The strongest oxidizing and reducing agents respectively are:

\(F_2\) and \(I^-\)

The correct answer is option 1. \(F_2\) and \(I^-\).

Let us delve into why \(F_2\) is the strongest oxidizing agent and \(I^-\) is the strongest reducing agent among the options provided.

Fluorine (\(F_2\)) as the Strongest Oxidizing Agent:

Fluorine (\(F_2\)) has the highest electronegativity among the halogens. Electronegativity is the tendency of an atom to attract electrons towards itself in a chemical bond. As a strong oxidizing agent, fluorine readily accepts electrons to achieve a stable, noble gas electron configuration (similar to the noble gas neon). Its high electronegativity means that it strongly attracts electrons, making it highly reactive and eager to accept electrons from other atoms. This property enables it to oxidize other species by gaining electrons itself.

Iodide Ion (\(I^-\)) as the Strongest Reducing Agent:

Iodide ion (\(I^-\)) has the largest atomic size among the halogens, and consequently, the lowest electronegativity. Being the least electronegative, iodide ion is the most easily oxidized (or, conversely, the strongest reducing agent) among the halide ions. It readily loses electrons to other species in redox reactions, enabling it to reduce other substances by donating electrons.

In summary, fluorine (\(F_2\)) is the strongest oxidizing agent because of its high electronegativity, while iodide ion (\(I^-\)) is the strongest reducing agent due to its low electronegativity. These properties make \(F_2\) an excellent electron acceptor (oxidizing agent) and \(I^-\) an efficient electron donor (reducing agent) in redox reactions.