Which of the following correctly ranks the nucleophilicity of the species $F_{-}$, ${\mathrm{Cl}}_{-}$, ${\mathrm{Br}}_{-}$, and $I_{-}$ in water from greatest to least?

• A. ${\mathrm{Cl}}_{-}$ > ${\mathrm{Br}}_{-}$ > $I_{-}$ > $F_{-}$
• B. $F_{-}$ > ${\mathrm{Cl}}_{-}$ > ${\mathrm{Br}}_{-}$ > $I_{-}$
• C. $I_{-}$ > ${\mathrm{Br}}_{-}$ > ${\mathrm{Cl}}_{-}$ > $F_{-}$
• D. ${\mathrm{Br}}_{-}$ > ${\mathrm{Cl}}_{-}$ > $F_{-}$ > $I_{-}$

Answer: (C)

Verified step by step guidance

1. Understand that nucleophilicity refers to the ability of a species to donate a pair of electrons to an electrophile, and it depends on factors such as charge, electronegativity, solvent effects, and polarizability.
2. Recognize that the species given are halide ions: fluoride (F\textsuperscript{\-}), chloride (Cl\textsuperscript{\-}), bromide (Br\textsuperscript{\-}), and iodide (I\textsuperscript{\-}), all negatively charged and differing mainly in size and polarizability.
3. Consider the solvent effect: in a protic solvent like water, nucleophilicity trends differ from basicity trends because smaller ions like F\textsuperscript{\-} are strongly solvated (surrounded by solvent molecules), which hinders their nucleophilicity.
4. Recall that in water, nucleophilicity increases down the group because larger ions like I\textsuperscript{\-} are less solvated and more polarizable, making them better nucleophiles despite being less basic.
5. Therefore, the correct ranking of nucleophilicity in water is: \(\mathrm{I}^- > \mathrm{Br}^- > \mathrm{Cl}^- > \mathrm{F}^-\), reflecting the balance of solvation and polarizability effects.