two nonmetals, nothing fixes it, and $CO$ and $CO_{2}$ both exist. Hence the sorting rule: a metal in the formula, no prefix.

2. Writing the ionic formula without simplifying it

1 mark, and every molar mass calculated afterward

What not to write

“Calcium oxide: $Ca^{2+}$ and $O^{2-}$ , I cross the charges, so $Ca_{2}O_{2}$ .”

What to write

“Cross the charges, THEN simplify: the simplest ratio is one to one, so $CaO$ .”

Cross the charges, then simplify:

Ca_{2}O_{2}

does not exist, it is

CaO

. For

Al^{3+}

and

O^{2-}

Al_{2}O_{3}

is already the simplest ratio.

Why: An ionic formula describes a LATTICE, not a molecule: there is no isolated $Ca_{2}O_{2}$ entity, only a crystal in which there is one calcium for each oxygen. The formula is therefore always empirical, and simplifying is not a nicety.

3. Confusing the suffixes -ic and -ous

1 to 2 marks, and a mix-up of reagents in the laboratory

What not to write

“ $H_{2}SO_{3}$ : it is sulfuric acid.”

What to write

“ $SO_{3}^{2-}$ is sulfite, ending in -ite, so the acid ends in -ous: sulfurous acid. It is $H_{2}SO_{4}$ , derived from sulfate, that gives sulfuric acid.”

Go through the anion: sulfate gives sulfuric, sulfite gives sulfurous. One oxygen fewer in the formula, one letter fewer in the name.

Why: One oxygen of difference in the formula, one letter of difference in the name. The way never to go wrong is to go through the anion: name the anion first, then apply -ate gives -ic, -ite gives -ous.

4. Applying the oxygen rule without looking at the exceptions

the whole exercise, since the entire electron balance depends on it

What not to write

“ $H_{2}O_{2}$ : oxygen is at $-2$ , so the two hydrogens should be $+2$ each.”

What to write

“The rules apply IN ORDER, and hydrogen comes first: $2(+1)+2x=0$ gives $x=-1$ . It is a peroxide, oxygen is at $-1$ .”

In water, oxygen takes the electrons of its two bonds with H:

-2

. In the peroxide, the O-O bond gives an electron to nobody: each oxygen is at

-1

Why: Three exceptions to know for oxygen: $0$ in the element $O_{2}$ , $-1$ in a peroxide, positive next to fluorine, the only element more electronegative than it, hence $+2$ in $OF_{2}$ . Hydrogen has one: $-1$ in a metal hydride such as $NaH$ , where the metal can only give up electrons.

5. Forgetting that the sum equals the charge in an ion

the rest of the balance, and a wrong conclusion about oxidizing power

What not to write

“ $Cr_{2}O_{7}^{2-}$ : the sum of the oxidation numbers is zero, so $2x-14=0$ and chromium is at $+7$ .”

What to write

“The sum equals the CHARGE of the ion, that is $-2$ : $2x+7(-2)=-2$ gives $2x=12$ and $x=+6$ .”

Why: The sum rule has two versions and only one gets remembered. The check is immediate: in $MnO_{4}^{-}$ , the right version gives $+7$ , the maximum of group 7, consistent with a powerful oxidizing agent. The wrong one would give $+8$ , impossible for an element with seven valence electrons.

6. Reading the formula without reading the charge

1 mark per species, and a wrong ranking of oxidizing agents in the next question

What not to write

“ $ClO_{2}$ : it is chlorite, I already calculated, chlorine is at $+3$ .”

What to write

“ $ClO_{2}$ is a NEUTRAL molecule, chlorine dioxide: $x+2(-2)=0$ gives $+4$ . It is the ion $ClO_{2}^{-}$ , with sum $-1$ , that puts chlorine at $+3$ .”

Why: The same string of letters names two different species depending on the superscript, and the sum rule changes with it: zero for the molecule, the charge for the ion. The reflex: write the target sum BEFORE setting up the equation. It also applies to $NO_{2}$ and $NO_{2}^{-}$ , $+4$ and $+3$ , or $SO_{3}$ and $SO_{3}^{2-}$ , $+6$ and $+4$ .

