Why doesn't fluoride ion have highest nucleophilic power (nucleophilicity)?

Question

As you can see from the image above,

• '$\ce{R3C}$' anion has 1 lone pair
• '$\ce{R2N}$' anion has 2 lone pairs
• $\ce{RO}$ anion has 3 lone pairs
• $\ce{F}$ anion has 4 lone pairs

According to VSEPR theory,

lone pair-lone pair repulsion > bond pair-lone pair repulsion > bond pair-bond pair repulsion

As fluoride ion has the greatest number of lone pairs (= 4), it should suffer from highest number of lone pair-lone pair repulsions.

In order to reduce its lone pair-lone pair repulsions and thus its unstability, shouldn't it donate its lone pair more readily than other anions? Thereby, shouldn't fluoride ion have more nucleophilic tendency than the other (mentioned) anions?

Answer 1

Nucleophilicity of an ion does not solely depend on the number of lone pairs it has. The solvent in which the ions are dissolved matters, too.

If you are using a polar protic solvent, then flouride ion would behave as a very bad nucleophile (it would be solvated in the case of water; water will form hydrogen bonds with it).

Among the nucleophiles you have stated, $\ce{R3C}$ anion would be the best nucleophile in a polar protic solvent. However, if you use a polar aprotic solvent, flouride ion would behave as a very good nucleophile (because aprotic solvents do not solvate anions).

Remember, nucleophilicity is a kinetic concept and basicity is a thermodynamic one. Your reasoning for comparing the nucleophilicities was based somewhat on basicity of the ions. (This would work well in an aprotic solvent.)

In very layman terms:

• Nucleophilicity: How good is something at reacting (doesn't concern stability of the ion)
• Basicity: How badly something would want to react (concerns stabilities)