Intuition The big picture (WHY care?)An atom's oxidation state tells you how many electrons it has "given up" or "grabbed" in a compound. Because chemistry is basically electron bookkeeping , knowing which oxidation states an element likes lets you predict its formulas (NaCl \text{NaCl} NaCl NaCl 2 \text{NaCl}_2 NaCl 2 map of electron availability , so oxidation states must vary in a regular, predictable pattern — and that pattern is what this note decodes.
Definition Oxidation state (oxidation number)The hypothetical charge an atom would carry if all its bonds were treated as 100% ionic — i.e. every shared electron pair is given entirely to the more electronegative atom.
Key bookkeeping rules (used to derive every number below):
Free element = 0 0 0 O 2 \text{O}_2 O 2 Na \text{Na} Na
Sum of oxidation states in a neutral molecule = 0 = 0 = 0 = = =
Common "anchors": O = − 2 \text{O} = -2 O = − 2 H = + 1 \text{H} = +1 H = + 1 = + 1 = +1 = + 1 = + 2 = +2 = + 2 = − 1 = -1 = − 1
Intuition Why does a maximum exist?An atom can only give away the electrons it actually has in its valence shell . Once you strip all valence electrons, the next ones sit in a full, low-energy inner shell — far too tightly bound to remove chemically. So the maximum positive oxidation state = number of valence electrons available for sharing .
First-principles derivation for main-group elements:
Group number (old 1–8 style / new 1,2,13–18) tells you valence electron count n v n_v n v
Give all of them to more-EN partners (usually O or F).
Maximum positive state = + n v = +n_v = + n v
Max positive OS = + n v = ( group valence electrons ) \boxed{\text{Max positive OS} = +\,n_v = (\text{group valence electrons})} Max positive OS = + n v = ( group valence electrons )
Check: for group 17 (Cl, n v = 7 n_v=7 n v = 7 + 7 +7 + 7 HClO 4 \text{HClO}_4 HClO 4 + 1 + x + 4 ( − 2 ) = 0 ⇒ x = + 7. ✓ +1 + x + 4(-2) = 0 \;\Rightarrow\; x = +7.\ \checkmark + 1 + x + 4 ( − 2 ) = 0 ⇒ x = + 7. ✓
Intuition Why a negative floor?A non-metal completes its octet by gaining electrons. It can grab only as many as it needs to fill the shell: 8 − n v 8 - n_v 8 − n v − ( 8 − n v ) -(8 - n_v) − ( 8 − n v )
Min OS = − ( 8 − n v ) \boxed{\text{Min OS} = -(8 - n_v)} Min OS = − ( 8 − n v )
For nitrogen (n v = 5 n_v = 5 n v = 5 = − ( 8 − 5 ) = − 3 = -(8-5) = -3 = − ( 8 − 5 ) = − 3 NH 3 \text{NH}_3 NH 3 = + 5 = +5 = + 5 HNO 3 \text{HNO}_3 HNO 3
Metals (left): only positive states, equal to group number (Na + 1 +1 + 1 + 2 +2 + 2 + 3 +3 + 3 e − e^- e − Non-metals (right): show BOTH a high positive (with O/F) and a negative state.Middle (p-block): widest variety, because they can lose valence p p p s s s or gain a few.
Intuition Why heavier elements preferlower oxidation states
Going down a group, the n s 2 ns^2 n s 2 reluctant to bond (poor shielding by d d d f f f − 2 -2 − 2
That is why PbCl 2 \text{PbCl}_2 PbCl 2 PbCl 4 \text{PbCl}_4 PbCl 4
Intuition Why so many states?The 3 d 3d 3 d 4 s 4s 4 s very close in energy . Electrons can be removed from both in small steps, so many oxidation states differ by 1 and are all accessible. Main-group s s s p p p
Manganese is the poster child: + 2 +2 + 2 + 3 +3 + 3 + 4 +4 + 4 MnO 2 \text{MnO}_2 MnO 2 + 6 +6 + 6 MnO 4 2 − \text{MnO}_4^{2-} MnO 4 2 − + 7 +7 + 7 MnO 4 − \text{MnO}_4^- MnO 4 − = + 7 = 3 d 5 4 s 2 =+7 = 3d^5 4s^2 = + 7 = 3 d 5 4 s 2
Worked example (a) Sulfur inH 2 SO 4 \text{H}_2\text{SO}_4 H 2 SO 4 = x =x = x 2 ( + 1 ) + x + 4 ( − 2 ) = 0 2(+1) + x + 4(-2) = 0 2 ( + 1 ) + x + 4 ( − 2 ) = 0 Why this step? H is + 1 +1 + 1 − 2 -2 − 2 0 0 0 ⇒ x = + 6 \Rightarrow x = +6 ⇒ x = + 6 3 s 2 3 p 4 = 6 3s^23p^4 = 6 3 s 2 3 p 4 = 6
Worked example (b) Chromium inCr 2 O 7 2 − \text{Cr}_2\text{O}_7^{2-} Cr 2 O 7 2 − 2 x + 7 ( − 2 ) = − 2 2x + 7(-2) = -2 2 x + 7 ( − 2 ) = − 2 Why this step? Sum equals the ion charge, not zero.
