For (i) $I^-$ (ii) $Cl^-$ (iii) $Br^-$, the usual increasing order of nucleophilicities would be

$Cl^-<Br^-<I^-$

The correct answer is Option (1) → $Cl^-<Br^-<I^-$

The nucleophilicity of halide ions \((I^-, Br^-, Cl^-)\) depends on several factors, such as the solvent used. However, for protic solvents (like water, alcohols, etc.), which are commonly considered, the nucleophilicity generally increases down the group of the periodic table. This is due to the following reasons:

Nucleophilicity in Protic Solvents: In protic solvents, nucleophiles are solvated by hydrogen bonding with the solvent molecules. Larger anions, like \(I^-\), are less strongly solvated than smaller anions, like \(Cl^-\). Therefore, \(I^-\) is more "free" to attack electrophiles, making it a stronger nucleophile compared to \(Br^-\) and \(Cl^-\).

Basicity and Nucleophilicity: While basicity decreases as you go down the group (because larger halide ions hold onto protons less tightly), nucleophilicity in protic solvents actually increases due to the decreased solvation.

Increasing Order of Nucleophilicity in Protic Solvents: \(I^- > Br^- > Cl^-\)

Thus, the correct increasing order of nucleophilicity for \(I^-\), \(Br^-\), and \(Cl^-\) in protic solvents is: \(Cl^- < Br^- < I^-\)

This is because \(I^-\) is the least solvated and thus the strongest nucleophile, while \(Cl^-\) is the most solvated and thus the weakest nucleophile.