1.3.6 · D5Chemical Reactions & Stoichiometry
Question bank — Oxidation number rules — assigning, change
True or false — justify
An oxidation number is the real electrical charge sitting on an atom.
False. It is a pretend charge got by handing every shared electron pair fully to the more electronegative atom; real molecules only share electrons partly. Only for monatomic ions like does the pretend charge equal the real charge.
Oxygen always has oxidation number .
False. It is in the vast majority of compounds, but in peroxides (), in superoxides (), and in because fluorine is even greedier than oxygen.
Hydrogen always has oxidation number .
False. With nonmetals H is , but in metal hydrides like or the metal is less electronegative, so H takes the electron and becomes .
Oxidation numbers can never be fractional because charges are whole numbers.
False. A single atom's true charge is not fractional, but the reported value is an average over non-equivalent atoms, e.g. for S in and for Fe in .
In the neutral molecule , the oxidation numbers must add to zero.
True. The sum rule says the pretend charges add to the species' net charge, and a neutral molecule has net charge : . ✓
Fluorine can be positive in some strong oxidiser.
False. Fluorine is the most electronegative element, so it always keeps the shared electrons and is in every compound (and only as free ).
If two atoms in a molecule have the same oxidation number, they must be chemically identical.
False. The value can be an average; the four sulfurs in share one printed number () yet occupy different bonding positions with different actual charges.
Adding electrons to an atom lowers its oxidation number.
True. Gaining electrons makes the pretend charge more negative — that is exactly the definition of reduction (OIL RIG: Reduction Is Gain).
Every atom in an odd-electron (radical) molecule still gets a whole-number oxidation state.
True usually. In the sum rule with O (net charge ) gives N , a clean integer — the unpaired electron sits in a molecular orbital and does not force a fractional score by itself. Fractions come from non-equivalent atoms, not from radicals.
Spot the error
"For I set the sum of oxidation numbers to because oxidation numbers are just bookkeeping." — find the mistake.
The sum must equal the species' net charge , not . Term by term: is two chromiums each of unknown value ; is seven oxygens each fixed at ; they add to the ion charge . So . Wrongly using would give , i.e. Cr .
"O is bonded to F in , and O is usually , so O is and each F is ." — find the mistake.
Fluorine is more electronegative than oxygen, so F wins the tug-of-war and is ; the sum rule (here = oxygen's value) then forces O . The rule "F is " outranks "O is ".
"In , both Na and H are electron-losers, so both are ." — find the mistake.
Only one atom in a bond can take the shared pair. Na (a group-1 metal) is far less electronegative, so it loses () and H gains (); the sum rule (here = hydrogen's value) gives H .
"S in has a real charge of , so it strongly attracts a nearby anion." — find the mistake.
is the hypothetical charge from full electron-transfer accounting; the actual charge on S is far smaller because the S–O bonds are covalent, only partly polar. The pretend score is not a real electric field.
" molecule: oxygen is as usual." — find the mistake.
is a free element (rule 1 outranks everything), so each O is . In the O–O bond both atoms are equally greedy, so the shared electrons split evenly and neither is pushed positive or negative.
"Zn goes from to , so Zn is reduced because its number changed." — find the mistake.
A change alone doesn't decide direction. The number increased, which is oxidation (loss of electrons); Zn is therefore the reducing agent here.
"For a peroxide I'll just use O ; the sum rule will fix any error." — find the mistake.
The sum rule can't detect a wrong assumption if you feed it the wrong O value — it will silently hand you a wrong answer for the other atom. You must first spot the O–O bond and set O before writing the equation.
Why questions
Why do we pretend bonds are fully ionic when we know they aren't?
In real oxygen only partly pulls the shared pairs, but if we exaggerate the same way on both sides of every bond, the exaggeration cancels — so the change in electron ownership during a reaction is still tracked correctly.
Why is the sum rule called the "master key"?
