Chemistry · Redox Equilibrium · NEET
An oxidising agent is a species that takes electrons from another species and gets reduced itself. For a halogen this is the reaction X2 + 2e- -> 2X-. Fluorine does this most easily, so F2 is reduced most readily and is the strongest oxidiser. In numbers, the F2/F- couple has the highest standard reduction potential (E° = +2.87 V), higher than Cl2 (+1.36 V), Br2 (+1.07 V) and I2 (+0.54 V). The bigger the positive E°, the stronger the oxidising agent.
This is the number-one trap. Electron gain enthalpy is only ONE step (gas atom + e- -> gas ion). Oxidising power in water depends on the WHOLE cycle: breaking the F-F bond, adding the electron, and then hydrating the ion. Fluorine wins on two of these: its F-F bond is very weak (small atoms, lone-pair repulsion) so it breaks easily, and the small F- ion has a very high hydration energy (a lot of energy released). These two large terms more than cancel fluorine's slightly poorer electron gain enthalpy, so overall F2 is still the strongest oxidiser.
Quote the standard reduction potential E°. For F2/F- it is +2.87 V, the highest of any halogen (and one of the highest of any common non-metal). E° already combines bond enthalpy, electron gain enthalpy and hydration energy into one value, so it is the safest and most complete answer for NEET. Higher positive E° = stronger oxidising agent.
For halogens the two orders happen to match: oxidising power F2 > Cl2 > Br2 > I2, and electronegativity F > Cl > Br > I. Both decrease down Group 17 because atomic size increases and the pull on incoming electrons weakens. But do not treat them as the same thing. Electronegativity is about attracting shared electrons inside a bond; oxidising power (E°) is about completely gaining electrons in solution. They agree here, but the correct 'why' for oxidising power is the high positive E°, not electronegativity alone.
Down the group the atoms get bigger, the nucleus is farther from and more shielded from the incoming electron, so the tendency to gain an electron falls. Hydration energy of the larger ions also falls. Both make E° smaller as you go down. So the same X2 + 2e- -> 2X- reaction becomes less favourable, and iodine ends up the weakest oxidiser while iodide is the strongest reducing halide.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Fluorine (F2). It has the highest standard reduction potential, E°(F2/F-) = +2.87 V, so it accepts electrons most readily and is reduced most easily.
F2 > Cl2 > Br2 > I2. Oxidising power decreases down Group 17 as atomic size increases and E° values fall.
No. Chlorine has a more negative electron gain enthalpy than fluorine, because fluorine's very small 2p subshell causes strong electron-electron repulsion. Despite this, fluorine is still the strongest oxidiser due to its weak F-F bond and high F- hydration energy.
Because oxidising power in solution is decided by the full cycle: bond dissociation + electron gain + hydration. Fluorine's weak F-F bond and the large hydration energy of the tiny F- ion outweigh its slightly poorer electron gain enthalpy, giving F2 the higher E°.
Electronegativity supports the trend but the correct reason for NEET is the highest positive standard reduction potential E°. E° already includes bond breaking, electron gain and hydration, so it is the complete measure of oxidising power.
