Why is fluoride the most nucleophilic halide in aprotic solvents? [duplicate]

Question

I am confused on the topic of electronegative nucleophiles, and polar aprotic solvents.

From my understanding, the more electronegative a nucleophile is, the weaker of a nucleophile it will be, because atoms will be less likely to share its electron density.

However, in a polar aprotic solvent, because H-bonding cannot occur, it seems as if the trend of nucleophilicity is such that $\ce{F- > Cl- > Br- > I-}$.

Can someone please clarify how this is possible? I would have expected $\ce{F-}$ to be the LEAST nucleophilic nucleophile in a polar aprotic solvent, because it is the most electronegative atom. Therefore according to the first point I learned, I thought it should be the weakest nucleophile.

All this is what I've learned through reading Kaplan's Organic Chemistry Review.

Answer 1

When talking about elements of a group, the size of the atom is more effective than the electronegativity factor. As $\ce{F-}$ size is too short so its electron density will be much higher than other hallogens.

In aprotic solvent the order of nucleophillicity: $\ce{F- > Cl- > Br- >I-}$(No hydration occurs) In protic solvant(like $\ce{H2O}$) its reversed because of hydration of ions having much electron density.