The decreasing order of the oxidising power of the halogens is

Step 1: Analyze the question You are asked to arrange the halogens (F, Cl, Br, I) from the strongest to the weakest oxidising agent.

Step 2: Explain the concept An oxidising agent gains electrons (it is itself reduced). The halogens do this by forming X⁻ ions:

X₂ + 2e⁻ → 2X⁻

The easier it is for a halogen to gain an electron, the stronger its oxidising power. This depends on:

• Atomic size: going down Group 17, atoms get bigger, so the nucleus attracts an incoming electron less strongly.
• Electronegativity and electron affinity: these decrease down the group, with fluorine the most electronegative element.
• Standard electrode potentials (E°): F₂ = +2.87 V, Cl₂ = +1.36 V, Br₂ = +1.07 V, I₂ = +0.54 V. A more positive E° means a stronger oxidising agent.

Step 3: Solution Oxidising power decreases down the group:

F₂ > Cl₂ > Br₂ > I₂

This is also shown by displacement reactions. Chlorine displaces bromine and iodine from their salts, for example:

Cl₂ + 2KBr → 2KCl + Br₂

Iodine cannot displace chlorine or bromine from their salts.

Step 4: Identify the answer The correct order is F > Cl > Br > I, which matches option C.

Step 5: Tip Remember that oxidising power decreases down Group 17 (F is the strongest), while the reducing power of the halide ions increases down the group (I⁻ is the strongest reducing agent). A quick check is that fluorine is the most reactive and most electronegative halogen, so it must be at the top of the list.