7. Declaring a decomposition redox without calculating

2 marks, and an electron balance invented for a reaction that has none

What not to write

“ $CaCO_{3}\rightarrow CaO+CO_{2}$ : bonds are broken and formed, so it is a redox reaction.”

What to write

“Calculate on both sides: calcium $+2$ , carbon $+4$ , oxygen $-2$ , before and after. Nothing moves, it is NOT a redox reaction.”

Under each species, its oxidation numbers: the decomposition of limestone changes nothing, that of the peroxide sends one oxygen down and another up.

Why: You do not guess whether a reaction is redox, you calculate. Two types settle it without calculation, single displacement which always is one and double displacement which never is, but a synthesis or a decomposition needs checking. A useful counterexample: $2\,H_{2}O_{2}\rightarrow 2\,H_{2}O+O_{2}$ IS a redox reaction, and even a disproportionation.

8. Swapping the oxidizing and reducing agents

1 mark each time, and the concluding sentence becomes incomprehensible

What not to write

“Zinc is oxidized, so it is the oxidizing agent of the reaction.”

What to write

“Zinc is OXIDIZED, it loses electrons: it is the REDUCING AGENT. The oxidizing agent is the copper ion, which is reduced.”

The electrons go from the zinc to the copper ion. Zinc, which LOSES them, is oxidized: it is the reducing agent. The copper ion, which takes them up, is reduced: it is the oxidizing agent.

Why: A species is named by what it CAUSES the other to undergo, not by what it undergoes. The oxidizing agent oxidizes the other, so it gets reduced itself. The wording that never fails: the oxidizing agent takes up electrons, and taking up lowers the oxidation number.

9. Believing that every nitric acid molecule is reduced

the whole balancing, and a balance that never closes

What not to write

“ $3\,Cu+8\,HNO_{3}\rightarrow \dots$ : eight nitrogens go from $+5$ to $+2$ , that is $24$ electrons gained.”

What to write

“Only two end up as $NO$ and change oxidation number. The other six stay at $+5$ in the nitrates: $2\times 3=6$ electrons gained, which equals the $3\times 2=6$ lost by the copper.”

Six electrons lost, six gained. The LCM of

2

and

3

gives the coefficients

3

and

2

directly; the rest follows from hydrogen and then oxygen.

Why: Nitric acid plays TWO roles at once: oxidizing agent for the molecules that get reduced, simple acid supplying nitrate for the others. This double role is what makes balancing by inspection fail, and it is the oxidation number method that makes it visible.

Which method to choose

How to name an unknown compound

The formula, and the single question of whether it contains a metal

If a metal with a FIXED charge, groups 1 and 2, plus $Al$ , $Zn$ , $Ag$ → name the cation first, then the anion ending in -ide or oxide, with no prefix and no Roman numeral

Example: $K_{2}S$ gives potassium sulfide
If a metal with a VARIABLE charge, most transition metals → the same, plus the charge of the metal in Roman numerals, deduced from the anion

Example: $Cu_{2}O$ gives copper(I) oxide
If a polyatomic ion in the formula → it keeps its name unchanged, and takes parentheses if there are several

Example: $Ca(NO_{3})_{2}$ gives calcium nitrate
If two nonmetals only → Greek prefixes on both, mono- omitted on the first one named

Example: $N_{2}O_{4}$ gives dinitrogen tetroxide
If a hydrogen at the front and an aqueous solution → it is an acid: -ide gives hydro-...-ic, -ate gives -ic, -ite gives -ous

Example: $H_{2}SO_{3}$ gives sulfurous acid

Sort by first looking for a metal, then for a polyatomic ion, then for a hydrogen at the front: these questions point to one of the five branches, and none of the three methods mixes with the others.

Is this reaction a redox reaction?

The type of the reaction, read from the diagram of reactants and products

Three cases settle it without calculation: an element, an exchange of partners, a combustion. A synthesis or a decomposition with no element forces you to calculate.

