The correct order of increasing nucleophilicity is

The correct option is Cl^- < Br^- < I^-.

To understand the order of increasing nucleophilicity for halide ions (${\text{Cl}}^{-}$, ${\text{Br}}^{-}$, and ${\text{I}}^{-}$), we must look at their behavior in polar protic solvents, which is the standard context for comparing these nucleophiles.

1. Size and Polarizability:
As we move down the halogen group in the periodic table ($\text{Cl}\to \text{Br}\to \text{I}$), the size of the halide ion increases. The iodide ion (${\text{I}}^{-}$) is the largest, followed by bromide (${\text{Br}}^{-}$), and chloride (${\text{Cl}}^{-}$) is the smallest.
Larger ions have more diffuse electron clouds, making them more polarizable. A highly polarizable nucleophile can begin bonding with a electrophilic carbon from a greater distance, making it a stronger nucleophile.

2. Solvation Effects:
In polar protic solvents (such as water or alcohols), hydrogen bonding occurs between the solvent molecules and the halide anions.
Smaller ions like ${\text{Cl}}^{-}$ have a higher charge density. This allows them to be strongly solvated (surrounded by a tight cage of solvent molecules via strong hydrogen bonds).
For the nucleophile to attack an electrophile, this solvent shell must be partially broken. Because the chloride ion is strongly solvated, it requires more energy to strip away the solvent molecules, making it less reactive (less nucleophilic).
Conversely, the iodide ion (${\text{I}}^{-}$) has a very low charge density due to its large size. It is weakly solvated by polar protic solvents, allowing it to easily approach and attack the electrophilic center.

Conclusion:
Because ${\text{I}}^{-}$ is the least solvated and most polarizable, it is the strongest nucleophile among the three. Conversely, ${\text{Cl}}^{-}$ is the most heavily solvated and least polarizable, making it the weakest. Therefore, the nucleophilicity increases in the order:
Cl^- < Br^- < I^-