For the species $(I^{-}, C l^{-}, B r^{-})$ , the increasing order of nucleophilicity in water would be:

I^{-} < C l^{-} < B r^{-}

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B r^{-} < C l^{-} < I^{-}

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I^{-} < B r^{-} < C l^{-}

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C l^{-} < B r^{-} < I^{-}

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Solution

The correct option is D $C l^{-} < B r^{-} < I^{-}$
$H_{2} O$ is a polar solvent.
$Nucleophilicity \propto size$
Smaller the size of anion, more it will be hydrated by $H_{2} O$ and less will be the nucleophilicity.
Hence, the order is: $C l^{-} < B r^{-} < I^{-}$

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For the species (I−,Cl−,Br−), the increasing order of nucleophilicity in water would be:

The correct option is D Cl–<Br–<I–H2O is a polar solvent. Nucleophilicity∝size Smaller the size of anion, more it will be hydrated by H2O and less will be the nucleophilicity. Hence, the order is: Cl–<Br–<I–

For the species $(I^{-}, C l^{-}, B r^{-})$ , the increasing order of nucleophilicity in water would be:

The correct option is D $C l^{-} < B r^{-} < I^{-}$
$H_{2} O$ is a polar solvent.
$Nucleophilicity \propto size$
Smaller the size of anion, more it will be hydrated by $H_{2} O$ and less will be the nucleophilicity.
Hence, the order is: $C l^{-} < B r^{-} < I^{-}$