⇒ 2 x = 12 ⇒ x = + 6 \Rightarrow 2x = 12 \Rightarrow x = +6 ⇒ 2 x = 12 ⇒ x = + 6 3 d 5 4 s 1 3d^54s^1 3 d 5 4 s 1
Worked example (c) Nitrogen inN 2 O \text{N}_2\text{O} N 2 O 2 x + ( − 2 ) = 0 ⇒ x = + 1 2x + (-2) = 0 \Rightarrow x = +1 2 x + ( − 2 ) = 0 ⇒ x = + 1 Why this step? An average oxidation state can be fractional/low even between the − 3 -3 − 3 + 5 +5 + 5
Common mistake Steel-man: "Oxygen isalways − 2 -2 − 2 Why it feels right: it's true in 99% of school compounds, so the shortcut usually works.
The fix: In peroxides (H 2 O 2 \text{H}_2\text{O}_2 H 2 O 2 − 1 -1 − 1 OF 2 \text{OF}_2 OF 2 + 2 +2 + 2 more electronegative; in superoxides (KO 2 \text{KO}_2 KO 2 − 1 2 -\tfrac12 − 2 1 more electronegative atom always gets the negative number , and F beats O.
Common mistake Steel-man: "Max positive OS = group number always."Why it feels right: works beautifully for Na→Cl in period 3.
The fix: F and O never reach their "group max" (F is only ever − 1 -1 − 1 0 0 0 to . And the inert-pair effect makes heavy p-block elements avoid their group max.
Recall Feynman: explain to a 12-year-oldImagine every atom is a kid with some marbles (electrons) in their pocket. A "bully" atom (very electronegative, like oxygen) snatches marbles; a "generous" atom (a metal) hands them over. The oxidation state is just how many marbles you gave away (plus) or grabbed (minus) . On the left of the table live generous kids who give a few marbles (+1, +2). On the right live grabbers. As you go right, kids have more marbles to give and are closer to a full pocket, so they can do both. Transition-metal kids have marbles in two nearby pockets, so they can hand them out one at a time — that's why they show so many numbers!
Mnemonic Remember the pattern"MAX = valence, MIN = valence minus eight."
And for heavy metals: "Big and Heavy? Be Low" → inert-pair, prefer the lower state.
What is an oxidation state? The hypothetical charge on an atom if every bond were treated as fully ionic, giving shared electrons to the more electronegative atom.
Formula for maximum positive oxidation state of a main-group element? + n v +n_v + n v , equal to its number of valence electrons (group valence).
Formula for the most negative oxidation state of a non-metal? − ( 8 − n v ) -(8 - n_v) − ( 8 − n v ) .
Why do transition metals show many oxidation states differing by 1? Because
3 d 3d 3 d and
4 s 4s 4 s orbitals are close in energy, so electrons can be removed one at a time from both.
What is the inert-pair effect? Down a group the
n s 2 ns^2 n s 2 pair becomes reluctant to bond, so the lower oxidation state (group max − 2) becomes more stable, e.g. Pb prefers
+ 2 +2 + 2 over
+ 4 +4 + 4 .
Oxidation state of S in H 2 SO 4 \text{H}_2\text{SO}_4 H 2 SO 4 + 6 +6 + 6 ; from
2 ( + 1 ) + x + 4 ( − 2 ) = 0 2(+1)+x+4(-2)=0 2 ( + 1 ) + x + 4 ( − 2 ) = 0 , matching S's 6 valence electrons.
Why can't oxygen usually be positive? Almost nothing is more electronegative than O (except F), so nothing donates electrons to it.
Max oxidation state of Mn and its electron basis? + 7 +7 + 7 , equal to its
3 d 5 4 s 2 = 7 3d^5 4s^2 = 7 3 d 5 4 s 2 = 7 removable electrons.
Across period 3 how does max positive OS change? Increases by
+ 1 +1 + 1 each step, from Na (
+ 1 +1 + 1 ) to Cl (
+ 7 +7 + 7 ).
Electronegativity — decides who "gets" the shared electrons.Ionisation Energy — sets how easily positive states form.Electron Affinity — sets how readily negative states form.Electronic Configuration — n v n_v n v d d d s s s Inert Pair Effect — controls heavy p-block behaviour.Transition Metal Chemistry — variable states drive redox and colour.Redox Reactions — application: balancing via oxidation-number change.
Hypothetical ionic charge
Non-metals: high + and negative
Intuition Hinglish mein samjhoDekho, oxidation state ka matlab simple hai: agar hum maan lein ki har bond poori tarah ionic hai, to atom pe kitna charge aayega — bas wahi number hai. Jo atom zyada electronegative hota hai (jaise oxygen, fluorine) woh electrons "cheen" leta hai, isliye usko negative number milta hai; jo dega usko positive.
Ab table mein pattern kyun banta hai? Kisi bhi main-group element ka maximum positive state uske valence electrons ke barabar hota hai (+ n v +n_v + n v minimum (sabse negative) state hota hai − ( 8 − n v ) -(8-n_v) − ( 8 − n v ) + 1 , + 2 , . . . + 7 +1, +2, ... +7 + 1 , + 2 , ... + 7
Transition metals ka funda alag hai: unke 3 d 3d 3 d 4 s 4s 4 s + 2 +2 + 2 + 7 +7 + 7 inert pair effect aata hai: bhaari elements (jaise Pb) apna n s 2 ns^2 n s 2 + 2 +2 + 2 + 4 +4 + 4
