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For (i) $I^-$, (ii) $Cl^-$, (iii) $Br^-$, the increasing order of nucleophilicity would be
A
$Cl^- \lt Br^- \lt I^-$
B
$I^- \lt Cl^- \lt Br^-$
C
$Br^- \lt Cl^- \lt I^-$
D
$I^- \lt Br^- \lt Cl^-$
Detailed Solution
Nucleophilicity increases down a group of the periodic table (in protic solvents).
As the size of the halide ion increases from $Cl^-$ to $I^-$, its electron cloud becomes more polarisable, so it can donate its electron pair to carbon more easily.
A larger ion is also less strongly solvated (hydrogen bonded) by the solvent, so it is freer to attack.
$I^- \gt Br^- \gt Cl^- \gt F^-$
So the increasing order of nucleophilicity is: $Cl^- \lt Br^- \lt I^-$
As the size of the halide ion increases from $Cl^-$ to $I^-$, its electron cloud becomes more polarisable, so it can donate its electron pair to carbon more easily.
A larger ion is also less strongly solvated (hydrogen bonded) by the solvent, so it is freer to attack.
$I^- \gt Br^- \gt Cl^- \gt F^-$
So the increasing order of nucleophilicity is: $Cl^- \lt Br^- \lt I^-$