If an element appears on one side or the other → it is a redox reaction, with no calculation: an element is at $0$ and does not stay there

Example: $Zn+CuSO_{4}\rightarrow ZnSO_{4}+Cu$
If two ionic compounds exchange partners → it is NEVER a redox reaction: each ion keeps its charge

Example: $AgNO_{3}+NaCl\rightarrow AgCl+NaNO_{3}$
If a compound reacts with oxygen gas → it is a combustion, so always a redox reaction

Example: $CH_{4}+2\,O_{2}\rightarrow CO_{2}+2\,H_{2}O$
If a synthesis or a decomposition with no element → you MUST calculate the oxidation numbers on both sides

Example: $CaCO_{3}\rightarrow CaO+CO_{2}$ is not one

The first two branches settle the question with no calculation at all and cover most cases on a test. The last one is where marks are lost, because you think you can guess.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Balancing by the oxidation number method

When to use it: Balancing by inspection fails, usually because one element plays two roles, like the nitrogen of nitric acid.

1 Calculate the oxidation number of each element on both sides and spot the two that change.
2 For each one, write the number of electrons exchanged PER ATOM, with the direction, lost or gained.
3 Take the least common multiple of the two numbers: it is the total number of electrons exchanged.
4 Deduce the coefficients of the oxidized species and of the reduced species, then carry them over.
5 Finish with the elements that do not change, hydrogen first, oxygen last, and check element by element.

Concluding sentence

“Copper goes from $0$ to $+2$ , it loses $2$ electrons; the nitrogen that ends up as $NO$ goes from $+5$ to $+2$ , it gains $3$ . The LCM is $6$ , hence $3\,Cu$ and $2\,NO$ , then $8\,HNO_{3}$ from nitrogen and $4\,H_{2}O$ from hydrogen.”

The trap: Counting all eight nitric acids as reduced. Only the two that end up as nitrogen monoxide change oxidation number; the other six pass through the reaction keeping $+5$ .

Marking: Typically 2 marks for the oxidation numbers, 2 for the electrons exchanged with their direction, 2 for the LCM and the two coefficients, 2 for finishing the balancing, 2 for the element-by-element check.

Handling a combustion, complete or incomplete

When to use it: A hydrocarbon or an organic compound reacts with oxygen gas.

1 Write the products: $CO_{2}$ and $H_{2}O$ if the combustion is complete, $CO$ and $H_{2}O$ if it is incomplete.
2 Balance carbon, then hydrogen, and keep oxygen for last.
3 Double all the coefficients if oxygen lands on a half molecule.
4 Check the electron balance from both sides: each $O_{2}$ takes up $4$ electrons, and the carbons give up as many.

Concluding sentence

“ $C_{3}H_{8}+5\,O_{2}\rightarrow 3\,CO_{2}+4\,H_{2}O$ . Carbon goes from $-\dfrac{8}{3}$ to $+4$ , that is $20$ electrons for three carbons, and the ten oxygen atoms take up $20$ : the balance closes.”

The trap: Believing that a fractional oxidation number is a calculation error. It is an average, and it is perfectly usable for an overall balance.

Marking: 1 mark for the correct products, 2 for the balancing, 1 for doubling when it is needed, 2 for the electron balance checked from both sides.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The sum of the oxidation numbers

It must equal zero for a neutral compound and the charge for an ion. Recalculate it over the whole formula, subscripts included.

$Ca(NO_{3})_{2}$ : $(+2)+2\left[(+5)+3(-2)\right]=(+2)+2(-1)=0$ .
Electrons lost equal electrons gained

In every redox reaction, the two counts must coincide. If they differ, the coefficients are wrong, and the mistake is almost always having counted spectators.

$3\times 2=6$ lost by the coppers, $2\times 3=6$ gained by the reduced nitrogens.
Balancing element by element, oxygen last

Recount each element on both sides. Oxygen last, because it appears in the most species and so reveals every earlier mistake.

$8\times 3=24$ oxygens on the left, $3\times 6+2+4=24$ on the right.
The maximum oxidation number of an element

It cannot exceed the number of valence electrons. A result beyond that signals a rule error, usually the sum taken as zero instead of the charge.

$+7$ for manganese in group 7 is its maximum; finding $+8$ is impossible.
Consistency of name and formula

Start again from the name you found and rewrite the formula. If you do not land on the one in the question, the name is wrong.

Potassium sulfide gives back $K^{+}$ and $S^{2-}$ , so $K_{2}S$ : you do land on it.

The typical problem, taken apart

A salt identified by its molar mass, then followed to its decomposition

A white ionic salt of formula $M(NO_{3})_{2}$ has a molar mass of $164.10$ g/mol.

