Arrange the increasing order of nucleophilicity.
$I^{-}, {B r}^{-}, F^{-}$

I^{-} < {B r}^{-} < F^{-}

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I^{-} > {B r}^{-} > F^{-}

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I^{-} = {B r}^{-} > F^{-}

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I^{-} = {B r}^{-} = F^{-}

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Solution

The correct option is B $I^{-} < {B r}^{-} < F^{-}$
In the polar aprotic solvent, the increasing order of nucleophilicity is $I^{-} < {B r}^{-} < F^{-}$ .
Iodide ion has large size. The negative charge is dispersed on larger anion. Hence, it can be less readily donated.
In case of fluoride ion, the negative charge is on small fluoride ion and can be readily donated.

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Arrange the increasing order of nucleophilicity.I−, Br−, F−

Arrange the increasing order of nucleophilicity.
$I^{-}, {B r}^{-}, F^{-}$