The strongest oxidising and reducing agents respectively are.

The correct answer is Option (1) → F₂ and I⁻

F₂ has the highest standard reduction potential of all elements (E° \( \ce{F2} \)/\( \ce{F^{-}} \) = +2.87 V), so it accepts electrons most readily: it is the strongest oxidising agent and oxidises every other halide ion.

Among the halide ions, I⁻ belongs to the couple with the lowest potential (E° \( \ce{I2} \)/\( \ce{I^{-}} \) = +0.54 V). It is the largest halide ion and holds its electron most loosely, so it loses it most easily: it is the strongest reducing agent of the halides.

Why the other options are incorrect

• Option (2) Br₂ and Cl⁻ — Incorrect. \( \ce{Br2} \) is weaker than \( \ce{F2} \) and \( \ce{Cl2} \); \( \ce{Cl^{-}} \) is a weaker reducing agent than \( \ce{Br^{-}} \) and \( \ce{I^{-}} \).

• Option (3) Cl₂ and Br⁻ — Incorrect. \( \ce{Cl2} \) is weaker than \( \ce{F2} \), and \( \ce{Br^{-}} \) is weaker than \( \ce{I^{-}} \).

• Option (4) Cl₂ and I₂ — Incorrect. \( \ce{I2} \) is an oxidising agent (a weak one), not the reducing agent; the reduced form is \( \ce{I^{-}} \).

NCERT (2026-27 edition):
Measurement of Electrode Potential – “… It can be seen that the standard electrode potential for fluorine is the highest in the Table indicating that fluorine gas (F₂) has the maximum tendency to get reduced to fluoride ions (F^–) and therefore fluorine gas is the strongest oxidising agent and fluoride ion is the weakest reducing agent. … It may be seen that as we go from top to bottom in Table 2.1 the standard electrode potential decreases and with this, decreases the oxidising power of the species on the left and increases the reducing power of the species on the right hand side of the reaction.”

Table 2.1: Standard Electrode Potentials at 298 K (halogen rows)