Every other rule pins a fixed value onto most atoms (e.g. in : H , O ), leaving a single unknown for sulfur; the sum rule then gives in one line.
Why does the total oxidation-number change in a balanced redox reaction sum to zero?
Electrons are only moved, never created. In , Zn rises by (loses 2 e⁻) and Cu falls by (gains those same 2 e⁻); .
Why does fluorine outrank oxygen when both are in a compound?
The rule assigns shared electrons to the more electronegative atom, and F sits one rung above O on the electronegativity ladder (, shown in the second figure), so in fluorine takes both bonding pairs ( each) and oxygen is left short ().
Why can the same element (e.g. Fe) show a fractional average in one formula but an integer in another?
In all irons are equivalent, so the sum rule gives a clean ; in two irons are and one is , and the printed is just their average .
Why is oxidation-number bookkeeping better than measuring real charges for redox?
Real partial charges are messy and continuous — you can't say in water is exactly . The pretend integer score is clean and reproducible, yet it still tells you Zn lost 2 e⁻ and Cu gained 2 e⁻.
Edge cases
What is the oxidation number of C in elemental graphite or diamond?
, because it is a free element — rule 1 applies regardless of how the C atoms bond to each other; identical atoms split every shared pair evenly.
Oxidation number of O in the peroxide ion by itself?
each. With = each oxygen's value, gives . The O–O bond is a tug between equals, so its electrons split evenly and neither O gets extra from it.
In versus , is oxygen's number the same?
No. In each O bonds only to F, so gives O . In there is an O–O bond, so gives an average O . The O–O bond changes the accounting.
For a bond between two atoms of the same electronegativity (a C–C bond), how are the shared electrons split?
Evenly — neither atom is greedier, so the pair is divided equally and contributes nothing to either atom's oxidation number.
Can a single element in one compound have two different oxidation states at once?
Yes. In iron is genuinely a mix of and ; the reported is the sum-rule average, not the value on any one atom.
Work out the "middle atom" case with no O, H, or F: nitrogen in the thiocyanate ion (net charge ).
Anchor from the electronegativity ladder, not from O/H/F. The structure is (see the third figure). Step 1 — sulfur: in the S–C bond, S ( on the ladder) wins both shared electrons; sulfur bonds only to carbon here and behaves like sulfide, so S . Step 2 — nitrogen: N () also wins both electrons in the C–N bond, giving N a full extra pair, so N . Step 3 — carbon: carbon loses to S on one side and to N on the other, i.e. it gives away four electrons (two to each greedier neighbour), so C . Check with the sum rule: , exactly the ion's charge. ✓
How is the oxidation number of N in the radical found, and does the odd electron cause a fraction?
With = nitrogen's value and O each, net charge : . Still a whole number — the unpaired electron lives in a delocalised orbital and does not by itself force a fractional score.
What oxidation number does oxygen take in the superoxide radical ion ?
With = each oxygen's value and net charge : . This is a genuine fraction, but it is an average — one oxygen effectively carries the extra electron more than the other, so the printed is bookkeeping, not a real per-atom charge.
Recall Self-test summary
Meaning of in these equations ::: the oxidation number of the single atom we are solving for; all other atoms are fixed known values. Two-atom bond, equal electronegativity ::: split electrons evenly — contributes to each. Net charge to use for an ion in the sum rule ::: the ion's own charge, not . Oxidation number increased ::: oxidation, loss of electrons, the reducing agent. Why fractional values appear ::: they are averages over non-equivalent atoms, not the effect of odd electrons. How SCN⁻ is solved without O/H/F ::: rank on the electronegativity ladder (), give each greedier neighbour the shared pairs (S , N ), leaving C ; sum .
Connections
- Redox reactions — balancing (half-reaction & ion-electron)
- Electronegativity & periodic trends
- Oxidising and reducing agents
- Stoichiometry — mole conservation in reactions
- Electrochemistry — cell potentials