Strongly heated, it gives $2\,M(NO_{3})_{2}\rightarrow 2\,MO+4\,NO_{2}+O_{2}$ .

Atomic masses: $N$ 14.01; $O$ 16.00; $Ca$ 40.08; $Mg$ 24.31.

The unknown nitrate heated in the tube: it decomposes into a solid oxide, reddish-brown nitrogen dioxide and oxygen gas.

Step 1

Nitrate ion: $14.01+3(16.00)=62.01$ g/mol. Two nitrates weigh $124.02$ g/mol.

Why

Isolate the known part before looking for the unknown. The formula gives the number of nitrates, so there is nothing to guess.

Step 2

Metal: $164.10-124.02=40.08$ g/mol, that is calcium. The compound is calcium nitrate $Ca(NO_{3})_{2}$ .

Why

Magnesium weighs $24.31$ , very far off: the identification leaves nothing to discuss. A free check: calcium is in group 2 and does give $Ca^{2+}$ , the charge that two nitrates require.

Step 3

Oxidation numbers: calcium $+2$ , oxygen $-2$ , and in the nitrate $x+3(-2)=-1$ puts nitrogen at $+5$ .

Why

The sum equals the CHARGE of the nitrate ion, $-1$ , and not zero. It is the most frequent error with polyatomic ions.

Step 4

Check over the whole formula: $(+2)+2\left[(+5)+3(-2)\right]=(+2)+2(-1)=0$ .

Why

The compound is neutral, so the sum must be zero. This check catches any mistake made in the previous step.

Step 5

In the decomposition, nitrogen goes from $+5$ to $+4$ in $NO_{2}$ : it gains $1$ electron, and there are $4$ of them, that is $4$ electrons gained.

Why

First spot what changes. Nitrogen goes down: it is reduced, so it is the oxidizing agent.

Step 6

Oxygen goes from $-2$ in the nitrate to $0$ in $O_{2}$ : it loses $2$ electrons, and there are $2$ of them, that is $4$ electrons lost.

Why

Oxygen plays the role of reducing agent here, which always comes as a surprise. Only the TWO oxygens that end up as oxygen gas are involved, the others stay at $-2$ in the oxide.

Step 7

Balance: $4$ gained and $4$ lost, the equality holds and the coefficients in the statement are therefore correct.

Why

It is an INTERNAL redox reaction, where the oxidizing agent and the reducing agent are two elements of the same compound. Calcium remains a spectator at $+2$ from start to finish.

Step 8

For $10.00$ g of salt: $n=\dfrac{10.00}{164.10}=6.094\times 10^{-2}$ mol, and the ratio of oxide to nitrate is $1$ .

Why

The equation has $2$ and $2$ : the ratio simplifies, and there is no need to drag the coefficients along.

Step 9

$M(CaO)=40.08+16.00=56.08$ g/mol, so $m=6.094\times 10^{-2}\times 56.08=3.417$ g.

Why

The mass falls to a third of the starting mass, which is consistent: all the nitrogen and most of the oxygen left in the gas phase.

The conclusion, written out

“The salt is calcium nitrate $Ca(NO_{3})_{2}$ . Its decomposition is an internal redox reaction in which nitrogen, reduced from $+5$ to $+4$ , gains $4$ electrons that oxygen, oxidized from $-2$ to $0$ , gives up. From $10.00$ g of salt we obtain $3.417$ g of calcium oxide.”

The classic mistake on this problem: Declaring that the decomposition is not redox because no element appears among the reactants. The oxygen gas formed IS an element, and it alone guarantees that some element changed oxidation number.

Learn by heart

• A metal in the formula: no prefix. Two nonmetals: prefixes.
• Variable charge: Roman numerals, deduced from the anion, whose charge is fixed.
• Cross the charges THEN simplify: $CaO$ , never $Ca_{2}O_{2}$ .
• -ate gives -ic, -ite gives -ous, -ide gives hydro-...-ic.
• One oxygen fewer in an oxyanion: two units fewer of oxidation number.
• The oxidation number rules apply IN ORDER, and hydrogen comes before oxygen.
• Peroxide $-1$ , positive next to fluorine, metal hydride $-1$ : the three exceptions.
• Zero sum for a neutral compound, equal to the CHARGE for an ion.
• The oxidizing agent gets reduced, the reducing agent gets oxidized.
• Disproportionation: one element goes both up AND down. Comproportionation: two states converge on the same one.
• Single displacement and combustion: always redox. Double displacement: never.
• Electrons lost equal electrons gained: that is what balances the equation.
• Incomplete combustion: carbon stops at $+2$ , energy is left over and the gas is toxic.

