When iron rusts, in which oxidation state is iron present in the final product (rust)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → +3
Rust is hydrated ferric oxide, \( \ce{Fe2O3.xH2O} \).
Water of hydration: neutral, so it does not enter the charge balance.
Charge balance in \( \ce{Fe2O3} \): 2x + 3(−2) = 0, so x = +3.
At the anodic spot iron is first oxidised to \( \ce{Fe^2+} \); these ferrous ions are then oxidised by atmospheric oxygen to ferric ions, which come out as rust:
\( \displaystyle \ce{2Fe(s) -> 2Fe^2+(aq) + 4e^-} \)
\( \displaystyle \ce{2Fe^2+(aq) + 2H2O(l) + 1/2O2(g) -> Fe2O3(s) + 4H^+(aq)} \)
Why the other options are incorrect
Option (1) +2 — Incorrect. \( \ce{Fe^2+} \) is only the intermediate formed at the anode; air oxidises it further to \( \ce{Fe^3+} \).
Option (3) 0 (elemental state) — Incorrect. Metallic iron is the starting material; rusting is its oxidation.
Option (4) Both +2 and +3 — Incorrect. The ferrous ions do not stay in the product; rust is ferric oxide, so iron in it is only +3.
NCERT (2026-27 edition):
2.8 Corrosion – “… The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide (Fe2O3. x H2O) and with further production of hydrogen ions.”