Frequently asked questions

When should you use prefixes like di or tri in a name?

Only between two nonmetals. In an ionic compound, the charges of the ions force a single possible formula, so stating the number would be redundant and counted as an error. Between two nonmetals, on the contrary, nothing forces a ratio: carbon and oxygen give the monoxide as well as the dioxide. The test comes down to one question: is there a metal in the formula? If so, no prefix.

Is an oxidation number the real charge of the atom?

No, except for a monatomic ion such as the sodium ion or the chloride ion. Everywhere else it is an accounting convention: you act as if every covalent bond were entirely ionic, assigning its two electrons to the more electronegative atom. The proof that it is not a real charge is that it can be fractional, like the carbon of propane. Its only function is to track who gave up electrons and who received them.

How do you tell whether a reaction is a redox reaction?

Calculate the oxidation numbers of each element on both sides, and see whether any of them has changed. Two cases are settled without calculation: a single displacement always is one, because an element appears or disappears, and a double displacement never is, because the ions exchange partners without changing charge. A synthesis or a decomposition needs checking, because it can be either.

Why does balancing copper in nitric acid resist inspection?

Because nitrogen plays two roles at once. Of the eight nitric acid molecules consumed, only two are reduced and end up as nitrogen monoxide; the other six pass through the reaction without changing oxidation number and end up as nitrate ions around the copper. Balancing by inspection does not see this split, whereas the oxidation number method makes it visible immediately.

Why is incomplete combustion dangerous?

Because carbon rises only to oxidation number two instead of four: the oxidation is unfinished and the product formed is carbon monoxide instead of carbon dioxide. Two consequences follow. Part of the chemical energy stays locked in that gas instead of being released, which lowers the efficiency. And carbon monoxide, colourless and odourless, binds to hemoglobin far better than oxygen does. The single cause is lack of air, and the remedy is ventilation.

How do you find the name of an oxyanion you have not learned?

Memorize only the reference ion of the column, the one whose name ends in -ate, with its formula. The rest follows from a single rule: one oxygen fewer turns -ate into -ite, two fewer add the prefix hypo-, and one more adds the prefix per-. For chlorine, starting from chlorate, you get chlorite, hypochlorite and perchlorate with nothing more to learn, and each oxygen removed lowers the oxidation number by two units.

Practise it

Corrected exercises: Redox, oxidation numbers and nomenclature, 202-SN1-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

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Revision sheet: redox, oxidation numbers and nomenclature (202-SN1-RE)

The essentials

Naming a compound: three families, three methods

The oxygen series and the acids

Oxidation numbers and the electron balance

The mistakes that cost marks

1. Putting a numerical prefix on an ionic compound

2. Writing the ionic formula without simplifying it

3. Confusing the suffixes -ic and -ous

4. Applying the oxygen rule without looking at the exceptions

5. Forgetting that the sum equals the charge in an ion

6. Reading the formula without reading the charge

7. Declaring a decomposition redox without calculating

8. Swapping the oxidizing and reducing agents

9. Believing that every nitric acid molecule is reduced

Which method to choose

How to name an unknown compound

Is this reaction a redox reaction?

How the answer is expected to be written

Balancing by the oxidation number method

Handling a combustion, complete or incomplete

Check before you hand in

The typical problem, taken apart

A salt identified by its molar mass, then followed to its decomposition

Learn by heart

Frequently asked questions

When should you use prefixes like di or tri in a name?

Is an oxidation number the real charge of the atom?

How do you tell whether a reaction is a redox reaction?

Why does balancing copper in nitric acid resist inspection?

Why is incomplete combustion dangerous?

How do you find the name of an oxyanion you have not learned?

Corrected exercises: Redox, oxidation numbers and nomenclature, 202-SN1-RE

See also

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